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so that they cancel each other out. Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case the best coordinate system has one axis horizontal and the other vertical. We call the horizontal the -axis and the vertical the -axis. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 163 Solution First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system. Figure 4.18 When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is stationary. The small angle results in being much greater than. Consider the horizontal components of the forces (denoted with a subscript ): net = L − R. The net external horizontal force net = 0, since the person is stationary. Thus, net = 0 = L − R L = R. Now, observe Figure 4.18. You can use trigonometry to determine the magnitude of L and R. Notice that: cos (5.0º) = L L L = L cos (5.0º) cos (5.0º) = R R R = R cos (5.0º). L cos (5.0º) = R cos (5.0º). L = R =, Equating L and R : Thus, (4.44) (4.45) (4.46) (4.47) (4.48) as predicted. Now, considering the vertical components (denoted by a subscript ), we can solve for. Again, since the person is stationary, Newton’s second law implies that net = 0. Thus, as illustrated in the free-body diagram in Figure 4.18, net = L + R − = 0. Observing Figure 4.18, we can use trigonometry to determine the relationship between L, R, and. As we determined from the analysis in the horizontal direction, L = R = : sin (5.0º) = L L L = L sin (5.0º) = sin (5.0º) sin (5.0º) = R R
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R = R sin (5.0º) = sin (5.0º). (4.49) (4.50) 164 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Now, we can substitute the values for L and R, into the net force equation in the vertical direction: = L + R − = 0 = sin (5.0º) + sin (5.0º) − = 0 net net 2 sin (5.0º) − = 0 = 2 sin (5.0º) and so that and the tension is Discussion = 2 sin (5.0º) = 2 sin (5.0º), = (70.0 kg)(9.80 m/s2) 2(0.0872), = 3900 N. (4.51) (4.52) (4.53) (4.54) Note that the vertical tension in the wire acts as a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and so most of the tension in the wire is not used to support the weight of the tightrope walker. If we wish to create a very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in Figure 4.19. As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way: = 2 sin (). (4.55) We can extend this expression to describe the tension created when a perpendicular force ( F⊥ ) is exerted at the middle of a flexible connector: = ⊥ 2 sin (). (4.56) Note that is the angle between the horizontal and the bent connector. In this case, becomes very large as approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., = 0 and sin = 0 ). (See Figure 4.19.) Figure 4.19 We can create a very large tension in the chain by
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pushing on it perpendicular to its length, as shown. Suppose we wish to pull a car out of the mud when no tow truck is available. Each time the car moves forward, the chain is tightened to keep it as nearly straight as possible. The tension in the chain is given by = ⊥ 2 sin () ; since is small, is very large. This situation is analogous to the tightrope walker shown in Figure 4.17, except that the tensions shown here are those transmitted to the car and the tree rather than those acting at the point where F⊥ is applied. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 165 Figure 4.20 Unless an infinite tension is exerted, any flexible connector—such as the chain at the bottom of the picture—will sag under its own weight, giving a characteristic curve when the weight is evenly distributed along the length. Suspension bridges—such as the Golden Gate Bridge shown in this image—are essentially very heavy flexible connectors. The weight of the bridge is evenly distributed along the length of flexible connectors, usually cables, which take on the characteristic shape. (credit: Leaflet, Wikimedia Commons) Extended Topic: Real Forces and Inertial Frames There is another distinction among forces in addition to the types already mentioned. Some forces are real, whereas others are not. Real forces are those that have some physical origin, such as the gravitational pull. Contrastingly, fictitious forces are those that arise simply because an observer is in an accelerating frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear acceleration (like a car slowing down). For example, if a satellite is heading due north above Earth’s northern hemisphere, then to an observer on Earth it will appear to experience a force to the west that has no physical origin. Of course, what is happening here is that Earth is rotating toward the east and moves east under the satellite. In Earth’s frame this looks like a westward force on the satellite, or it can be interpreted as a violation of Newton’s first law (the law of inertia). An inertial frame of reference is one in which all forces are real and, equivalently, one in which Newton’s laws have the simple forms given in this chapter. Earth’s rotation is slow enough that Earth is nearly an inertial frame.
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You ordinarily must perform precise experiments to observe fictitious forces and the slight departures from Newton’s laws, such as the effect just described. On the large scale, such as for the rotation of weather systems and ocean currents, the effects can be easily observed. The crucial factor in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless stated otherwise, all phenomena discussed in this text are considered in inertial frames. All the forces discussed in this section are real forces, but there are a number of other real forces, such as lift and thrust, that are not discussed in this section. They are more specialized, and it is not necessary to discuss every type of force. It is natural, however, to ask where the basic simplicity we seek to find in physics is in the long list of forces. Are some more basic than others? Are some different manifestations of the same underlying force? The answer to both questions is yes, as will be seen in the next (extended) section and in the treatment of modern physics later in the text. PhET Explorations: Forces in 1 Dimension Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces). Figure 4.21 Forces in 1 Dimension (http://cnx.org/content/m54857/1.4/forces-1d_en.jar) 4.6 Problem-Solving Strategies By the end of this section, you will be able to: Learning Objectives 166 Chapter 4 | Dynamics: Force and Newton's Laws of Motion • Apply a problem-solving procedure to solve problems using Newton's laws of motion The information presented in this section supports the following AP® learning objectives and science practices: • 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) • 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P. 1.4) • 3.B.1.1 The student is able to predict the motion of
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an object subject to forces exerted by several objects using an application of Newton's second law in a variety of physical situations with acceleration in one dimension. (S.P. 6.4, 7.2) • 3.B.1.3 The student is able to re-express a free-body diagram representation into a mathematical representation and solve the mathematical representation for the acceleration of the object. (S.P. 1.5, 2.2) • 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) Success in problem solving is obviously necessary to understand and apply physical principles, not to mention the more immediate need of passing exams. The basics of problem solving, presented earlier in this text, are followed here, but specific strategies useful in applying Newton’s laws of motion are emphasized. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, and so the following techniques should reinforce skills you have already begun to develop. Problem-Solving Strategy for Newton’s Laws of Motion Step 1. As usual, it is first necessary to identify the physical principles involved. Once it is determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 4.22(a). Then, as in Figure 4.22(b), use arrows to represent all forces, label them carefully, and make their lengths and directions correspond to the forces they represent (whenever sufficient information exists). Figure 4.22 (a) A sketch of Tarzan hanging from a vine. (b) Arrows are used to represent all forces. T is the tension in the vine above Tarzan, FT is the force he exerts on the vine, and w is his weight. All other forces, such as the nudge of a breeze, are assumed negligible. (c) Suppose we are given the ape man’s mass and asked to find the tension in the vine. We then define the system of interest as shown and draw a free-body diagram. FT is no longer shown, because it is not a force acting on the system of interest; rather, FT acts on the outside world. (d) Showing
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only the arrows, the head-to-tail method of addition is used. It is apparent that T = - w, if Tarzan is stationary. Step 2. Identify what needs to be determined and what is known or can be inferred from the problem as stated. That is, make a list of knowns and unknowns. Then carefully determine the system of interest. This decision is a crucial step, since Newton’s second law involves only external forces. Once the system of interest has been identified, it becomes possible to determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 4.22(c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated earlier in this chapter, the system of interest depends on what This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 167 question we need to answer. This choice becomes easier with practice, eventually developing into an almost unconscious process. Skill in clearly defining systems will be beneficial in later chapters as well. A diagram showing the system of interest and all of the external forces is called a free-body diagram. Only forces are shown on free-body diagrams, not acceleration or velocity. We have drawn several of these in worked examples. Figure 4.22(c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram. Step 3. Once a free-body diagram is drawn, Newton’s second law can be applied to solve the problem. This is done in Figure 4.22(d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then they add like scalars. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. This is done by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of
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axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Applying Newton’s Second Law Before you write net force equations, it is critical to determine whether the system is accelerating in a particular direction. If the acceleration is zero in a particular direction, then the net force is zero in that direction. Similarly, if the acceleration is nonzero in a particular direction, then the net force is described by the equation: net =. For example, if the system is accelerating in the horizontal direction, but it is not accelerating in the vertical direction, then you will have the following conclusions: net =, net = 0. (4.57) (4.58) You will need this information in order to determine unknown forces acting in a system. Step 4. As always, check the solution to see whether it is reasonable. In some cases, this is obvious. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving, and with experience it becomes progressively easier to judge whether an answer is reasonable. Another way to check your solution is to check the units. If you are solving for force and end up with units of m/s, then you have made a mistake. 4.7 Further Applications of Newton's Laws of Motion Learning Objectives By the end of this section, you will be able to: • Apply problem-solving techniques to solve for quantities in more complex systems of forces. • Integrate concepts from kinematics to solve problems using Newton's laws of motion. The information presented in this section supports the following AP® learning objectives and science practices: • 3.A.2.1 The student is able to represent forces in diagrams or mathematically using appropriately labeled vectors with magnitude, direction, and units during the analysis of a situation. (S.P. 1.1) • 3.A.3.1 The student is able to analyze a scenario and make claims (develop arguments, justify assertions) about the forces exerted on an object by other objects for different types of forces or components of forces. (S.P. 6.4, 7.2) • 3.A.3.3 The student is able to describe a force as an interaction between two objects and identify both objects for any force. (S.P.
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1.4) • 3.B.1.1 The student is able to predict the motion of an object subject to forces exerted by several objects using an application of Newton's second law in a variety of physical situations with acceleration in one dimension. (S.P. 6.4, 7.2) • 3.B.1.3 The student is able to re-express a free-body diagram representation into a mathematical representation and solve the mathematical representation for the acceleration of the object. (S.P. 1.5, 2.2) • 3.B.2.1 The student is able to create and use free-body diagrams to analyze physical situations to solve problems with motion qualitatively and quantitatively. (S.P. 1.1, 1.4, 2.2) There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. 168 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Example 4.7 Drag Force on a Barge Suppose two tugboats push on a barge at different angles, as shown in Figure 4.23. The first tugboat exerts a force of 2.7×105 N in the x-direction, and the second tugboat exerts a force of 3.6×105 N in the y-direction. Figure 4.23 (a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the ship contains only forces acting in the plane of the water. It omits the two vertical forces—the weight of the barge and the buoyant force of the water supporting it cancel and are not shown. Since the applied forces are perpendicular, the x- and y-axes are in the same direction as F and F. The problem quickly becomes a one-dimensional problem along the direction of Fapp, since friction is in the direction opposite to Fapp. If the mass of the barge is 5.0×106 kg and its acceleration is observed to be 7.5×10−2 m/s2 in the direction shown, what is the drag force of the water on the barge resisting the motion? (Note: drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object.) Strategy The directions
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and magnitudes of acceleration and the applied forces are given in Figure 4.23(a). We will define the total force of the tugboats on the barge as Fapp so that: Fapp =F + F (4.59) Since the barge is flat bottomed, the drag of the water FD will be in the direction opposite to Fapp, as shown in the freebody diagram in Figure 4.23(b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Our strategy is to find the magnitude and direction of the net applied force Fapp, and then apply Newton’s second law to solve for the drag force FD. Solution Since F and F are perpendicular, the magnitude and direction of Fapp are easily found. First, the resultant magnitude is given by the Pythagorean theorem: 2 2 + F app = F app = (2.7×105 N)2 + (3.6×105 N)2 = 4.5×105 N. The angle is given by = tan−1 3.6×105 N = tan−1 2.7×105 N = 53º (4.60) (4.61) which we know, because of Newton’s first law, is the same direction as the acceleration. FD is in the opposite direction of Fapp, since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as Fapp, but its magnitude is slightly less than Fapp. The problem is now one-dimensional. From Figure 4.23(b), we can see that net = app − D. (4.62) But Newton’s second law states that This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Thus, net =. app − D =. This can be solved for the magnitude of the drag force of the water D in terms of known quantities: D = app −. 169 (4.63) (4.64) (4.65) Substituting known values gives FD = (4.5×105 N) − (5.0×106 kg)(7.5×10–2 m/s2 ) = 7.5×104 N. The direction of FD has already been determined to be in the direction opposite to Fapp, or at
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an angle of 53º south of west. (4.66) Discussion The numbers used in this example are reasonable for a moderately large barge. It is certainly difficult to obtain larger accelerations with tugboats, and small speeds are desirable to avoid running the barge into the docks. Drag is relatively small for a well-designed hull at low speeds, consistent with the answer to this example, where D is less than 1/600th of the weight of the ship. In the earlier example of a tightrope walker we noted that the tensions in wires supporting a mass were equal only because the angles on either side were equal. Consider the following example, where the angles are not equal; slightly more trigonometry is involved. Example 4.8 Different Tensions at Different Angles Consider the traffic light (mass 15.0 kg) suspended from two wires as shown in Figure 4.24. Find the tension in each wire, neglecting the masses of the wires. 170 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.24 A traffic light is suspended from two wires. (b) Some of the forces involved. (c) Only forces acting on the system are shown here. The free-body diagram for the traffic light is also shown. (d) The forces projected onto vertical (y) and horizontal (x) axes. The horizontal components of the tensions must cancel, and the sum of the vertical components of the tensions must equal the weight of the traffic light. (e) The free-body diagram shows the vertical and horizontal forces acting on the traffic light. Strategy The system of interest is the traffic light, and its free-body diagram is shown in Figure 4.24(c). The three forces involved are not parallel, and so they must be projected onto a coordinate system. The most convenient coordinate system has one axis vertical and one horizontal, and the vector projections on it are shown in part (d) of the figure. There are two unknowns in this problem ( 1 and 2 ), so two equations are needed to find them. These two equations come from applying Newton’s second law along the vertical and horizontal axes, noting that the net external force is zero along each axis because acceleration is zero. Solution First consider the horizontal or x-axis: Thus, as you might expect, net = 2 − 1 = 0. (4.67) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter
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4 | Dynamics: Force and Newton's Laws of Motion This gives us the following relationship between 1 and 2 : 1 = 2. Thus, 1 cos (30º) = 2 cos (45º). 2 = (1.225)1. 171 (4.68) (4.69) (4.70) Note that 1 and 2 are not equal in this case, because the angles on either side are not equal. It is reasonable that 2 ends up being greater than 1, because it is exerted more vertically than 1. Now consider the force components along the vertical or y-axis: This implies net = 1 + 2 − = 0. 1 + 2 =. Substituting the expressions for the vertical components gives 1 sin (30º) + 2 sin (45º) =. There are two unknowns in this equation, but substituting the expression for 2 in terms of 1 reduces this to one equation with one unknown: which yields 1(0.500) + (1.2251)(0.707) = =, (1.366)1 = (15.0 kg)(9.80 m/s2). Solving this last equation gives the magnitude of 1 to be 1 = 108 N. (4.71) (4.72) (4.73) (4.74) (4.75) (4.76) Finally, the magnitude of 2 is determined using the relationship between them, 2 = 1.225 1, found above. Thus we obtain 2 = 132 N. (4.77) Discussion Both tensions would be larger if both wires were more horizontal, and they will be equal if and only if the angles on either side are the same (as they were in the earlier example of a tightrope walker). The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed: will the scale still read more than your weight at rest? Consider the following example. Example 4.9 What Does the Bathroom Scale Read in an Elevator? Figure 4.25 shows a 75.0-kg man (weight of about 165 lb) standing on a bathroom scale in
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an elevator. Calculate the scale reading: (a) if the elevator accelerates upward at a rate of 1.20 m/s2, and (b) if the elevator moves upward at a constant speed of 1 m/s. 172 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.25 (a) The various forces acting when a person stands on a bathroom scale in an elevator. The arrows are approximately correct for when the elevator is accelerating upward—broken arrows represent forces too large to be drawn to scale. T is the tension in the supporting cable, w is the weight of the person, ws is the weight of the scale, we is the weight of the elevator, Fs is the force of the scale on the person, Fp is the force of the person on the scale, Ft is the force of the scale on the floor of the elevator, and N is the force of the floor upward on the scale. (b) The free-body diagram shows only the external forces acting on the designated system of interest—the person. Strategy If the scale is accurate, its reading will equal p, the magnitude of the force the person exerts downward on it. Figure 4.25(a) shows the numerous forces acting on the elevator, scale, and person. It makes this one-dimensional problem look much more formidable than if the person is chosen to be the system of interest and a free-body diagram is drawn as in Figure 4.25(b). Analysis of the free-body diagram using Newton’s laws can produce answers to both parts (a) and (b) of this example, as well as some other questions that might arise. The only forces acting on the person are his weight w and the upward force of the scale Fs. According to Newton’s third law Fp and Fs are equal in magnitude and opposite in direction, so that we need to find s in order to find what the scale reads. We can do this, as usual, by applying Newton’s second law, net =. From the free-body diagram we see that net = s −, so that s − =. Solving for s gives an equation with only one unknown: or, because =, simply s = +, s = +. (4.78) (4.79) (4.80) (4.81) No assumptions were made about the acceleration, and so this solution should be valid for a variety of acceler
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ations in addition to the ones in this exercise. Solution for (a) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 173 In this part of the problem, = 1.20 m/s2, so that s = (75.0 kg)(1.20 m/s2 ) + (75.0 kg)(9.80 m/s2), yielding Discussion for (a) s = 825 N. (4.82) (4.83) This is about 185 lb. What would the scale have read if he were stationary? Since his acceleration would be zero, the force of the scale would be equal to his weight: net = = 0 = s − s = = = (75.0 kg)(9.80 m/s2) = 735 N. s s (4.84) So, the scale reading in the elevator is greater than his 735-N (165 lb) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. Solution for (b) Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because = Δ Δ, and Δ = 0. Thus, Now which gives Discussion for (b75.0 kg)(9.80 m/s2), s = 735 N. (4.85) (4.86) (4.87) The scale reading is 735 N, which equals the person’s weight. This will be the case whenever the elevator has a constant velocity—moving up, moving down, or stationary. The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, is negative, and the scale reading is less than the weight of the person, until a constant downward velocity is reached, at which time the scale reading again becomes equal to the person’s weight. If the elevator is in free-fall and accelerating downward at, then the scale reading will be zero and the person will appear to be weightless. Integ
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rating Concepts: Newton’s Laws of Motion and Kinematics Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics, and hence the relevance of earlier chapters. When approaching problems that involve various types of forces, acceleration, velocity, and/or position, use the following steps to approach the problem: Problem-Solving Strategy Step 1. Identify which physical principles are involved. Listing the givens and the quantities to be calculated will allow you to identify the principles involved. Step 2. Solve the problem using strategies outlined in the text. If these are available for the specific topic, you should refer to them. You should also refer to the sections of the text that deal with a particular topic. The following worked example illustrates how these strategies are applied to an integrated concept problem. 174 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Example 4.10 What Force Must a Soccer Player Exert to Reach Top Speed? A soccer player starts from rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s. (a) What was his average acceleration? (b) What average force did he exert backward on the ground to achieve this acceleration? The player’s mass is 70.0 kg, and air resistance is negligible. Strategy 1. To solve an integrated concept problem, we must first identify the physical principles involved and identify the chapters in which they are found. Part (a) of this example considers acceleration along a straight line. This is a topic of kinematics. Part (b) deals with force, a topic of dynamics found in this chapter. 2. The following solutions to each part of the example illustrate how the specific problem-solving strategies are applied. These involve identifying knowns and unknowns, checking to see if the answer is reasonable, and so forth. Solution for (a) We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ = 8.00 m/s. We are given the elapsed time, and so Δ = 2.50 s. The unknown is acceleration, which can be found from its definition: = Δ Δ. (4.88) Subst
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ituting the known values yields = 8.00 m/s 2.50 s = 3.20 m/s2. (4.89) Discussion for (a) This is an attainable acceleration for an athlete in good condition. Solution for (b) Here we are asked to find the average force the player exerts backward to achieve this forward acceleration. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes his acceleration. Since we now know the player’s acceleration and are given his mass, we can use Newton’s second law to find the force exerted. That is, Substituting the known values of and gives net =. net = (70.0 kg)(3.20 m/s2) = 224 N. Discussion for (b) This is about 50 pounds, a reasonable average force. (4.90) (4.91) This worked example illustrates how to apply problem-solving strategies to situations that include topics from different chapters. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknown using familiar problem-solving strategies. These strategies are found throughout the text, and many worked examples show how to use them for single topics. You will find these techniques for integrated concept problems useful in applications of physics outside of a physics course, such as in your profession, in other science disciplines, and in everyday life. The following problems will build your skills in the broad application of physical principles. 4.8 Extended Topic: The Four Basic Forces—An Introduction By the end of this section, you will be able to: • Understand the four basic forces that underlie the processes in nature. Learning Objectives The information presented in this section supports the following AP® learning objectives and science practices: • 3.C.4.1 The student is able to make claims about various contact forces between objects based on the microscopic cause of those forces. (S.P. 6.1) • 3.C.4.2 The student is able to explain contact forces (tension, friction, normal, buoyant, spring) as arising from interatomic electric forces and that they therefore have certain directions. (S.P. 6.2) • 3.G.1.1 The student is able to articulate situations when the gravitational force is the dominant force and when the electromagnetic, weak, and strong forces can be ignored. (S.P
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. 7.1) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 175 One of the most remarkable simplifications in physics is that only four distinct forces account for all known phenomena. In fact, nearly all of the forces we experience directly are due to only one basic force, called the electromagnetic force. (The gravitational force is the only force we experience directly that is not electromagnetic.) This is a tremendous simplification of the myriad of apparently different forces we can list, only a few of which were discussed in the previous section. As we will see, the basic forces are all thought to act through the exchange of microscopic carrier particles, and the characteristics of the basic forces are determined by the types of particles exchanged. Action at a distance, such as the gravitational force of Earth on the Moon, is explained by the existence of a force field rather than by “physical contact.” The four basic forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force. Their properties are summarized in Table 4.2. Since the weak and strong nuclear forces act over an extremely short range, the size of a nucleus or less, we do not experience them directly, although they are crucial to the very structure of matter. These forces determine which nuclei are stable and which decay, and they are the basis of the release of energy in certain nuclear reactions. Nuclear forces determine not only the stability of nuclei, but also the relative abundance of elements in nature. The properties of the nucleus of an atom determine the number of electrons it has and, thus, indirectly determine the chemistry of the atom. More will be said of all of these topics in later chapters. Concept Connections: The Four Basic Forces The four basic forces will be encountered in more detail as you progress through the text. The gravitational force is defined in Uniform Circular Motion and Gravitation, electric force in Electric Charge and Electric Field, magnetic force in Magnetism, and nuclear forces in Radioactivity and Nuclear Physics. On a macroscopic scale, electromagnetism and gravity are the basis for all forces. The nuclear forces are vital to the substructure of matter, but they are not directly experienced on the macroscopic scale. Table 4.2 Properties of the Four Basic Forces[1] Force Approximate Relative Strengths Range Attraction/Repulsion Carrier Particle Gravitational 10−38 Electrom
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agnetic 10 – 2 Weak nuclear 10 – 13 Strong nuclear 1 ∞ ∞ attractive only Graviton attractive and repulsive Photon < 10–18 m attractive and repulsive W+, W –, Z0 < 10–15 m attractive and repulsive gluons The gravitational force is surprisingly weak—it is only because gravity is always attractive that we notice it at all. Our weight is the gravitational force due to the entire Earth acting on us. On the very large scale, as in astronomical systems, the gravitational force is the dominant force determining the motions of moons, planets, stars, and galaxies. The gravitational force also affects the nature of space and time. As we shall see later in the study of general relativity, space is curved in the vicinity of very massive bodies, such as the Sun, and time actually slows down near massive bodies. Take a good look at the ranges for the four fundamental forces listed in Table 4.2. The range of the strong nuclear force, 10−15 m, is approximately the size of the nucleus of an atom; the weak nuclear force has an even shorter range. At scales on the order of 10−10 m, approximately the size of an atom, both nuclear forces are completely dominated by the electromagnetic force. Notice that this scale is still utterly tiny compared to our everyday experience. At scales that we do experience daily, electromagnetism tends to be negligible, due to its attractive and repulsive properties canceling each other out. That leaves gravity, which is usually the strongest of the forces at scales above ~10−4 m, and hence includes our everyday activities, such as throwing, climbing stairs, and walking. Electromagnetic forces can be either attractive or repulsive. They are long-range forces, which act over extremely large distances, and they nearly cancel for macroscopic objects. (Remember that it is the net external force that is important.) If they did not cancel, electromagnetic forces would completely overwhelm the gravitational force. The electromagnetic force is a combination of electrical forces (such as those that cause static electricity) and magnetic forces (such as those that affect a compass needle). These two forces were thought to be quite distinct until early in the 19th century, when scientists began to discover that they are different manifestations of the same force. This discovery is a classical case of the unification of forces. Similarly, friction, tension, and all of the other classes of forces we experience directly (except gravity, of course) are due to electromagnetic interactions of atoms and molecules. It
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is still convenient to consider these forces separately in specific applications, however, because of the ways they manifest themselves. 1. The graviton is a proposed particle, though it has not yet been observed by scientists. See the discussion of gravitational waves later in this section. The particles W+ are called vector bosons; these were predicted by theory and first observed in 1983. There are eight types of gluons proposed by scientists, and their existence is indicated by meson exchange in the nuclei of atoms., W−, and Z0 176 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Concept Connections: Unifying Forces Attempts to unify the four basic forces are discussed in relation to elementary particles later in this text. By “unify” we mean finding connections between the forces that show that they are different manifestations of a single force. Even if such unification is achieved, the forces will retain their separate characteristics on the macroscopic scale and may be identical only under extreme conditions such as those existing in the early universe. Physicists are now exploring whether the four basic forces are in some way related. Attempts to unify all forces into one come under the rubric of Grand Unified Theories (GUTs), with which there has been some success in recent years. It is now known that under conditions of extremely high density and temperature, such as existed in the early universe, the electromagnetic and weak nuclear forces are indistinguishable. They can now be considered to be different manifestations of one force, called the electroweak force. So the list of four has been reduced in a sense to only three. Further progress in unifying all forces is proving difficult—especially the inclusion of the gravitational force, which has the special characteristics of affecting the space and time in which the other forces exist. While the unification of forces will not affect how we discuss forces in this text, it is fascinating that such underlying simplicity exists in the face of the overt complexity of the universe. There is no reason that nature must be simple—it simply is. Action at a Distance: Concept of a Field All forces act at a distance. This is obvious for the gravitational force. Earth and the Moon, for example, interact without coming into contact. It is also true for all other forces. Friction, for example, is an electromagnetic force between atoms that may not actually touch. What is it that carries forces between objects? One way to answer this question is to imagine that a force field surrounds whatever object creates the force.
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A second object (often called a test object) placed in this field will experience a force that is a function of location and other variables. The field itself is the “thing” that carries the force from one object to another. The field is defined so as to be a characteristic of the object creating it; the field does not depend on the test object placed in it. Earth’s gravitational field, for example, is a function of the mass of Earth and the distance from its center, independent of the presence of other masses. The concept of a field is useful because equations can be written for force fields surrounding objects (for gravity, this yields = at Earth’s surface), and motions can be calculated from these equations. (See Figure 4.26.) Figure 4.26 The electric force field between a positively charged particle and a negatively charged particle. When a positive test charge is placed in the field, the charge will experience a force in the direction of the force field lines. Concept Connections: Force Fields The concept of a force field is also used in connection with electric charge and is presented in Electric Charge and Electric Field. It is also a useful idea for all the basic forces, as will be seen in Particle Physics. Fields help us to visualize forces and how they are transmitted, as well as to describe them with precision and to link forces with subatomic carrier particles. Making Connections: Vector and Scalar Fields These fields may be either scalar or vector fields. Gravity and electromagnetism are examples of vector fields. A test object placed in such a field will have both the magnitude and direction of the resulting force on the test object completely defined by the object’s location in the field. We will later cover examples of scalar fields, which have a magnitude but no direction. The field concept has been applied very successfully; we can calculate motions and describe nature to high precision using field equations. As useful as the field concept is, however, it leaves unanswered the question of what carries the force. It has been proposed in recent decades, starting in 1935 with Hideki Yukawa’s (1907–1981) work on the strong nuclear force, that all forces are transmitted by the exchange of elementary particles. We can visualize particle exchange as analogous to macroscopic phenomena such as two people passing a basketball back and forth, thereby exerting a repulsive force without touching one another. (See Figure 4.27.) This content is available for free at http://cnx
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.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 177 Figure 4.27 The exchange of masses resulting in repulsive forces. (a) The person throwing the basketball exerts a force Fp1 on it toward the other person and feels a reaction force FB away from the second person. (b) The person catching the basketball exerts a force Fp2 on it to stop the ball and feels a reaction force F′B away from the first person. (c) The analogous exchange of a meson between a proton and a neutron carries the strong nuclear forces Fexch and F′exch between them. An attractive force can also be exerted by the exchange of a mass—if person 2 pulled the basketball away from the first person as he tried to retain it, then the force between them would be attractive. This idea of particle exchange deepens rather than contradicts field concepts. It is more satisfying philosophically to think of something physical actually moving between objects acting at a distance. Table 4.2 lists the exchange or carrier particles, both observed and proposed, that carry the four forces. But the real fruit of the particle-exchange proposal is that searches for Yukawa’s proposed particle found it and a number of others that were completely unexpected, stimulating yet more research. All of this research eventually led to the proposal of quarks as the underlying substructure of matter, which is a basic tenet of GUTs. If successful, these theories will explain not only forces, but also the structure of matter itself. Yet physics is an experimental science, so the test of these theories must lie in the domain of the real world. As of this writing, scientists at the CERN laboratory in Switzerland are starting to test these theories using the world’s largest particle accelerator: the Large Hadron Collider. This accelerator (27 km in circumference) allows two high-energy proton beams, traveling in opposite directions, to collide. An energy of 14 trillion electron volts will be available. It is anticipated that some new particles, possibly force carrier particles, will be found. (See Figure 4.28.) One of the force carriers of high interest that researchers hope to detect is the Higgs boson. The observation of its properties might tell us why different particles have different masses. Figure 4.28 The world’s largest particle accelerator spans the border between Switzerland and France. Two beams, traveling in opposite directions close to the speed of light
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, collide in a tube similar to the central tube shown here. External magnets determine the beam’s path. Special detectors will analyze particles created in these collisions. Questions as broad as what is the origin of mass and what was matter like the first few seconds of our universe will be explored. This accelerator began preliminary operation in 2008. (credit: Frank Hommes) 178 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Tiny particles also have wave-like behavior, something we will explore more in a later chapter. To better understand force-carrier particles from another perspective, let us consider gravity. The search for gravitational waves has been going on for a number of years. Almost 100 years ago, Einstein predicted the existence of these waves as part of his general theory of relativity. Gravitational waves are created during the collision of massive stars, in black holes, or in supernova explosions—like shock waves. These gravitational waves will travel through space from such sites much like a pebble dropped into a pond sends out ripples—except these waves move at the speed of light. A detector apparatus has been built in the U.S., consisting of two large installations nearly 3000 km apart—one in Washington state and one in Louisiana! The facility is called the Laser Interferometer Gravitational-Wave Observatory (LIGO). Each installation is designed to use optical lasers to examine any slight shift in the relative positions of two masses due to the effect of gravity waves. The two sites allow simultaneous measurements of these small effects to be separated from other natural phenomena, such as earthquakes. Initial operation of the detectors began in 2002, and work is proceeding on increasing their sensitivity. Similar installations have been built in Italy (VIRGO), Germany (GEO600), and Japan (TAMA300) to provide a worldwide network of gravitational wave detectors. International collaboration in this area is moving into space with the joint EU/US project LISA (Laser Interferometer Space Antenna). Earthquakes and other Earthly noises will be no problem for these monitoring spacecraft. LISA will complement LIGO by looking at much more massive black holes through the observation of gravitational-wave sources emitting much larger wavelengths. Three satellites will be placed in space above Earth in an equilateral triangle (with 5,000,000-km sides) (Figure 4.29). The system will measure the relative positions of each satellite to detect passing gravitational waves. Accuracy to within 10% of the size of an atom will be needed to detect any
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waves. The launch of this project might be as early as 2018. “I’m sure LIGO will tell us something about the universe that we didn’t know before. The history of science tells us that any time you go where you haven’t been before, you usually find something that really shakes the scientific paradigms of the day. Whether gravitational wave astrophysics will do that, only time will tell.” —David Reitze, LIGO Input Optics Manager, University of Florida Figure 4.29 Space-based future experiments for the measurement of gravitational waves. Shown here is a drawing of LISA’s orbit. Each satellite of LISA will consist of a laser source and a mass. The lasers will transmit a signal to measure the distance between each satellite’s test mass. The relative motion of these masses will provide information about passing gravitational waves. (credit: NASA) The ideas presented in this section are but a glimpse into topics of modern physics that will be covered in much greater depth in later chapters. Glossary acceleration: the rate at which an object’s velocity changes over a period of time carrier particle: a fundamental particle of nature that is surrounded by a characteristic force field; photons are carrier particles of the electromagnetic force dynamics: the study of how forces affect the motion of objects and systems external force: a force acting on an object or system that originates outside of the object or system force: a push or pull on an object with a specific magnitude and direction; can be represented by vectors; can be expressed as a multiple of a standard force force field: a region in which a test particle will experience a force free-body diagram: a sketch showing all of the external forces acting on an object or system; the system is represented by a dot, and the forces are represented by vectors extending outward from the dot free-fall: a situation in which the only force acting on an object is the force due to gravity friction: a force past each other of objects that are touching; examples include rough surfaces and air resistance This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 179 inertia: the tendency of an object to remain at rest or remain in motion inertial frame of reference: a coordinate system that is not accelerating; all forces acting in an inertial frame of reference are real forces, as opposed to fictitious forces that are
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observed due to an accelerating frame of reference law of inertia: see Newton’s first law of motion mass: the quantity of matter in a substance; measured in kilograms net external force: the vector sum of all external forces acting on an object or system; causes a mass to accelerate Newton’s first law of motion: in an inertial frame of reference, a body at rest remains at rest, or, if in motion, remains in motion at a constant velocity unless acted on by a net external force; also known as the law of inertia Newton’s second law of motion: the net external force Fnet on an object with mass is proportional to and in the same direction as the acceleration of the object, a, and inversely proportional to the mass; defined mathematically as a = Fnet Newton’s third law of motion: whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts normal force: the force that a surface applies to an object to support the weight of the object; acts perpendicular to the surface on which the object rests system: defined by the boundaries of an object or collection of objects being observed; all forces originating from outside of the system are considered external forces tension: the pulling force that acts along a medium, especially a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the force on the object due to the rope is called a tension force thrust: a reaction force that pushes a body forward in response to a backward force; rockets, airplanes, and cars are pushed forward by a thrust reaction force weight: the force w due to gravity acting on an object of mass ; defined mathematically as: w g, where g is the magnitude and direction of the acceleration due to gravity Section Summary 4.1 Development of Force Concept • Dynamics is the study of how forces affect the motion of objects. • Force is a push or pull that can be defined in terms of various standards, and it is a vector having both magnitude and direction. • External forces are any outside forces that act on a body. A free-body diagram is a drawing of all external forces acting on a body. 4.2 Newton's First Law of Motion: Inertia • Newton’s first law of motion states that in an inertial frame of reference a body at rest remains at rest, or, if in motion, remains in motion
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at a constant velocity unless acted on by a net external force. This is also known as the law of inertia. Inertia is the tendency of an object to remain at rest or remain in motion. Inertia is related to an object’s mass. • • Mass is the quantity of matter in a substance. 4.3 Newton's Second Law of Motion: Concept of a System • Acceleration, a, is defined as a change in velocity, meaning a change in its magnitude or direction, or both. • An external force is one acting on a system from outside the system, as opposed to internal forces, which act between components within the system. • Newton’s second law of motion states that the acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass. • In equation form, Newton’s second law of motion is a = Fnet. • This is often written in the more familiar form: Fnet = a. • The weight w of an object is defined as the force of gravity acting on an object of mass. The object experiences an acceleration due to gravity g : • If the only force acting on an object is due to gravity, the object is in free fall. w = g. 180 Chapter 4 | Dynamics: Force and Newton's Laws of Motion • Friction is a force that opposes the motion past each other of objects that are touching. 4.4 Newton's Third Law of Motion: Symmetry in Forces • Newton’s third law of motion represents a basic symmetry in nature. It states: Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the force that the first body exerts. • A thrust is a reaction force that pushes a body forward in response to a backward force. Rockets, airplanes, and cars are pushed forward by a thrust reaction force. 4.5 Normal, Tension, and Other Examples of Force • When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts perpendicular to and away from the surface. It is called a normal force, N. • When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object: • When objects rest on an inclined plane that makes an angle with the horizontal surface, the
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weight of the object can be resolved into components that act perpendicular ( w⊥ ) and parallel ( w ∥ ) to the surface of the plane. These components can be calculated using: =. • The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T. When a rope ∥ = sin () = sin () ⊥ = cos () = cos (). supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object: • =. In any inertial frame of reference (one that is not accelerated or rotated), Newton’s laws have the simple forms given in this chapter and all forces are real forces having a physical origin. 4.6 Problem-Solving Strategies • To solve problems involving Newton’s laws of motion, follow the procedure described: 1. Draw a sketch of the problem. 2. Identify known and unknown quantities, and identify the system of interest. Draw a free-body diagram, which is a sketch showing all of the forces acting on an object. The object is represented by a dot, and the forces are represented by vectors extending in different directions from the dot. If vectors act in directions that are not horizontal or vertical, resolve the vectors into horizontal and vertical components and draw them on the free-body diagram. 3. Write Newton’s second law in the horizontal and vertical directions and add the forces acting on the object. If the object does not accelerate in a particular direction (for example, the -direction) then net = 0. If the object does accelerate in that direction, net =. 4. Check your answer. Is the answer reasonable? Are the units correct? 4.7 Further Applications of Newton's Laws of Motion • Newton’s laws of motion can be applied in numerous situations to solve problems of motion. • Some problems will contain multiple force vectors acting in different directions on an object. Be sure to draw diagrams, resolve all force vectors into horizontal and vertical components, and draw a free-body diagram. Always analyze the direction in which an object accelerates so that you can determine whether net = or net = 0. • The normal force on an object is not always equal in magnitude to the weight of the object. If an object is accelerating, the normal force will be less than or greater than the weight of the object. Also, if the object is on an inclined plane, the normal force will always be less than the full
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weight of the object. • Some problems will contain various physical quantities, such as forces, acceleration, velocity, or position. You can apply concepts from kinematics and dynamics in order to solve these problems of motion. 4.8 Extended Topic: The Four Basic Forces—An Introduction • The various types of forces that are categorized for use in many applications are all manifestations of the four basic forces in nature. • The properties of these forces are summarized in Table 4.2. • Everything we experience directly without sensitive instruments is due to either electromagnetic forces or gravitational forces. The nuclear forces are responsible for the submicroscopic structure of matter, but they are not directly sensed because of their short ranges. Attempts are being made to show all four forces are different manifestations of a single unified force. • A force field surrounds an object creating a force and is the carrier of that force. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 181 Conceptual Questions 4.1 Development of Force Concept 1. Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly. 2. What properties do forces have that allow us to classify them as vectors? 4.2 Newton's First Law of Motion: Inertia 3. How are inertia and mass related? 4. What is the relationship between weight and mass? Which is an intrinsic, unchanging property of a body? 4.3 Newton's Second Law of Motion: Concept of a System 5. Which statement is correct? (a) Net force causes motion. (b) Net force causes change in motion. Explain your answer and give an example. 6. Why can we neglect forces such as those holding a body together when we apply Newton’s second law of motion? 7. Explain how the choice of the “system of interest” affects which forces must be considered when applying Newton’s second law of motion. 8. Describe a situation in which the net external force on a system is not zero, yet its speed remains constant. 9. A system can have a nonzero velocity while the net external force on it is zero. Describe such a situation. 10. A rock is thrown straight up. What is the net external force acting on the rock when it is at the top of its trajectory? 11. (a) Give an example
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of different net external forces acting on the same system to produce different accelerations. (b) Give an example of the same net external force acting on systems of different masses, producing different accelerations. (c) What law accurately describes both effects? State it in words and as an equation. 12. If the acceleration of a system is zero, are no external forces acting on it? What about internal forces? Explain your answers. 13. If a constant, nonzero force is applied to an object, what can you say about the velocity and acceleration of the object? 14. The gravitational force on the basketball in Figure 4.6 is ignored. When gravity is taken into account, what is the direction of the net external force on the basketball—above horizontal, below horizontal, or still horizontal? 4.4 Newton's Third Law of Motion: Symmetry in Forces 15. When you take off in a jet aircraft, there is a sensation of being pushed back into the seat. Explain why you move backward in the seat—is there really a force backward on you? (The same reasoning explains whiplash injuries, in which the head is apparently thrown backward.) 16. A device used since the 1940s to measure the kick or recoil of the body due to heart beats is the “ballistocardiograph.” What physics principle(s) are involved here to measure the force of cardiac contraction? How might we construct such a device? 17. Describe a situation in which one system exerts a force on another and, as a consequence, experiences a force that is equal in magnitude and opposite in direction. Which of Newton’s laws of motion apply? 18. Why does an ordinary rifle recoil (kick backward) when fired? The barrel of a recoilless rifle is open at both ends. Describe how Newton’s third law applies when one is fired. Can you safely stand close behind one when it is fired? 19. An American football lineman reasons that it is senseless to try to out-push the opposing player, since no matter how hard he pushes he will experience an equal and opposite force from the other player. Use Newton’s laws and draw a free-body diagram of an appropriate system to explain how he can still out-push the opposition if he is strong enough. 20. Newton’s third law of motion tells us that forces always occur in pairs of equal and opposite magnitude. Explain how the choice of the “system of interest” affects
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whether one such pair of forces cancels. 4.5 Normal, Tension, and Other Examples of Force 21. If a leg is suspended by a traction setup as shown in Figure 4.30, what is the tension in the rope? 182 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Figure 4.30 A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the force T without changing its magnitude. 22. In a traction setup for a broken bone, with pulleys and rope available, how might we be able to increase the force along the tibia using the same weight? (See Figure 4.30.) (Note that the tibia is the shin bone shown in this image.) 4.7 Further Applications of Newton's Laws of Motion 23. To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at. Why will they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft? 24. A cartoon shows the toupee coming off the head of an elevator passenger when the elevator rapidly stops during an upward ride. Can this really happen without the person being tied to the floor of the elevator? Explain your answer. 4.8 Extended Topic: The Four Basic Forces—An Introduction 25. Explain, in terms of the properties of the four basic forces, why people notice the gravitational force acting on their bodies if it is such a comparatively weak force. 26. What is the dominant force between astronomical objects? Why are the other three basic forces less significant over these very large distances? 27. Give a detailed example of how the exchange of a particle can result in an attractive force. (For example, consider one child pulling a toy out of the hands of another.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 183 Problems & Exercises 4.3 Newton's Second Law of Motion: Concept of a System You may assume data taken from illustrations is accurate to three digits. 1. A 63.0-kg sprinter starts a race with an acceleration of 4.20 m/s2. What is the net external force on him? Figure 4.32 2. If the sprinter from the previous problem acceler
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ates at that rate for 20 m, and then maintains that velocity for the remainder of the 100-m dash, what will be his time for the race? 8. What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.) 3. A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate the magnitude of its acceleration. 4. Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be 0.893 m/s2. (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided. 5. In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force is removed. How far will the mower go before stopping? 6. The same rocket sled drawn in Figure 4.31 is decelerated at a rate of 196 m/s2. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg. 9. Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg. (a) What is the system of interest if the acceleration of the child in the wagon is to be calculated? (b) Draw a free-body diagram, including all forces acting on the system. (c) Calcul
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ate the acceleration. (d) What would the acceleration be if friction were 15.0 N? 10. A powerful motorcycle can produce an acceleration of 3.50 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force the motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg? 11. The rocket sled shown in Figure 4.33 accelerates at a rate of 49.0 m/s2. Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight by using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body. Figure 4.31 7. (a) If the rocket sled shown in Figure 4.32 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4×104 N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not onefourth of what it is with all rockets burning? Figure 4.33 12. Repeat the previous problem for the situation in which the rocket sled decelerates at a rate of 201 m/s2. In this problem, the forces are exerted by the seat and restraining belts. 13. The weight of an astronaut plus his space suit on the Moon is only 250 N. How much do they weigh on Earth? What is the mass on the Moon? On Earth? 14. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 10,000 kg. The thrust of its engines is 30,000 N. (a) Calculate its the magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration. 184 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 4.4 Newton's Third Law of Motion: Symmetry in Forces 15. What net external force is exerted on a 1100-kg artillery shell fired from a battleship if the shell is accelerated at 2
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.40104 m/s2? What is the magnitude of the force exerted on the ship by the artillery shell? 16. A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 1.20 m/s2 backward. (a) What is the force of friction between the losing player’s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation. 4.5 Normal, Tension, and Other Examples of Force 17. Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope between the teams? 18. What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s2? Note that the answer is independent of the velocity of the gymnast—she can be moving either up or down, or be stationary. 19. (a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00×10−5 kg hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 4.17. The strand sags at an angle of 12º below the horizontal. Compare this with the tension in the vertical strand (find their ratio). 20. Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he climbs at a constant speed? (b) What is the tension in the rope if he accelerates upward at a rate of 1.50 m/s2? 21. Show that, as stated in the
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text, a force F⊥ exerted on a flexible medium at its center and perpendicular to its length (such as on the tightrope wire in Figure 4.17) gives rise to a ⊥ 2 sin () tension of magnitude =. 22. Consider the baby being weighed in Figure 4.34. (a) What is the mass of the child and basket if a scale reading of 55 N is observed? (b) What is the tension 1 in the cord attaching the baby to the scale? (c) What is the tension 2 in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible. This content is available for free at http://cnx.org/content/col11844/1.13 Figure 4.34 A baby is weighed using a spring scale. 4.6 Problem-Solving Strategies 23. A 5.00×105-kg rocket is accelerating straight up. Its engines produce 1.250×107 N of thrust, and air resistance is 4.50×106 N. What is the rocket’s acceleration? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. 24. The wheels of a midsize car exert a force of 2100 N backward on the road to accelerate the car in the forward direction. If the force of friction including air resistance is 250 N and the acceleration of the car is 1.80 m/s2, what is the mass of the car plus its occupants? Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. For this situation, draw a free-body diagram and write the net force equation. 25. Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. 26. When landing after a spectacular somersault, a 40.0-kg gymnast decelerates by pushing straight down on the mat. Calculate the force she must exert if her deceleration is 7.00 times the acceleration due to gravity. Explicitly show how you follow the steps in the Problem-Solving Strategy for
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Newton’s laws of motion. Chapter 4 | Dynamics: Force and Newton's Laws of Motion 185 27. A freight train consists of two 8.00×104 -kg engines and 45 cars with average masses of 5.50×104 kg. (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00×10–2 m/s2 if the force of friction is 7.50×105 N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines? 28. Commercial airplanes are sometimes pushed out of the passenger loading area by a tractor. (a) An 1800-kg tractor exerts a force of 1.75×104 N backward on the pavement, and the system experiences forces resisting motion that total 2400 N. If the acceleration is 0.150 m/s2, what is the mass of the airplane? (b) Calculate the force exerted by the tractor on the airplane, assuming 2200 N of the friction is experienced by the airplane. (c) Draw two sketches showing the systems of interest used to solve each part, including the free-body diagrams for each. 29. A 1100-kg car pulls a boat on a trailer. (a) What total force resists the motion of the car, boat, and trailer, if the car exerts a 1900-N force on the road and produces an acceleration of 0.550 m/s2? The mass of the boat plus trailer is 700 kg. (b) What is the force in the hitch between the car and the trailer if 80% of the resisting forces are experienced by the boat and trailer? 30. (a) Find the magnitudes of the forces F1 and F2 that add to give the total force Ftot shown in Figure 4.35. This may be done either graphically or by using trigonometry. (b) Show graphically that the same total force is obtained independent of the order of addition of F1 and F2. (c) Find the direction and magnitude of some other pair of vectors that add to give Ftot. Draw these to scale
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on the same drawing used in part (b) or a similar picture. Figure 4.35 31. Two children pull a third child on a snow saucer sled exerting forces F1 and F2 as shown from above in Figure 4.36. Find the acceleration of the 49.00-kg sled and child system. Note that the direction of the frictional force is unspecified; it will be in the opposite direction of the sum of F1 and F2. Figure 4.36 An overhead view of the horizontal forces acting on a child’s snow saucer sled. 32. Suppose your car was mired deeply in the mud and you wanted to use the method illustrated in Figure 4.37 to pull it out. (a) What force would you have to exert perpendicular to the center of the rope to produce a force of 12,000 N on the car if the angle is 2.00°? In this part, explicitly show how you follow the steps in the Problem-Solving Strategy for Newton’s laws of motion. (b) Real ropes stretch under such forces. What force would be exerted on the car if the angle increases to 7.00° and you still apply the force found in part (a) to its center? Figure 4.37 33. What force is exerted on the tooth in Figure 4.38 if the tension in the wire is 25.0 N? Note that the force applied to the tooth is smaller than the tension in the wire, but this is necessitated by practical considerations of how force can be applied in the mouth. Explicitly show how you follow steps in the Problem-Solving Strategy for Newton’s laws of motion. Figure 4.38 Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, Fapp, points straight toward the back of the mouth. 34. Figure 4.39 shows Superhero and Trusty Sidekick hanging motionless from a rope. Superhero’s mass is 90.0 kg, while Trusty Sidekick’s is 55.0 kg, and the mass of the rope is negligible. (a) Draw a free-body diagram of the situation showing all forces acting on Superhero, Trusty Sidekick, and the rope. (b) Find the tension in the rope above Superhero. (c) Find the tension in the rope between
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186 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Superhero and Trusty Sidekick. Indicate on your free-body diagram the system of interest used to solve each part. is unreasonable about the result? (c) Which premise is unreasonable, and why is it unreasonable? 39. Unreasonable Results (a) What is the initial acceleration of a rocket that has a mass of 1.50×106 kg at takeoff, the engines of which produce a thrust of 2.00×106 N? Do not neglect gravity. (b) What is unreasonable about the result? (This result has been unintentionally achieved by several real rockets.) (c) Which premise is unreasonable, or which premises are inconsistent? (You may find it useful to compare this problem to the rocket problem earlier in this section.) 4.7 Further Applications of Newton's Laws of Motion 40. A flea jumps by exerting a force of 1.20×10−5 N straight down on the ground. A breeze blowing on the flea parallel to the ground exerts a force of 0.500×10−6 N on the flea. Find the direction and magnitude of the acceleration of the flea if its mass is 6.00×10−7 kg. Do not neglect the gravitational force. 41. Two muscles in the back of the leg pull upward on the Achilles tendon, as shown in Figure 4.40. (These muscles are called the medial and lateral heads of the gastrocnemius muscle.) Find the magnitude and direction of the total force on the Achilles tendon. What type of movement could be caused by this force? Figure 4.40 Achilles tendon 42. A 76.0-kg person is being pulled away from a burning building as shown in Figure 4.41. Calculate the tension in the two ropes if the person is momentarily motionless. Include a free-body diagram in your solution. Figure 4.39 Superhero and Trusty Sidekick hang motionless on a rope as they try to figure out what to do next. Will the tension be the same everywhere in the rope? 35. A nurse pushes a cart by exerting a force on the handle at a downward angle 35.0º below the horizontal. The loaded cart has a mass of 28.0 kg, and the force of friction is 60.0 N. (a) Draw a free-body diagram for the system of interest. (b) What force must the nurse exert to move at a constant velocity? 36
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. Construct Your Own Problem Consider the tension in an elevator cable during the time the elevator starts from rest and accelerates its load upward to some cruising velocity. Taking the elevator and its load to be the system of interest, draw a free-body diagram. Then calculate the tension in the cable. Among the things to consider are the mass of the elevator and its load, the final velocity, and the time taken to reach that velocity. 37. Construct Your Own Problem Consider two people pushing a toboggan with four children on it up a snowcovered slope. Construct a problem in which you calculate the acceleration of the toboggan and its load. Include a freebody diagram of the appropriate system of interest as the basis for your analysis. Show vector forces and their components and explain the choice of coordinates. Among the things to be considered are the forces exerted by those pushing, the angle of the slope, and the masses of the toboggan and children. 38. Unreasonable Results (a) Repeat Exercise 4.29, but assume an acceleration of 1.20 m/s2 is produced. (b) What This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 4 | Dynamics: Force and Newton's Laws of Motion 187 average force on the shell in the mortar? Express your answer in newtons and as a ratio to the weight of the shell. 48. Integrated Concepts Repeat Exercise 4.47 for a shell fired at an angle 10.0º from the vertical. 49. Integrated Concepts An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of 0.600 m/s2 for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity? 50. Unreasonable Results (a) What is the final velocity of a car originally traveling at 50.0 km/h that decelerates at a rate of 0.400 m/s2 for 50.0 s? (b) What is unreasonable about the result? (c) Which
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premise is unreasonable, or which premises are inconsistent? 51. Unreasonable Results A 75.0-kg man stands on a bathroom scale in an elevator that accelerates from rest to 30.0 m/s in 2.00 s. (a) Calculate the scale reading in newtons and compare it with his weight. (The scale exerts an upward force on him equal to its reading.) (b) What is unreasonable about the result? (c) Which premise is unreasonable, or which premises are inconsistent? 4.8 Extended Topic: The Four Basic Forces—An Introduction 52. (a) What is the strength of the weak nuclear force relative to the strong nuclear force? (b) What is the strength of the weak nuclear force relative to the electromagnetic force? Since the weak nuclear force acts at only very short distances, such as inside nuclei, where the strong and electromagnetic forces also act, it might seem surprising that we have any knowledge of it at all. We have such knowledge because the weak nuclear force is responsible for beta decay, a type of nuclear decay not explained by other forces. 53. (a) What is the ratio of the strength of the gravitational force to that of the strong nuclear force? (b) What is the ratio of the strength of the gravitational force to that of the weak nuclear force? (c) What is the ratio of the strength of the gravitational force to that of the electromagnetic force? What do your answers imply about the influence of the gravitational force on atomic nuclei? 54. What is the ratio of the strength of the strong nuclear force to that of the electromagnetic force? Based on this ratio, you might expect that the strong force dominates the nucleus, which is true for small nuclei. Large nuclei, however, have sizes greater than the range of the strong nuclear force. At these sizes, the electromagnetic force begins to affect nuclear stability. These facts will be used to explain nuclear fusion and fission later in this text. Figure 4.41 The force T2 needed to hold steady the person being rescued from the fire is less than her weight and less than the force T1 in the other rope, since the more vertical rope supports a greater part of her weight (a vertical force). 43. Integrated Concepts A 35.0-kg dolphin decelerates from 12.0 to 7.50 m/s in 2.30 s to join another dolphin in play. What average force was exerted to slow him if he was moving horizontally? (The gravitational force is balanced
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by the buoyant force of the water.) 44. Integrated Concepts When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. (a) What is his final speed? (b) How far does he travel? 45. Integrated Concepts A large rocket has a mass of 2.00×106 kg at takeoff, and its engines produce a thrust of 3.50×107 N. (a) Find its initial acceleration if it takes off vertically. (b) How long does it take to reach a velocity of 120 km/h straight up, assuming constant mass and thrust? (c) In reality, the mass of a rocket decreases significantly as its fuel is consumed. Describe qualitatively how this affects the acceleration and time for this motion. 46. Integrated Concepts A basketball player jumps straight up for a ball. To do this, he lowers his body 0.300 m and then accelerates through this distance by forcefully straightening his legs. This player leaves the floor with a vertical velocity sufficient to carry him 0.900 m above the floor. (a) Calculate his velocity when he leaves the floor. (b) Calculate his acceleration while he is straightening his legs. He goes from zero to the velocity found in part (a) in a distance of 0.300 m. (c) Calculate the force he exerts on the floor to do this, given that his mass is 110 kg. 47. Integrated Concepts A 2.50-kg fireworks shell is fired straight up from a mortar and reaches a height of 110 m. (a) Neglecting air resistance (a poor assumption, but we will make it for this example), calculate the shell’s velocity when it leaves the mortar. (b) The mortar itself is a tube 0.450 m long. Calculate the average acceleration of the shell in the tube as it goes from zero to the velocity found in (a). (c) What is the 188 Chapter 4 | Dynamics: Force and Newton's Laws of Motion Test Prep for AP® Courses 4.1 Development of Force Concept 1. Figure 4.42 The figure above represents a racetrack with semicircular sections connected by straight sections. Each section has length d, and markers along the track are spaced d/4 apart. Two people drive cars counterclockwise around the track, as shown. Car X goes around the curves at constant speed vc, increases speed at
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constant acceleration for half of each straight section to reach a maximum speed of 2vc, then brakes at constant acceleration for the other half of each straight section to return to speed vc. Car Y also goes around the curves at constant speed vc, increases its speed at constant acceleration for one-fourth of each straight section to reach the same maximum speed 2vc, stays at that speed for half of each straight section, then brakes at constant acceleration for the remaining fourth of each straight section to return to speed vc. (a) On the figures below, draw an arrow showing the direction of the net force on each of the cars at the positions noted by the dots. If the net force is zero at any position, label the dot with 0. Figure 4.43 The position of the six dots on the Car Y track on the right are as follows: The first dot on the left center of the track is at the same position as it is on the Car X track. The second dot is just slight to the right of the Car X dot (less than a dash) past three perpendicular hash marks moving to the right. The third dot is about one and two-thirds perpendicular hash marks to the right of the center top perpendicular has mark. The fourth dot is in the same position as the Car X figure (one perpendicular hash mark above the center right perpendicular hash mark). The fifth dot is about one and two-third perpendicular hash marks to the right of the center bottom perpendicular hash mark. The sixth dot is in the same position as the Car Y dot (one and two third perpendicular hash marks to the left of the center bottom hash mark). (b) i. Indicate which car, if either, completes one trip around the track in less time, and justify your answer qualitatively without using equations. This content is available for free at http://cnx.org/content/col11844/1.13 ii. Justify your answer about which car, if either, completes one trip around the track in less time quantitatively with appropriate equations. 2. Which of the following is an example of a body exerting a force on itself? a. a person standing up from a seated position b. a car accelerating while driving c. both of the above d. none of the above 3. A hawk accelerates as it glides in the air. Does the force causing the acceleration come from the hawk itself? Explain. 4. What causes the force that moves a boat forward when someone rows it? a.
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The force is caused by the rower’s arms. b. The force is caused by an interaction between the oars and gravity. c. The force is caused by an interaction between the oars and the water the boat is traveling in. d. The force is caused by friction. 4.4 Newton's Third Law of Motion: Symmetry in Forces 5. What object or objects commonly exert forces on the following objects in motion? (a) a soccer ball being kicked, (b) a dolphin jumping, (c) a parachutist drifting to Earth. 6. A ball with a mass of 0.25 kg hits a gym ceiling with a force of 78.0 N. What happens next? a. The ball accelerates downward with a force of 80.5 N. b. The ball accelerates downward with a force of 78.0 N. c. The ball accelerates downward with a force of 2.45 N. d. It depends on the height of the ceiling. 7. Which of the following is true? a. Earth exerts a force due to gravity on your body, and your body exerts a smaller force on the Earth, because your mass is smaller than the mass of the Earth. b. The Moon orbits the Earth because the Earth exerts a force on the Moon and the Moon exerts a force equal in magnitude and direction on the Earth. c. A rocket taking off exerts a force on the Earth equal to the force the Earth exerts on the rocket. d. An airplane cruising at a constant speed is not affected by gravity. 8. Stationary skater A pushes stationary skater B, who then accelerates at 5.0 m/s2. Skater A does not move. Since forces act in action-reaction pairs, explain why Skater A did not move? 9. The current in a river exerts a force of 9.0 N on a balloon floating in the river. A wind exerts a force of 13.0 N on the balloon in the opposite direction. Draw a free-body diagram to show the forces acting on the balloon. Use your free-body diagram to predict the effect on the balloon. 10. A force is applied to accelerate an object on a smooth icy surface. When the force stops, which of the following will be true? (Assume zero friction.) a. The object’s acceleration becomes zero. b. The object’s speed becomes zero. c. The object
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’s acceleration continues to increase at a constant rate. d. The object accelerates, but in the opposite direction. 11. A parachutist’s fall to Earth is determined by two opposing forces. A gravitational force of 539 N acts on the parachutist. After 2 s, she opens her parachute and experiences an air resistance of 615 N. At what speed is the parachutist falling after 10 s? Chapter 4 | Dynamics: Force and Newton's Laws of Motion 189 12. A flight attendant pushes a cart down the aisle of a plane in flight. In determining the acceleration of the cart relative to the plane, which factor do you not need to consider? a. The friction of the cart’s wheels. b. The force with which the flight attendant’s feet push on the floor. c. The velocity of the plane. d. The mass of the items in the cart. 13. A landscaper is easing a wheelbarrow full of soil down a hill. Define the system you would analyze and list all the forces that you would need to include to calculate the acceleration of the wheelbarrow. 14. Two water-skiers, with masses of 48 kg and 61 kg, are preparing to be towed behind the same boat. When the boat accelerates, the rope the skiers hold onto accelerates with it and exerts a net force of 290 N on the skiers. At what rate will the skiers accelerate? a. 10.8 m/s2 b. 2.7 m/s2 c. 6.0 m/s2 and 4.8 m/s2 d. 5.3 m/s2 15. A figure skater has a mass of 40 kg and her partner's mass is 50 kg. She pushes against the ice with a force of 120 N, causing her and her partner to move forward. Calculate the pair’s acceleration. Assume that all forces opposing the motion, such as friction and air resistance, total 5.0 N. 4.5 Normal, Tension, and Other Examples of Force 16. An archer shoots an arrow straight up with a force of 24.5 N. The arrow has a mass of 0.4 kg. What is the force of gravity on the arrow? a. 9.8 m/s2 b. 9.8 N c. 61.25 N d. 3.9 N 17. A cable raises a mass of 120.0
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kg with an acceleration of 1.3 m/s2. What force of tension is in the cable? 18. A child pulls a wagon along a grassy field. Define the system, the pairs of forces at work, and the results. 19. Two teams are engaging in a tug–of-war. The rope suddenly snaps. Which statement is true about the forces involved? a. The forces exerted by the two teams are no longer equal; the teams will accelerate in opposite directions as a result. b. The forces exerted by the players are no longer balanced by the force of tension in the rope; the teams will accelerate in opposite directions as a result. c. The force of gravity balances the forces exerted by the players; the teams will fall as a result d. The force of tension in the rope is transferred to the players; the teams will accelerate in opposite directions as a result. 20. The following free-body diagram represents a toboggan on a hill. What acceleration would you expect, and why? Figure 4.44 a. Acceleration down the hill; the force due to being pushed, together with the downhill component of gravity, overcomes the opposing force of friction. b. Acceleration down the hill; friction is less than the opposing component of force due to gravity. c. No movement; friction is greater than the force due to being pushed. It depends on how strong the force due to friction is. p d. 21. Draw a free-body diagram to represent the forces acting on a kite on a string that is floating stationary in the air. Label the forces in your diagram. 22. A car is sliding down a hill with a slope of 20°. The mass of the car is 965 kg. When a cable is used to pull the car up the slope, a force of 4215 N is applied. What is the car’s acceleration, ignoring friction? 4.6 Problem-Solving Strategies 23. A toboggan with two riders has a total mass of 85.0 kg. A third person is pushing the toboggan with a force of 42.5 N at the top of a hill with an angle of 15°. The force of friction on the toboggan is 31.0 N. Which statement describes an accurate free-body diagram to represent the situation? a. An arrow of magnitude 10.5 N points down the slope of the hill. b. An arrow of magnitude 833 N points straight down. c. An
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arrow of magnitude 833 N points perpendicular to the slope of the hill. d. An arrow of magnitude 73.5 N points down the slope of the hill. 24. A mass of 2.0 kg is suspended from the ceiling of an elevator by a rope. What is the tension in the rope when the elevator (i) accelerates upward at 1.5 m/s2? (ii) accelerates downward at 1.5 m/s2? a. b. Because the mass is hanging from the elevator itself, the tension in the rope will not change in either case. (i) 22.6 N; (ii) 19.6 N (i) 16.6 N; (ii) 19.6 N (i) 22.6 N; (ii) 16.6 N c. d. 25. Which statement is true about drawing free-body diagrams? 190 Chapter 4 | Dynamics: Force and Newton's Laws of Motion a. Drawing a free-body diagram should be the last step in solving a problem about forces. b. Drawing a free-body diagram helps you compare forces 30. Explain which of the four fundamental forces is responsible for a ball bouncing off the ground after it hits, and why this force has this effect. quantitatively. c. The forces in a free-body diagram should always balance. d. Drawing a free-body diagram can help you determine the net force. 4.7 Further Applications of Newton's Laws of Motion 26. A basketball player jumps as he shoots the ball. Describe the forces that are acting on the ball and on the basketball player. What are the results? 27. Two people push on a boulder to try to move it. The mass of the boulder is 825 kg. One person pushes north with a force of 64 N. The other pushes west with a force of 38 N. Predict the magnitude of the acceleration of the boulder. Assume that friction is negligible. 28. 31. Which of the basic forces best explains tension in a rope being pulled between two people? Is the acting force causing attraction or repulsion in this instance? a. gravity; attraction b. electromagnetic; attraction c. weak and strong nuclear; attraction d. weak and strong nuclear; repulsion 32. Explain how interatomic electric forces produce the normal force, and why it has the direction it does. 33. The gravitational force is the weakest of the four basic forces. In which case can the electromagnetic, strong, and weak forces be ignored because the gravitational
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force is so strongly dominant? a. a person jumping on a trampoline b. a rocket blasting off from Earth c. a log rolling down a hill d. all of the above 34. Describe a situation in which gravitational force is the dominant force. Why can the other three basic forces be ignored in the situation you described? Figure 4.45 The figure shows the forces exerted on a block that is sliding on a horizontal surface: the gravitational force of 40 N, the 40 N normal force exerted by the surface, and a frictional force exerted to the left. The coefficient of friction between the block and the surface is 0.20. The acceleration of the block is most nearly a. 1.0 m/s2 to the right b. 1.0 m/s2 to the left c. 2.0 m/s2 to the right d. 2.0 m/s2 to the left 4.8 Extended Topic: The Four Basic Forces—An Introduction 29. Which phenomenon correctly describes the direction and magnitude of normal forces? a. electromagnetic attraction b. electromagnetic repulsion c. gravitational attraction d. gravitational repulsion This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 191 FURTHER APPLICATIONS OF NEWTON'S 5 LAWS: FRICTION, DRAG, AND ELASTICITY Figure 5.1 Total hip replacement surgery has become a common procedure. The head (or ball) of the patient's femur fits into a cup that has a hard plastic-like inner lining. (credit: National Institutes of Health, via Wikimedia Commons) Chapter Outline 5.1. Friction 5.2. Drag Forces 5.3. Elasticity: Stress and Strain Connection for AP® Courses Have you ever wondered why it is difficult to walk on a smooth surface like ice? The interaction between you and the surface is a result of forces that affect your motion. In the previous chapter, you learned Newton's laws of motion and examined how net force affects the motion, position and shape of an object. Now we will look at some interesting and common forces that will provide further applications of Newton's laws of motion. The information presented in this chapter supports learning objectives covered under Big Idea 3 of the AP Physics Curriculum Framework, which refer to the nature of forces and their roles in interactions among objects. The chapter discusses examples
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of specific contact forces, such as friction, air or liquid drag, and elasticity that may affect the motion or shape of an object. It also discusses the nature of forces on both macroscopic and microscopic levels (Enduring Understanding 3.C and Essential Knowledge 3.C.4). In addition, Newton's laws are applied to describe the motion of an object (Enduring Understanding 3.B) and to examine relationships between contact forces and other forces exerted on an object (Enduring Understanding 3.A, 3.A.3 and 192 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Essential Knowledge 3.A.4). The examples in this chapter give you practice in using vector properties of forces (Essential Knowledge 3.A.2) and free-body diagrams (Essential Knowledge 3.B.2) to determine net force (Essential Knowledge 3.B.1). Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.2 Forces are described by vectors. Essential Knowledge 3.A.3 A force exerted on an object is always due to the interaction of that object with another object. Essential Knowledge 3.A.4 If one object exerts a force on a second object, the second object always exerts a force of equal magnitude on the first object in the opposite direction. Enduring Understanding 3.B Classically, the acceleration of an object interacting with other objects can be predicted by using → = → ∑. Essential Knowledge 3.B.1 If an object of interest interacts with several other objects, the net force is the vector sum of the individual forces. Essential Knowledge 3.B.2 Free-body diagrams are useful tools for visualizing forces being exerted on a single object and writing the equations that represent a physical situation. Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. Essential Knowledge 3.C.4 Contact forces result from the interaction of one object touching another object, and they arise from interatomic electric forces. These forces include tension, friction, normal, spring (Physics 1), and buoyant (Physics 2). 5.1 Friction Learning Objectives By the end of this section, you will be able
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to: • Discuss the general characteristics of friction. • Describe the various types of friction. • Calculate the magnitudes of static and kinetic frictional forces. The information presented in this section supports the following AP® learning objectives and science practices: • 3.C.4.1 The student is able to make claims about various contact forces between objects based on the microscopic cause of those forces. (S.P. 6.1) • 3.C.4.2 The student is able to explain contact forces (tension, friction, normal, buoyant, spring) as arising from interatomic electric forces and that they therefore have certain directions. (S.P. 6.2) Friction is a force that is around us all the time that opposes relative motion between systems in contact but also allows us to move (which you have discovered if you have ever tried to walk on ice). While a common force, the behavior of friction is actually very complicated and is still not completely understood. We have to rely heavily on observations for whatever understandings we can gain. However, we can still deal with its more elementary general characteristics and understand the circumstances in which it behaves. Friction Friction is a force that opposes relative motion between systems in contact. One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative to each other. If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice. But when objects are stationary, static friction can act between them; the static friction is usually greater than the kinetic friction between the objects. Kinetic Friction If two systems are in contact and moving relative to one another, then the friction between them is called kinetic friction. Imagine, for example, trying to slide a heavy crate across a concrete floor—you may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion it is easier to keep it in motion than it was to get it started, indicating that the kinetic friction force is less than the static friction This content is available for free at http://cnx.org/content/col11844/1
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.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 193 force. If you add mass to the crate, say by placing a box on top of it, you need to push even harder to get it started and also to keep it moving. Furthermore, if you oiled the concrete you would find it to be easier to get the crate started and keep it going (as you might expect). Figure 5.2 is a crude pictorial representation of how friction occurs at the interface between two objects. Close-up inspection of these surfaces shows them to be rough. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. A considerable force can be resisted by friction with no apparent motion. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. Part of the friction is due to adhesive forces between the surface molecules of the two objects, which explain the dependence of friction on the nature of the substances. Adhesion varies with substances in contact and is a complicated aspect of surface physics. Once an object is moving, there are fewer points of contact (fewer molecules adhering), so less force is required to keep the object moving. At small but nonzero speeds, friction is nearly independent of speed. Figure 5.2 Frictional forces, such as, always oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view. In order for the object to move, it must rise to where the peaks can skip along the bottom surface. Thus a force is required just to set the object in motion. Some of the peaks will be broken off, also requiring a force to maintain motion. Much of the friction is actually due to attractive forces between molecules making up the two objects, so that even perfectly smooth surfaces are not friction-free. Such adhesive forces also depend on the substances the surfaces are made of, explaining, for example, why rubber-soled shoes slip less than those with leather soles. The magnitude of the frictional force has two forms: one for static situations (static friction), the other for when there is motion (kinetic friction). When there is no motion between the objects, the magnitude of static friction fs is s ≤ s
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, (5.1) where s is the coefficient of static friction and is the magnitude of the normal force (the force perpendicular to the surface). Magnitude of Static Friction Magnitude of static friction s is where s is the coefficient of static friction and is the magnitude of the normal force. s ≤ s, (5.2) The symbol ≤ means less than or equal to, implying that static friction can have a minimum and a maximum value of s. Static friction is a responsive force that increases to be equal and opposite to whatever force is exerted, up to its maximum limit. Once the applied force exceeds s(max), the object will move. Thus Once an object is moving, the magnitude of kinetic friction fk is given by k = k, s(max) = s. (5.3) (5.4) where k is the coefficient of kinetic friction. A system in which k = k is described as a system in which friction behaves simply. Magnitude of Kinetic Friction The magnitude of kinetic friction k is given by 194 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity where k is the coefficient of kinetic friction. k = k, (5.5) As seen in Table 5.1, the coefficients of kinetic friction are less than their static counterparts. That values of in Table 5.1 are stated to only one or, at most, two digits is an indication of the approximate description of friction given by the above two equations. Table 5.1 Coefficients of Static and Kinetic Friction System Static friction μs Kinetic friction μk Rubber on dry concrete Rubber on wet concrete Wood on wood Waxed wood on wet snow Metal on wood Steel on steel (dry) Steel on steel (oiled) Teflon on steel 1.0 0.7 0.5 0.14 0.5 0.6 0.05 0.04 Bone lubricated by synovial fluid 0.016 Shoes on wood Shoes on ice Ice on ice Steel on ice 0.9 0.1 0.1 0.4 0.7 0.5 0.3 0.1 0.3 0.3 0.03 0.04 0.015 0.7 0.05 0.03 0.02 The equations given earlier include the dependence of friction on materials and the normal force. The direction of friction is always opposite that of motion, parallel to the surface between objects, and perpendicular to the normal force. For
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example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force would be equal to its weight, = = (100 kg)(9.80 m/s2) = 980 N, perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than s(max) = s = (0.45)(980 N) = 440 N to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only 290 N ( k = k = (0.30)(980 N) = 290 N ) would keep it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unit less quantity with a magnitude usually between 0 and 1.0. The coefficient of the friction depends on the two surfaces that are in contact. Take-Home Experiment Find a small plastic object (such as a food container) and slide it on a kitchen table by giving it a gentle tap. Now spray water on the table, simulating a light shower of rain. What happens now when you give the object the same-sized tap? Now add a few drops of (vegetable or olive) oil on the surface of the water and give the same tap. What happens now? This latter situation is particularly important for drivers to note, especially after a light rain shower. Why? Many people have experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints, have much smaller coefficients of friction—often three or four times less than ice. A joint is formed by the ends of two bones, which are connected by thick tissues. The knee joint is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at the end of the femur) and socket (part of the pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, almost glassy surface. The joints also produce a fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint can be replaced by an artificial joint (Figure 5.3). These replacements can be made of metals (stainless steel or titanium) or plastic
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(polyethylene), also with very small coefficients of friction. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 195 Figure 5.3 Artificial knee replacement is a procedure that has been performed for more than 20 years. In this figure, we see the post-op x rays of the right knee joint replacement. (credit: Mike Baird, Flickr) Other natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus found between organs in the body, allowing them to move freely past each other during heartbeats, during breathing, and when a person moves. Artificial lubricants are also common in hospitals and doctor's clinics. For example, when ultrasonic imaging is carried out, the gel that couples the transducer to the skin also serves to to lubricate the surface between the transducer and the skin—thereby reducing the coefficient of friction between the two surfaces. This allows the transducer to mover freely over the skin. Example 5.1 Skiing Exercise A skier with a mass of 62 kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 N. Strategy The magnitude of kinetic friction was given in to be 45.0 N. Kinetic friction is related to the normal force N as k = k ; thus, the coefficient of kinetic friction can be found if we can find the normal force of the skier on a slope. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier's weight perpendicular to the slope. (See the skier and free-body diagram in Figure 5.4.) 196 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Figure 5.4 The motion of the skier and friction are parallel to the slope and so it is most convenient to project all forces onto a coordinate system where one axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N (the normal force) is perpendicular to the slope, and f (the friction) is parallel to the slope, but w (the skier's weight) has components along both axes, namely w⊥ and W//. N is equal in
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magnitude to w⊥, so there is no motion perpendicular to the slope. However, f is less than W// in magnitude, so there is acceleration down the slope (along the x-axis). That is, = ⊥ = cos 25º = cos 25º. Substituting this into our expression for kinetic friction, we get k = k cos 25º, which can now be solved for the coefficient of kinetic friction k. Solution Solving for k gives k = k = k = k cos 25º cos 25º. Substituting known values on the right-hand side of the equation, k = 45.0 N (62 kg)(9.80 m/s2)(0.906) = 0.082. Discussion (5.6) (5.7) (5.8) (5.9) This result is a little smaller than the coefficient listed in Table 5.1 for waxed wood on snow, but it is still reasonable since values of the coefficients of friction can vary greatly. In situations like this, where an object of mass slides down a slope that makes an angle with the horizontal, friction is given by k = k cos. All objects will slide down a slope with constant acceleration under these circumstances. Proof of this is left for this chapter's Problems and Exercises. Take-Home Experiment An object will slide down an inclined plane at a constant velocity if the net force on the object is zero. We can use this fact to measure the coefficient of kinetic friction between two objects. As shown in Example 5.1, the kinetic friction on a slope k = k cos. The component of the weight down the slope is equal to sin (see the free-body diagram in Figure 5.4). These forces act in opposite directions, so when they have equal magnitude, the acceleration is zero. Writing these out: Solving for k, we find that k = k cos = sin. k = sin cos = tan. (5.10) (5.11) (5.12) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 197 Put a coin on a book and tilt it until the coin slides at a constant velocity down the book. You might need to tap the book lightly to get the coin to move. Measure the angle of tilt relative to the horizontal and find k.
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Note that the coin will not start to slide at all until an angle greater than is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Discuss how this may affect the value for k and its uncertainty. We have discussed that when an object rests on a horizontal surface, there is a normal force supporting it equal in magnitude to its weight. Furthermore, simple friction is always proportional to the normal force. Making Connections: Submicroscopic Explanations of Friction The simpler aspects of friction dealt with so far are its macroscopic (large-scale) characteristics. Great strides have been made in the atomic-scale explanation of friction during the past several decades. Researchers are finding that the atomic nature of friction seems to have several fundamental characteristics. These characteristics not only explain some of the simpler aspects of friction—they also hold the potential for the development of nearly friction-free environments that could save hundreds of billions of dollars in energy which is currently being converted (unnecessarily) to heat. Figure 5.5 illustrates one macroscopic characteristic of friction that is explained by microscopic (small-scale) research. We have noted that friction is proportional to the normal force, but not to the area in contact, a somewhat counterintuitive notion. When two rough surfaces are in contact, the actual contact area is a tiny fraction of the total area since only high spots touch. When a greater normal force is exerted, the actual contact area increases, and it is found that the friction is proportional to this area. Figure 5.5 Two rough surfaces in contact have a much smaller area of actual contact than their total area. When there is a greater normal force as a result of a greater applied force, the area of actual contact increases as does friction. But the atomic-scale view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is now being determined. In other words, why do surfaces get warmer when rubbed? Essentially, atoms are linked with one another to form lattices. When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate—essentially creating sound waves that penetrate the material. The sound waves diminish with distance and their energy is converted into heat. Chemical reactions that are related to frictional wear can also occur between atoms and molecules on the surfaces. Figure 5.6 shows how the tip of a probe drawn across another material is deformed by atomic-scale friction. The force needed to drag the tip can be measured and is found to be
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related to shear stress, which will be discussed later in this chapter. The variation in shear stress is remarkable (more than a factor of 1012 ) and difficult to predict theoretically, but shear stress is yielding a fundamental understanding of a large-scale phenomenon known since ancient times—friction. 198 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Figure 5.6 The tip of a probe is deformed sideways by frictional force as the probe is dragged across a surface. Measurements of how the force varies for different materials are yielding fundamental insights into the atomic nature of friction. PhET Explorations: Forces and Motion Explore the forces at work when you try to push a filing cabinet. Create an applied force and see the resulting friction force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. Draw a free-body diagram of all the forces (including gravitational and normal forces). Figure 5.7 Forces and Motion (http://cnx.org/content/m54899/1.2/forces-and-motion_en.jar) 5.2 Drag Forces Learning Objectives By the end of this section, you will be able to: • Define drag force and model it mathematically. • Discuss the applications of drag force. • Define terminal velocity. • Perform calculations to find terminal velocity. Another interesting force in everyday life is the force of drag on an object when it is moving in a fluid (either a gas or a liquid). You feel the drag force when you move your hand through water. You might also feel it if you move your hand during a strong wind. The faster you move your hand, the harder it is to move. You feel a smaller drag force when you tilt your hand so only the side goes through the air—you have decreased the area of your hand that faces the direction of motion. Like friction, the drag force always opposes the motion of an object. Unlike simple friction, the drag force is proportional to some function of the velocity of the object in that fluid. This functionality is complicated and depends upon the shape of the object, its size, its velocity, and the fluid it is in. For most large objects such as bicyclists, cars, and baseballs not moving too slowly, the magnitude of the drag force D is found to be proportional to the square of the speed of the object. We can write this relationship mathematically as.
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When taking into account other factors, this relationship becomes D = 1 2 Cρ 2, (5.13) where is the drag coefficient, is the area of the object facing the fluid, and is the density of the fluid. (Recall that density is mass per unit volume.) This equation can also be written in a more generalized fashion as D = 2, where is a constant equivalent to 0.5. We have set the exponent for these equations as 2 because, when an object is moving at high velocity This content is available for free at http://cnx.org/content/col11844/1.13 212 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Although measurable, this is not a significant decrease in volume considering that the force per unit area is about 500 atmospheres (1 million pounds per square foot). Liquids and solids are extraordinarily difficult to compress. Conversely, very large forces are created by liquids and solids when they try to expand but are constrained from doing so—which is equivalent to compressing them to less than their normal volume. This often occurs when a contained material warms up, since most materials expand when their temperature increases. If the materials are tightly constrained, they deform or break their container. Another very common example occurs when water freezes. Water, unlike most materials, expands when it freezes, and it can easily fracture a boulder, rupture a biological cell, or crack an engine block that gets in its way. Other types of deformations, such as torsion or twisting, behave analogously to the tension, shear, and bulk deformations considered here. Glossary deformation: change in shape due to the application of force drag force: D, found to be proportional to the square of the speed of the object; mathematically D ∝ 2 D = 1 where is the drag coefficient, is the area of the object facing the fluid, and is the density of the fluid 2 2, friction: a force that opposes relative motion or attempts at motion between systems in contact Hooke's law: proportional relationship between the force on a material and the deformation Δ it causes, = Δ kinetic friction: a force that opposes the motion of two systems that are in contact and moving relative to one another magnitude of kinetic friction: k = k, where k is the coefficient of kinetic friction magnitude of static friction: s ≤ s, where s is the coefficient of static friction and is the magnitude of the normal force shear deformation:
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deformation perpendicular to the original length of an object static friction: a force that opposes the motion of two systems that are in contact and are not moving relative to one another Stokes' law: s = 6, where is the radius of the object, is the viscosity of the fluid, and is the object's velocity strain: ratio of change in length to original length stress: ratio of force to area tensile strength: measure of deformation for a given tension or compression Section Summary 5.1 Friction • Friction is a contact force between systems that opposes the motion or attempted motion between them. Simple friction is proportional to the normal force pushing the systems together. (A normal force is always perpendicular to the contact surface between systems.) Friction depends on both of the materials involved. The magnitude of static friction s between systems stationary relative to one another is given by s ≤ s, where s is the coefficient of static friction, which depends on both of the materials. • The kinetic friction force k between systems moving relative to one another is given by where k is the coefficient of kinetic friction, which also depends on both materials. k = k, 5.2 Drag Forces This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 213 • Drag forces acting on an object moving in a fluid oppose the motion. For larger objects (such as a baseball) moving at a velocity in air, the drag force is given by D = 1 2 2, where is the drag coefficient (typical values are given in Table 5.2), is the area of the object facing the fluid, and is the fluid density. • For small objects (such as a bacterium) moving in a denser medium (such as water), the drag force is given by Stokes' law, s = 6, where is the radius of the object, is the fluid viscosity, and is the object's velocity. 5.3 Elasticity: Stress and Strain • Hooke's law is given by = Δ, where Δ is the amount of deformation (the change in length), is the applied force, and is a proportionality constant that depends on the shape and composition of the object and the direction of the force. The relationship between the deformation and the applied force can also be written as Δ = 1 0, where is Young's modulus, which depends on the substance,
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is the cross-sectional area, and 0 is the original length. • The ratio of force to area,, is defined as stress, measured in N/m2. • The ratio of the change in length to length, Δ 0, is defined as strain (a unitless quantity). In other words, • The expression for shear deformation is stress = ×strain. Δ = 1 where is the shear modulus and is the force applied perpendicular to 0 and parallel to the cross-sectional area. 0, • The relationship of the change in volume to other physical quantities is given by Δ = 1 where is the bulk modulus, 0 is the original volume, and surfaces. 0, is the force per unit area applied uniformly inward on all Conceptual Questions 5.1 Friction 1. Define normal force. What is its relationship to friction when friction behaves simply? 2. The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings. 3. When you learn to drive, you discover that you need to let up slightly on the brake pedal as you come to a stop or the car will stop with a jerk. Explain this in terms of the relationship between static and kinetic friction. 4. When you push a piece of chalk across a chalkboard, it sometimes screeches because it rapidly alternates between slipping and sticking to the board. Describe this process in more detail, in particular explaining how it is related to the fact that kinetic friction is less than static friction. (The same slip-grab process occurs when tires screech on pavement.) 5.2 Drag Forces 5. Athletes such as swimmers and bicyclists wear body suits in competition. Formulate a list of pros and cons of such suits. 6. Two expressions were used for the drag force experienced by a moving object in a liquid. One depended upon the speed, while the other was proportional to the square of the speed. In which types of motion would each of these expressions be more applicable than the other one? 7. As cars travel, oil and gasoline leaks onto the road surface. If a light rain falls, what does this do to the control of the car? Does a heavy rain make any difference? 8. Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall? 214 Chapter 5 | Further Applications of Newton
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's Laws: Friction, Drag, and Elasticity 5.3 Elasticity: Stress and Strain 9. The elastic properties of the arteries are essential for blood flow. Explain the importance of this in terms of the characteristics of the flow of blood (pulsating or continuous). 10. What are you feeling when you feel your pulse? Measure your pulse rate for 10 s and for 1 min. Is there a factor of 6 difference? 11. Examine different types of shoes, including sports shoes and thongs. In terms of physics, why are the bottom surfaces designed as they are? What differences will dry and wet conditions make for these surfaces? 12. Would you expect your height to be different depending upon the time of day? Why or why not? 13. Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall? 14. Explain why pregnant women often suffer from back strain late in their pregnancy. 15. An old carpenter's trick to keep nails from bending when they are pounded into hard materials is to grip the center of the nail firmly with pliers. Why does this help? 16. When a glass bottle full of vinegar warms up, both the vinegar and the glass expand, but vinegar expands significantly more with temperature than glass. The bottle will break if it was filled to its tightly capped lid. Explain why, and also explain how a pocket of air above the vinegar would prevent the break. (This is the function of the air above liquids in glass containers.) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity 215 Problems & Exercises 5.1 Friction 1. A physics major is cooking breakfast when he notices that the frictional force between his steel spatula and his Teflon frying pan is only 0.200 N. Knowing the coefficient of kinetic friction between the two materials, he quickly calculates the normal force. What is it? 2. (a) When rebuilding her car's engine, a physics major must exert 300 N of force to insert a dry steel piston into a steel cylinder. What is the magnitude of the normal force between the piston and cylinder? (b) What is the magnitude of the force would she have to exert if the steel parts were oiled? 3. (a) What is the maximum frictional force in the
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knee joint of a person who supports 66.0 kg of her mass on that knee? (b) During strenuous exercise it is possible to exert forces to the joints that are easily ten times greater than the weight being supported. What is the maximum force of friction under such conditions? The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain. 4. Suppose you have a 120-kg wooden crate resting on a wood floor. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the crate starts to slip, what will the magnitude of its acceleration then be? 5. (a) If half of the weight of a small 1.00×103 kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both problems assuming the truck has four-wheel drive. 6. A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg. (a) Calculate the magnitude of the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow. (b) What is the magnitude of the acceleration once the sled starts to move? (c) For both situations, calculate the magnitude of the force in the coupling between the dogs and the sled. 7. Consider the 65.0-kg ice skater being pushed by two others shown in Figure 5.21. (a) Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes 1 and 2 are 26.4 N and 18.6 N, respectively. (b) What is her initial acceleration if she is initially stationary and wearing steel-bladed skates that point in the direction of Ftot? (c) What is her acceleration assuming she is already moving in the direction of Ftot? (Remember that friction always acts in the direction opposite that of motion or attempted motion between surfaces in contact.) Figure 5.21 8. Show that the acceleration of any object down a frictionless incline
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that makes an angle with the horizontal is = sin. (Note that this acceleration is independent of mass.) 9. Show that the acceleration of any object down an incline where friction behaves simply (that is, where k = k ) is = ( sin − kcos ). Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small ( k = 0). 10. Calculate the deceleration of a snow boarder going up a 5.0º, slope assuming the coefficient of friction for waxed wood on wet snow. The result of Exercise 5.9 may be useful, but be careful to consider the fact that the snow boarder is going uphill. Explicitly show how you follow the steps in Problem-Solving Strategies. 11. (a) Calculate the acceleration of a skier heading down a 10.0º slope, assuming the coefficient of friction for waxed wood on wet snow. (b) Find the angle of the slope down which this skier could coast at a constant velocity. You can neglect air resistance in both parts, and you will find the result of Exercise 5.9 to be useful. Explicitly show how you follow the steps in the Problem-Solving Strategies. 12. If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object parallel to the incline. This requires greater and greater friction for steeper slopes. Show that the maximum angle of an incline above the horizontal for which an object will not slide down is = tan–1 μs. You may use the result of the previous problem. Assume that = 0 and that static friction has reached its maximum value. 13. Calculate the maximum deceleration of a car that is heading down a 6º slope (one that makes an angle of 6º with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the deceleration. (Ignore rolling.) Calculate for a car: (a) On dry concrete. (b) On wet concrete. (c) On ice, assuming that s = 0.100, the same as for shoes on ice. 14. Calculate the maximum acceleration of a car that is heading up a 4º slope (one that makes an angle of 4º with the
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horizontal) under the following road conditions. Assume that only half the weight of the car is supported by the two drive wheels and that the coefficient of static friction is involved—that is, the tires are not allowed to slip during the acceleration. (Ignore rolling.) (a) On dry concrete. (b) On wet 216 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice. 15. Repeat Exercise 5.14 for a car with four-wheel drive. 16. A freight train consists of two 8.00×105-kg engines and 45 cars with average masses of 5.50×105 kg. (a) What force must each engine exert backward on the track to accelerate the train at a rate of 5.00×10−2 m / s2 if the force of friction is 7.50×105 N, assuming the engines exert identical forces? This is not a large frictional force for such a massive system. Rolling friction for trains is small, and consequently trains are very energy-efficient transportation systems. (b) What is the magnitude of the force in the coupling between the 37th and 38th cars (this is the force each exerts on the other), assuming all cars have the same mass and that friction is evenly distributed among all of the cars and engines? 17. Consider the 52.0-kg mountain climber in Figure 5.22. (a) Find the tension in the rope and the force that the mountain climber must exert with her feet on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. (b) What is the minimum coefficient of friction between her shoes and the cliff? Figure 5.22 Part of the climber's weight is supported by her rope and part by friction between her feet and the rock face. 18. A contestant in a winter sporting event pushes a 45.0-kg block of ice across a frozen lake as shown in Figure 5.23(a). (a) Calculate the minimum force he must exert to get the block moving. (b) What is the magnitude of its acceleration once it starts to move, if that force is maintained? 19. Repeat Exercise 5.18 with the contestant pulling the block of ice with a rope over his shoulder at the same angle above the horizontal as
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shown in Figure 5.23(b). This content is available for free at http://cnx.org/content/col11844/1.13 Figure 5.23 Which method of sliding a block of ice requires less force—(a) pushing or (b) pulling at the same angle above the horizontal? 5.2 Drag Forces 20. The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a pike (headfirst) position with a surface area of 0.140 m2. 21. A 60-kg and a 90-kg skydiver jump from an airplane at an altitude of 6000 m, both falling in the pike position. Make some assumption on their frontal areas and calculate their terminal velocities. How long will it take for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)? Assume all values are accurate to three significant digits. 22. A 560-g squirrel with a surface area of 930 cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a drag coefficient for a horizontal skydiver.) What will be the velocity of a 56-kg person hitting the ground, assuming no drag contribution in such a short distance? 23. To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the road (the rolling resistance). (a) What are the magnitudes of drag forces at 70 km/h and 100 km/h for a Toyota Camry? (Drag area is 0.70 m2 ) (b) What is the magnitude of drag force at 70 km/h and 100 km/h for a Hummer H2? (Drag area is 2.44 m2 ) Assume all values are accurate to three significant digits. 24. By what factor does the drag force on a car increase as it goes from 65 to 110 km/h? 25. Calculate the speed a spherical rain drop would achieve falling from 5.00 km (a) in the absence of air drag (b) with air drag. Take the size across of the drop to be 4 mm, the density to be 1.00×103 kg/m3, and the surface area to be 2. Chapter 5 |
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Further Applications of Newton's Laws: Friction, Drag, and Elasticity 217 26. Using Stokes' law, verify that the units for viscosity are kilograms per meter per second. 27. Find the terminal velocity of a spherical bacterium (diameter 2.00 μm ) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.10×103 kg/m3. 28. Stokes' law describes sedimentation of particles in liquids and can be used to measure viscosity. Particles in liquids achieve terminal velocity quickly. One can measure the time it takes for a particle to fall a certain distance and then use Stokes' law to calculate the viscosity of the liquid. Suppose a steel ball bearing (density 7.8×103 kg/m3 3.0 mm ) is dropped in a container of motor oil. It takes 12 s to fall a distance of 0.60 m. Calculate the viscosity of the oil., diameter 5.3 Elasticity: Stress and Strain 29. During a circus act, one performer swings upside down hanging from a trapeze holding another, also upside-down, performer by the legs. If the upward force on the lower performer is three times her weight, how much do the bones (the femurs) in her upper legs stretch? You may assume each is equivalent to a uniform rod 35.0 cm long and 1.80 cm in radius. Her mass is 60.0 kg. 30. During a wrestling match, a 150 kg wrestler briefly stands on one hand during a maneuver designed to perplex his already moribund adversary. By how much does the upper arm bone shorten in length? The bone can be represented by a uniform rod 38.0 cm in length and 2.10 cm in radius. 31. (a) The “lead” in pencils is a graphite composition with a Young's modulus of about 1×109 N / m2. Calculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long. (b) Is the answer reasonable? That is, does it seem to be consistent with what you have observed when using pencils? 32. TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72
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.0-kg physicist placed himself and 400 kg of equipment at the top of one 610-m high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius? 33. (a) By how much does a 65.0-kg mountain climber stretch her 0.800-cm diameter nylon rope when she hangs 35.0 m below a rock outcropping? (b) Does the answer seem to be consistent with what you have observed for nylon ropes? Would it make sense if the rope were actually a bungee cord? 34. A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 900 N exerted at the top would. How far to the side does the top of the pole flex? 35. As an oil well is drilled, each new section of drill pipe supports its own weight and that of the pipe and drill bit beneath it. Calculate the stretch in a new 6.00 m length of steel pipe that supports 3.00 km of pipe having a mass of 20.0 kg/m and a 100-kg drill bit. The pipe is equivalent in strength to a solid cylinder 5.00 cm in diameter. 36. Calculate the force a piano tuner applies to stretch a steel piano wire 8.00 mm, if the wire is originally 0.850 mm in diameter and 1.35 m long. 37. A vertebra is subjected to a shearing force of 500 N. Find the shear deformation, taking the vertebra to be a cylinder 3.00 cm high and 4.00 cm in diameter. 38. A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have the shear modulus of 1×109 N / m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter. 39. When using a pencil eraser, you exert a vertical force of 6.00 N at a distance of 2.00 cm from the hardwood-eraser joint. The pencil is 6.00 mm in diameter and is held at an angle of 20.0º to the horizontal. (a) By how much does the wood flex perpendicular to its length? (b) How much is it
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compressed lengthwise? 40. To consider the effect of wires hung on poles, we take data from Example 4.8, in which tensions in wires supporting a traffic light were calculated. The left wire made an angle 30.0º below the horizontal with the top of its pole and carried a tension of 108 N. The 12.0 m tall hollow aluminum pole is equivalent in strength to a 4.50 cm diameter solid cylinder. (a) How far is it bent to the side? (b) By how much is it compressed? 41. A farmer making grape juice fills a glass bottle to the brim and caps it tightly. The juice expands more than the glass when it warms up, in such a way that the volume increases by 0.2% (that is, Δ / 0 = 2×10−3 available. Calculate the magnitude of the normal force exerted by the juice per square centimeter if its bulk modulus is 1.8×109 N/m2, assuming the bottle does not break. In view of your answer, do you think the bottle will survive? ) relative to the space 42. (a) When water freezes, its volume increases by 9.05% (that is, Δ / 0 = 9.05×10−2 ). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water in this problem.) (b) Is it surprising that such forces can fracture engine blocks, boulders, and the like? 43. This problem returns to the tightrope walker studied in Example 4.6, who created a tension of 3.94×103 N in a wire making an angle 5.0º below the horizontal with each supporting pole. Calculate how much this tension stretches the steel wire if it was originally 15 m long and 0.50 cm in diameter. 44. The pole in Figure 5.24 is at a 90.0º bend in a power line and is therefore subjected to more shear force than poles in straight parts of the line. The tension in each line is 4.00×104 N, at the angles shown. The pole is 15.0 m tall, has an 18.0 cm diameter, and can be considered to have half the strength of hardwood. (a) Calculate the compression of the pole. (b) Find how much it bends and in what direction. (c) Find the tension in a guy wire used to
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keep the pole straight if it is attached to the top of the pole at an angle of 30.0º with the vertical. (Clearly, the guy wire must be in the opposite direction of the bend.) 218 Chapter 5 | Further Applications of Newton's Laws: Friction, Drag, and Elasticity Figure 5.24 This telephone pole is at a 90º bend in a power line. A guy wire is attached to the top of the pole at an angle of 30º with the vertical. Test Prep for AP® Courses 5.1 Friction 1. When a force of 20 N is applied to a stationary box weighing 40 N, the box does not move. This means the coefficient of static friction is equal to 0.5. is greater than 0.5. is less than 0.5. a. b. c. d. cannot be determined. 2. A 2-kg block slides down a ramp which is at an incline of 25º. If the frictional force is 4.86 N, what is the coefficient of friction? At what incline will the box slide at a constant velocity? Assume g = 10 m/s2. 3. A block is given a short push and then slides with constant friction across a horizontal floor. Which statement best explains the direction of the force that friction applies on the moving block? a. Friction will be in the same direction as the block's motion because molecular interactions between the block and the floor will deform the block in the direction of motion. b. Friction will be in the same direction as the block's motion because thermal energy generated at the interface between the block and the floor adds kinetic energy to the block. c. Friction will be in the opposite direction of the block's motion because molecular interactions between the block and the floor will deform the block in the opposite direction of motion. d. Friction will be in the opposite direction of the block's motion because thermal energy generated at the interface between the block and the floor converts some of the block's kinetic energy to potential energy. 4. A student pushes a cardboard box across a carpeted floor and afterwards notices that the bottom of the box feels warm. Explain how interactions between molecules in the cardboard and molecules in the carpet produced this heat. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 219 6 GRAVITATION AND UNIFORM CIRCULAR MOTION
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Figure 6.1 This Australian Grand Prix Formula 1 race car moves in a circular path as it makes the turn. Its wheels also spin rapidly—the latter completing many revolutions, the former only part of one (a circular arc). The same physical principles are involved in each. (credit: Richard Munckton) Chapter Outline 6.1. Rotation Angle and Angular Velocity 6.2. Centripetal Acceleration 6.3. Centripetal Force 6.4. Fictitious Forces and Non-inertial Frames: The Coriolis Force 6.5. Newton's Universal Law of Gravitation 6.6. Satellites and Kepler's Laws: An Argument for Simplicity Connection for AP® Courses Many motions, such as the arc of a bird's flight or Earth's path around the Sun, are curved. Recall that Newton's first law tells us that motion is along a straight line at constant speed unless there is a net external force. We will therefore study not only motion along curves, but also the forces that cause it, including gravitational forces. This chapter supports Big Idea 3 that interactions between objects are described by forces, and thus change in motion is a result of a net force exerted on an object. In this chapter, this idea is applied to uniform circular motion. In some ways, this chapter is a continuation of Dynamics: Newton's Laws of Motion as we study more applications of Newton's laws of motion. This chapter deals with the simplest form of curved motion, uniform circular motion, which is motion in a circular path at constant speed. As an object moves on a circular path, the magnitude of its velocity remains constant, but the direction of the velocity is changing. This means there is an acceleration that we will refer to as a “centripetal” acceleration caused by a net external force, also called the “centripetal” force (Enduring Understanding 3.B). The centripetal force is the net force totaling all 220 Chapter 6 | Gravitation and Uniform Circular Motion external forces acting on the object (Essential Knowledge 3.B.1). In order to determine the net force, a free-body diagram may be useful (Essential Knowledge 3.B.2). Studying this topic illustrates most of the concepts associated with rotational motion and leads to many new topics we group under the name rotation. This motion can be described using kinematics variables (Essential Knowledge 3.A.1), but
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in addition to linear variables, we will introduce angular variables. We use various ways to describe motion, namely, verbally, algebraically and graphically (Learning Objective 3.A.1.1). Pure rotational motion occurs when points in an object move in circular paths centered on one point. Pure translational motion is motion with no rotation. Some motion combines both types, such as a rotating hockey puck moving over ice. Some combinations of both types of motion are conveniently described with fictitious forces which appear as a result of using a non-inertial frame of reference (Enduring Understanding 3.A). Furthermore, the properties of uniform circular motion can be applied to the motion of massive objects in a gravitational field. Thus, this chapter supports Big Idea 1 that gravitational mass is an important property of an object or a system. We have experimental evidence that gravitational and inertial masses are equal (Enduring Understanding 1.C), and that gravitational mass is a measure of the strength of the gravitational interaction (Essential Knowledge 1.C.2). Therefore, this chapter will support Big Idea 2 that fields existing in space can be used to explain interactions, because any massive object creates a gravitational field in space (Enduring Understanding 2.B). Mathematically, we use Newton's universal law of gravitation to provide a model for the gravitational interaction between two massive objects (Essential Knowledge 2.B.2). We will discover that this model describes the interaction of one object with mass with another object with mass (Essential Knowledge 3.C.1), and also that gravitational force is a long-range force (Enduring Understanding 3.C). The concepts in this chapter support: Big Idea 1 Objects and systems have properties such as mass and charge. Systems may have internal structure. Enduring Understanding 1.C Objects and systems have properties of inertial mass and gravitational mass that are experimentally verified to be the same and that satisfy conservation principles. Essential Knowledge 1.C.2 Gravitational mass is the property of an object or a system that determines the strength of the gravitational interaction with other objects, systems, or gravitational fields. Essential Knowledge 1.C.3 Objects and systems have properties of inertial mass and gravitational mass that are experimentally verified to be the same and that satisfy conservation principles. Big Idea 2 Fields existing in space can be used to explain interactions. Enduring Understanding 2.B A gravitational field is caused by an object with mass. Essential Knowledge 2.B.2. The gravitational field caused by a spherically symm
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etric object with mass is radial and, outside the object, varies as the inverse square of the radial distance from the center of that object. Big Idea 3 The interactions of an object with other objects can be described by forces. Enduring Understanding 3.A All forces share certain common characteristics when considered by observers in inertial reference frames. Essential Knowledge 3.A.1. An observer in a particular reference frame can describe the motion of an object using such quantities as position, displacement, distance, velocity, speed, and acceleration. Essential Knowledge 3.A.3. A force exerted on an object is always due to the interaction of that object with another object. Enduring Understanding 3.B Classically, the acceleration of an object interacting with other objects can be predicted by using = ∑ /. Essential Knowledge 3.B.1 If an object of interest interacts with several other objects, the net force is the vector sum of the individual forces. Essential Knowledge 3.B.2 Free-body diagrams are useful tools for visualizing forces being exerted on a single object and writing the equations that represent a physical situation. Enduring Understanding 3.C At the macroscopic level, forces can be categorized as either long-range (action-at-a-distance) forces or contact forces. Essential Knowledge 3.C.1. Gravitational force describes the interaction of one object that has mass with another object that has mass. 6.1 Rotation Angle and Angular Velocity Learning Objectives By the end of this section, you will be able to: • Define arc length, rotation angle, radius of curvature, and angular velocity. • Calculate the angular velocity of a car wheel spin. In Kinematics, we studied motion along a straight line and introduced such concepts as displacement, velocity, and acceleration. Two-Dimensional Kinematics dealt with motion in two dimensions. Projectile motion is a special case of two-dimensional This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 221 kinematics in which the object is projected into the air, while being subject to the gravitational force, and lands a distance away. In this chapter, we consider situations where the object does not land but moves in a curve. We begin the study of uniform circular motion by defining two angular quantities needed to describe rotational motion. Rotation Angle When objects rotate about some axis—for example, when the CD (compact
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disc) in Figure 6.2 rotates about its center—each point in the object follows a circular arc. Consider a line from the center of the CD to its edge. Each pit used to record sound along this line moves through the same angle in the same amount of time. The rotation angle is the amount of rotation and is analogous to linear distance. We define the rotation angle Δ to be the ratio of the arc length to the radius of curvature: Δ = Δ. (6.1) Figure 6.2 All points on a CD travel in circular arcs. The pits along a line from the center to the edge all move through the same angle Δ in a time Δ. Figure 6.3 The radius of a circle is rotated through an angle Δ. The arc length Δs is described on the circumference. The arc length Δ is the distance traveled along a circular path as shown in Figure 6.3 Note that is the radius of curvature of the circular path. We know that for one complete revolution, the arc length is the circumference of a circle of radius. The circumference of a circle is 2π. Thus for one complete revolution the rotation angle is Δ = 2π = 2π. This result is the basis for defining the units used to measure rotation angles, Δ to be radians (rad), defined so that A comparison of some useful angles expressed in both degrees and radians is shown in Table 6.1. 2π rad = 1 revolution. (6.2) (6.3) 222 Chapter 6 | Gravitation and Uniform Circular Motion Table 6.1 Comparison of Angular Units Degree Measures Radian Measure 30º 60º 90º 120º 135º 180º 6 3 2 2π 3 3π 4 Figure 6.4 Points 1 and 2 rotate through the same angle ( Δ ), but point 2 moves through a greater arc length (Δ) because it is at a greater distance from the center of rotation (). If Δ = 2 rad, then the CD has made one complete revolution, and every point on the CD is back at its original position. Because there are 360º in a circle or one revolution, the relationship between radians and degrees is thus so that Angular Velocity 2 rad = 360º 1 rad = 360º 2π ≈ 57.3º. How fast is an object rotating? We define angular velocity as the rate of change of an angle. In symbols, this is = Δ Δ, (6.4) (6.5) (6.6
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) where an angular rotation Δ takes place in a time Δ. The greater the rotation angle in a given amount of time, the greater the angular velocity. The units for angular velocity are radians per second (rad/s). Angular velocity is analogous to linear velocity. To get the precise relationship between angular and linear velocity, we again consider a pit on the rotating CD. This pit moves an arc length Δ in a time Δ, and so it has a linear velocity = Δ Δ. From Δ = Δ we see that Δ = Δ. Substituting this into the expression for gives = Δ Δ =. (6.7) (6.8) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion We write this relationship in two different ways and gain two different insights: = or =. 223 (6.9) states that the linear velocity is proportional to the distance from the center of The first relationship in = or = rotation, thus, it is largest for a point on the rim (largest ), as you might expect. We can also call this linear speed of a point on the rim the tangential speed. The second relationship in = or = moving car. Note that the speed of a point on the rim of the tire is the same as the speed of the car. See Figure 6.5. So the faster the car moves, the faster the tire spins—large means a large, because =. Similarly, a larger-radius tire rotating at the same angular velocity ( ) will produce a greater linear speed ( ) for the car. can be illustrated by considering the tire of a Figure 6.5 A car moving at a velocity to the right has a tire rotating with an angular velocity.The speed of the tread of the tire relative to the axle is, the same as if the car were jacked up. Thus the car moves forward at linear velocity =, where is the tire radius. A larger angular velocity for the tire means a greater velocity for the car. Example 6.1 How Fast Does a Car Tire Spin? Calculate the angular velocity of a 0.300 m radius car tire when the car travels at 15.0 m/s (about 54 km/h ). See Figure 6.5. Strategy Because the linear speed of the tire rim is the same as the speed of the car, we have = 15.0 m/s. The radius of the tire is given to
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be = 0.300 m. Knowing and, we can use the second relationship in =, = to calculate the angular velocity. Solution To calculate the angular velocity, we will use the following relationship: Substituting the knowns, Discussion =. = 15.0 m/s 0.300 m = 50.0 rad/s. (6.10) (6.11) When we cancel units in the above calculation, we get 50.0/s. But the angular velocity must have units of rad/s. Because radians are actually unitless (radians are defined as a ratio of distance), we can simply insert them into the answer for the angular velocity. Also note that if an earth mover with much larger tires, say 1.20 m in radius, were moving at the same speed of 15.0 m/s, its tires would rotate more slowly. They would have an angular velocity = (15.0 m/s) / (1.20 m) = 12.5 rad/s. (6.12) 224 Chapter 6 | Gravitation and Uniform Circular Motion Both and have directions (hence they are angular and linear velocities, respectively). Angular velocity has only two directions with respect to the axis of rotation—it is either clockwise or counterclockwise. Linear velocity is tangent to the path, as illustrated in Figure 6.6. Take-Home Experiment Tie an object to the end of a string and swing it around in a horizontal circle above your head (swing at your wrist). Maintain uniform speed as the object swings and measure the angular velocity of the motion. What is the approximate speed of the object? Identify a point close to your hand and take appropriate measurements to calculate the linear speed at this point. Identify other circular motions and measure their angular velocities. Figure 6.6 As an object moves in a circle, here a fly on the edge of an old-fashioned vinyl record, its instantaneous velocity is always tangent to the circle. The direction of the angular velocity is clockwise in this case. PhET Explorations: Ladybug Revolution Figure 6.7 Ladybug Revolution (http://cnx.org/content/m54992/1.2/rotation_en.jar) Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's
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x,y position, velocity, and acceleration using vectors or graphs. 6.2 Centripetal Acceleration Learning Objectives By the end of this section, you will be able to: • Establish the expression for centripetal acceleration. • Explain the centrifuge. We know from kinematics that acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. (If you hold the wheel steady during a turn and move at constant speed, you are in uniform circular motion.) What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become. In this section we examine the direction and magnitude of that acceleration. Figure 6.8 shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). This pointing is shown with the vector diagram in the figure. We call the acceleration of This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 225 an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration( c ); centripetal means “toward the center” or “center seeking.” Figure 6.8 The directions of the velocity of an object at two different points are shown, and the change in velocity Δv is seen to point directly toward the center of curvature. (See small inset.) Because ac = Δv / Δ, the acceleration is also toward the center; a is called centripetal acceleration. (Because Δ is very small, the arc length Δ is equal to the chord length Δ for small time differences.) The direction of centripetal acceleration is toward the center of curvature, but what is its magnitude? Note that the triangle formed by the velocity vectors and the one formed by the radii and Δ are similar. Both the triangles ABC and PQR are isosceles triangles (two equal sides). The
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two equal sides of the velocity vector triangle are the speeds 1 = 2 =. Using the properties of two similar triangles, we obtain Acceleration is Δ / Δ, and so we first solve this expression for Δ : Δ = Δ. Then we divide this by Δ, yielding Δ = Δ. Δ Δ = × Δ Δ. (6.13) (6.14) (6.15) Finally, noting that Δ / Δ = c and that Δ / Δ =, the linear or tangential speed, we see that the magnitude of the centripetal acceleration is c = 2, which is the acceleration of an object in a circle of radius at a speed. So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. But it is a bit surprising that c is proportional to speed squared, implying, for example, that it is four times as hard to take a curve at 100 km/h than at 50 km/h. A sharp corner has a small radius, so that c is greater for tighter turns, as you have probably noticed. (6.16) It is also useful to express c in terms of angular velocity. Substituting = into the above expression, we find c = ()2 / = 2. We can express the magnitude of centripetal acceleration using either of two equations: c = 2 ; c = 2. (6.17) Recall that the direction of c is toward the center. You may use whichever expression is more convenient, as illustrated in examples below. 226 Chapter 6 | Gravitation and Uniform Circular Motion A centrifuge (see Figure 6.9b) is a rotating device used to separate specimens of different densities. High centripetal acceleration significantly decreases the time it takes for separation to occur, and makes separation possible with small samples. Centrifuges are used in a variety of applications in science and medicine, including the separation of single cell suspensions such as bacteria, viruses, and blood cells from a liquid medium and the separation of macromolecules, such as DNA and protein, from a solution. Centrifuges are often rated in terms of their centripetal acceleration relative to acceleration due to gravity () ; maximum centripetal acceleration of several hundred thousand is possible in a vacuum. Human centrifuges, extremely large centrifuges, have been used to test the tolerance of astronauts to the effects of accelerations larger than that
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of Earth's gravity. Example 6.2 How Does the Centripetal Acceleration of a Car Around a Curve Compare with That Due to Gravity? What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 6.9(a). Strategy Because and are given, the first expression in c = 2 ; c = 2 is the most convenient to use. Solution Entering the given values of = 25.0 m/s and = 500 m into the first expression for c gives c = 2 = (25.0 m/s)2 500 m = 1.25 m/s2. (6.18) Discussion To compare this with the acceleration due to gravity ( = 9.80 m/s2), we take the ratio of c / = seat belt. 9.80 m/s2 1.25 m/s2 / = 0.128. Thus, c = 0.128 g and is noticeable especially if you were not wearing a This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 227 Figure 6.9 (a) The car following a circular path at constant speed is accelerated perpendicular to its velocity, as shown. The magnitude of this centripetal acceleration is found in Example 6.2. (b) A particle of mass in a centrifuge is rotating at constant angular velocity. It must be accelerated perpendicular to its velocity or it would continue in a straight line. The magnitude of the necessary acceleration is found in Example 6.3. Example 6.3 How Big Is the Centripetal Acceleration in an Ultracentrifuge? Calculate the centripetal acceleration of a point 7.50 cm from the axis of an ultracentrifuge spinning at 7.5 × 104 rev/min. Determine the ratio of this acceleration to that due to gravity. See Figure 6.9(b). Strategy The term rev/min stands for revolutions per minute. By converting this to radians per second, we obtain the angular velocity. Because is given, we can use the second expression in the equation c = 2 c = 2 to calculate the centripetal acceleration. Solution To convert 7.50×104
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rev / min to radians per second, we use the facts that one revolution is 2πrad and one minute is 60.0 s. Thus, = 7.50×104 rev min × 2π rad 1 rev × 1 min 60.0 s = 7854 rad/s. Now the centripetal acceleration is given by the second expression in c = 2 ; c = 2 as Converting 7.50 cm to meters and substituting known values gives c = (0.0750 m)(7854 rad/s)2 = 4.63×106 m/s2. c = 2. (6.19) (6.20) (6.21) 228 Chapter 6 | Gravitation and Uniform Circular Motion Note that the unitless radians are discarded in order to get the correct units for centripetal acceleration. Taking the ratio of c to yields = 4.63×106 c 9.80 = 4.72×105. (6.22) Discussion This last result means that the centripetal acceleration is 472,000 times as strong as. It is no wonder that such high centrifuges are called ultracentrifuges. The extremely large accelerations involved greatly decrease the time needed to cause the sedimentation of blood cells or other materials. Of course, a net external force is needed to cause any acceleration, just as Newton proposed in his second law of motion. So a net external force is needed to cause a centripetal acceleration. In Centripetal Force, we will consider the forces involved in circular motion. PhET Explorations: Ladybug Motion 2D Learn about position, velocity and acceleration vectors. Move the ladybug by setting the position, velocity or acceleration, and see how the vectors change. Choose linear, circular or elliptical motion, and record and playback the motion to analyze the behavior. Figure 6.10 Ladybug Motion 2D (http://cnx.org/content/m54995/1.2/ladybug-motion-2d_en.jar) 6.3 Centripetal Force Learning Objectives By the end of this section, you will be able to: • Calculate coefficient of friction on a car tire. • Calculate ideal speed and angle of a car on a turn. Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force
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of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton's second law of motion, net force is mass times acceleration: net F =. For uniform circular motion, the acceleration is the centripetal acceleration— =. Thus, the magnitude of centripetal force Fc is Fc = c. (6.23) By using the expressions for centripetal acceleration from = 2 force Fc in terms of mass, velocity, angular velocity, and radius of curvature: = 2, we get two expressions for the centripetal = 2. You may use whichever expression for centripetal force is more convenient. Centripetal force c is always perpendicular to the path and pointing to the center of curvature, because a is perpendicular to the velocity and pointing to the center of curvature. = 2 (6.24) Note that if you solve the first expression for, you get This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve. = 2. (6.25) This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 229 Figure 6.11 The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the Fc, the smaller the radius of curvature and the sharper the curve. The second curve has the same, but a larger Fc produces a smaller ′. Example 6.4 What Coefficient of Friction Do Car Tires Need on a Flat Curve? (a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s. (b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure 6.12). Strategy and Solution for
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(a) We know that c = 2. Thus, Strategy for (b) c = 2 = (900 kg)(25.0 m/s)2 (500 m) = 1125 N. (6.26) Figure 6.12 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is s, where s is the static coefficient of friction and N is the normal force. The normal force equals the car's weight on level ground, so that =. Thus the centripetal force in this situation is c = = s = s. (6.27) Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for c from the equation c = 2 c = 2 2 = s. (6.28) (6.29) We solve this for s, noting that mass cancels, and obtain 230 Chapter 6 | Gravitation and Uniform Circular Motion Solution for (b) Substituting the knowns, s = 2. s = (25.0 m/s)2 (500 m)(9.80 m/s2) = 0.13. (6.30) (6.31) (Because coefficients of friction are approximate, the answer is given to only two digits.) Discussion We could also solve part (a) using the first expression in c = 2 c = 2 because and are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than s. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be
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less as will be discussed below. Figure 6.12 This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway. Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 6.13. The greater the angle, the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for for an ideally banked curve and consider an example related to it. For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions. Figure 6.13 shows a free body diagram for a car on a frictionless banked curve. If the angle is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight w and the normal force of the road N. (A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 231 magnitude mv2 /r. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is, sin = 2. (6.32) Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external
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forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is cos, and the only other vertical force is the car's weight. These must be equal in magnitude; thus, Now we can combine the last two equations to eliminate and get an expression for, as desired. Solving the second equation for = / (cos ), and substituting this into the first yields cos =. = 2 sin cos tan() = 2 2 tan = Taking the inverse tangent gives = tan−1 2 (ideally banked curve, no friction). (6.33) (6.34) (6.35) (6.36) This expression can be understood by considering how depends on and. A large will be obtained for a large and a small. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that does not depend on the mass of the vehicle. Figure 6.13 The car on this banked curve is moving away and turning to the left. Example 6.5 What Is the Ideal Speed to Take a Steeply Banked Tight Curve? Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless. Strategy We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities. Solution Starting with tan = 2 (6.37) 232 we get Noting that tan 65.0º = 2.14, we obtain = ( tan )1 / 2. = (100 m)(9.80 m/s2)(2.14) Chapter 6 | Gravitation and Uniform Circular Motion 1 / 2 (6.38) (6.39) Discussion = 45.8 m/s. This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly
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higher speeds. Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter's Problems and Exercises. Take-Home Experiment Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion. PhET Explorations: Gravity and Orbits Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it! Figure 6.14 Gravity and Orbits (http://cnx.org/content/m55002/1.2/gravity-and-orbits_en.jar) 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force Learning Objectives By the end of this section, you will be able to: • Discuss the inertial frame of reference. • Discuss the non-inertial frame of reference. • Describe the effects of the Coriolis force. What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone have in common? Each exhibits fictitious forces—unreal forces that arise from motion and may seem real, because the observer’s frame of reference is accelerating or rotating. When taking off in a jet, most people would agree it feels as if you are being pushed back into the seat as the airplane accelerates down the runway. Yet a physicist would say that you tend to remain stationary while the seat pushes forward on you, and there is no real force backward on you. An even more common experience occurs when you make a tight curve in your car—say, to the right. You feel as if you are thrown (that is, forced) toward the left relative to the car. Again, a physicist would say that you are going in a straight line but the car moves to the right, and there is no real force on you to the left. Recall Newton’s first law. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 233 Figure 6.
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15 (a) The car driver feels herself forced to the left relative to the car when she makes a right turn. This is a fictitious force arising from the use of the car as a frame of reference. (b) In the Earth’s frame of reference, the driver moves in a straight line, obeying Newton’s first law, and the car moves to the right. There is no real force to the left on the driver relative to Earth. There is a real force to the right on the car to make it turn. We can reconcile these points of view by examining the frames of reference used. Let us concentrate on people in a car. Passengers instinctively use the car as a frame of reference, while a physicist uses Earth. The physicist chooses Earth because it is very nearly an inertial frame of reference—one in which all forces are real (that is, in which all forces have an identifiable physical origin). In such a frame of reference, Newton’s laws of motion take the form given in Dynamics: Newton's Laws of Motion The car is a non-inertial frame of reference because it is accelerated to the side. The force to the left sensed by car passengers is a fictitious force having no physical origin. There is nothing real pushing them left—the car, as well as the driver, is actually accelerating to the right. Let us now take a mental ride on a merry-go-round—specifically, a rapidly rotating playground merry-go-round. You take the merry-go-round to be your frame of reference because you rotate together. In that non-inertial frame, you feel a fictitious force, named centrifugal force (not to be confused with centripetal force), trying to throw you off. You must hang on tightly to counteract the centrifugal force. In Earth’s frame of reference, there is no force trying to throw you off. Rather you must hang on to make yourself go in a circle because otherwise you would go in a straight line, right off the merry-go-round. Figure 6.16 (a) A rider on a merry-go-round feels as if he is being thrown off. This fictitious force is called the centrifugal force—it explains the rider’s motion in the rotating frame of reference. (b) In an inertial frame of reference and according to Newton’s laws, it is his inertia that carries him off and not a real force (the unshaded
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rider has net = 0 and heads in a straight line). A real force, centripetal, is needed to cause a circular path. This inertial effect, carrying you away from the center of rotation if there is no centripetal force to cause circular motion, is put to good use in centrifuges (see Figure 6.17). A centrifuge spins a sample very rapidly, as mentioned earlier in this chapter. Viewed from the rotating frame of reference, the fictitious centrifugal force throws particles outward, hastening their sedimentation. The greater the angular velocity, the greater the centrifugal force. But what really happens is that the inertia of the particles carries them along a line tangent to the circle while the test tube is forced in a circular path by a centripetal force. 234 Chapter 6 | Gravitation and Uniform Circular Motion Figure 6.17 Centrifuges use inertia to perform their task. Particles in the fluid sediment come out because their inertia carries them away from the center of rotation. The large angular velocity of the centrifuge quickens the sedimentation. Ultimately, the particles will come into contact with the test tube walls, which will then supply the centripetal force needed to make them move in a circle of constant radius. Let us now consider what happens if something moves in a frame of reference that rotates. For example, what if you slide a ball directly away from the center of the merry-go-round, as shown in Figure 6.18? The ball follows a straight path relative to Earth (assuming negligible friction) and a path curved to the right on the merry-go-round’s surface. A person standing next to the merry-go-round sees the ball moving straight and the merry-go-round rotating underneath it. In the merry-go-round’s frame of reference, we explain the apparent curve to the right by using a fictitious force, called the Coriolis force, that causes the ball to curve to the right. The fictitious Coriolis force can be used by anyone in that frame of reference to explain why objects follow curved paths and allows us to apply Newton’s Laws in non-inertial frames of reference. Figure 6.18 Looking down on the counterclockwise rotation of a merry-go-round, we see that a ball slid straight toward the edge follows a path curved to the right. The person slides the ball toward point B, starting at point A. Both points
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rotate to the shaded positions (A’ and B’) shown in the time that the ball follows the curved path in the rotating frame and a straight path in Earth’s frame. Up until now, we have considered Earth to be an inertial frame of reference with little or no worry about effects due to its rotation. Yet such effects do exist—in the rotation of weather systems, for example. Most consequences of Earth’s rotation can be qualitatively understood by analogy with the merry-go-round. Viewed from above the North Pole, Earth rotates counterclockwise, as does the merry-go-round in Figure 6.18. As on the merry-go-round, any motion in Earth’s northern hemisphere experiences a This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 235 Coriolis force to the right. Just the opposite occurs in the southern hemisphere; there, the force is to the left. Because Earth’s angular velocity is small, the Coriolis force is usually negligible, but for large-scale motions, such as wind patterns, it has substantial effects. The Coriolis force causes hurricanes in the northern hemisphere to rotate in the counterclockwise direction, while the tropical cyclones (what hurricanes are called below the equator) in the southern hemisphere rotate in the clockwise direction. The terms hurricane, typhoon, and tropical storm are regionally-specific names for tropical cyclones, storm systems characterized by low pressure centers, strong winds, and heavy rains. Figure 6.19 helps show how these rotations take place. Air flows toward any region of low pressure, and tropical cyclones contain particularly low pressures. Thus winds flow toward the center of a tropical cyclone or a low-pressure weather system at the surface. In the northern hemisphere, these inward winds are deflected to the right, as shown in the figure, producing a counterclockwise circulation at the surface for low-pressure zones of any type. Low pressure at the surface is associated with rising air, which also produces cooling and cloud formation, making low-pressure patterns quite visible from space. Conversely, wind circulation around high-pressure zones is clockwise in the northern hemisphere but is less visible because high pressure is associated with sinking air, producing clear skies. The rotation of tropical cyclones and the path of a ball on a merry-go-round can just as
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well be explained by inertia and the rotation of the system underneath. When non-inertial frames are used, fictitious forces, such as the Coriolis force, must be invented to explain the curved path. There is no identifiable physical source for these fictitious forces. In an inertial frame, inertia explains the path, and no force is found to be without an identifiable source. Either view allows us to describe nature, but a view in an inertial frame is the simplest and truest, in the sense that all forces have real origins and explanations. Figure 6.19 (a) The counterclockwise rotation of this northern hemisphere hurricane is a major consequence of the Coriolis force. (credit: NASA) (b) Without the Coriolis force, air would flow straight into a low-pressure zone, such as that found in tropical cyclones. (c) The Coriolis force deflects the winds to the right, producing a counterclockwise rotation. (d) Wind flowing away from a high-pressure zone is also deflected to the right, producing a clockwise rotation. (e) The opposite direction of rotation is produced by the Coriolis force in the southern hemisphere, leading to tropical cyclones. (credit: NASA) 6.5 Newton's Universal Law of Gravitation Learning Objectives By the end of this section, you will be able to: • Explain Earth's gravitational force. • Describe the gravitational effect of the Moon on Earth. • Discuss weightlessness in space. • Understand the Cavendish experiment. The information presented in this section supports the following AP® learning objectives and science practices: • 2.B.2.1 The student is able to apply = 2 to calculate the gravitational field due to an object with mass M, where the field is a vector directed toward the center of the object of mass M. (S.P. 2.2) • 2.B.2.2 The student is able to approximate a numerical value of the gravitational field (g) near the surface of an object from its radius and mass relative to those of the Earth or other reference objects. (S.P. 2.2) • 3.A.3.4. The student is able to make claims about the force on an object due to the presence of other objects with the same property: mass, electric charge. (S.P. 6.1, 6.4) 236 Chapter 6 | Gravitation and Uniform Circular Motion What do aching feet
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, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our feet are strained by supporting our weight—the force of Earth's gravity on us. An apple falls from a tree because of the same force acting a few meters above Earth's surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of meters. In fact, the same force causes planets to orbit the Sun, stars to orbit the center of the galaxy, and galaxies to cluster together. Gravity is another example of underlying simplicity in nature. It is the weakest of the four basic forces found in nature, and in some ways the least understood. It is a force that acts at a distance, without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and distances that vary from the tiny to the immense. Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See Figure 6.20. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the same cause. Some of Newton's contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph—it had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose a mechanism that caused them to follow these paths and not others. This was one of the earliest examples of a theory derived from empirical evidence doing more than merely describing those empirical results; it made claims about the fundamental workings of the universe. Figure 6.20 According to early accounts, Newton was inspired to make the connection between falling bodies and astronomical motions when he saw an apple fall from a tree and realized that if the gravitational force could extend above the ground to a tree, it might also reach the Sun. The inspiration of Newton's apple is a part of worldwide folklore and may even be based in fact. Great importance is attached to it because Newton's universal law of
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gravitation and his laws of motion answered very old questions about nature and gave tremendous support to the notion of underlying simplicity and unity in nature. Scientists still expect underlying simplicity to emerge from their ongoing inquiries into nature. The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated in modern language, Newton's universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 237 Figure 6.21 Gravitational attraction is along a line joining the centers of mass of these two bodies. The magnitude of the force is the same on each, consistent with Newton's third law. Misconception Alert The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton's third law. The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is concentrated at one specific point called the center of mass (CM), which will be further explored in Linear Momentum and Collisions. For two bodies having masses and with a distance between their centers of mass, the equation for Newton's universal law of gravitation is = 2, (6.40) where is the magnitude of the gravitational force and is a proportionality factor called the gravitational constant. is a universal gravitational constant—that is, it is thought to be the same everywhere in the universe. It has been measured experimentally to be = 6.673×10−11N ⋅ m2 kg2 (6.41) in SI units. Note that the units of are such that a force in newtons is obtained from = 2, when considering masses in kilograms and distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational attraction of 6.673×10−11 N. This is an extraordinarily small force. The small magnitude of the gravitational force is consistent with everyday experience. We are unaware that even large objects like mountains exert gravitational forces on us. In fact, our body weight is the force of attraction of the entire Earth on us with a mass of 6×1024 kg. The
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experiment to measure G was first performed by Cavendish, and is explained in more detail later. The fundamental concept it is based on is having a known mass on a spring with a known force (or spring) constant. Then, a second known mass is placed at multiple known distances from the first, and the amount of stretch in the spring resulting from the gravitational attraction of the two masses is measured. Recall that the acceleration due to gravity is about 9.80 m/s2 on Earth. We can now determine why this is so. The weight of an object mg is the gravitational force between it and Earth. Substituting mg for in Newton's universal law of gravitation gives = 2 where is the mass of the object, is the mass of Earth, and is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure 6.22. The mass of the object cancels, leaving an equation for :, (6.42) 238 Chapter 6 | Gravitation and Uniform Circular Motion = 2. Substituting known values for Earth's mass and radius (to three significant figures), 5.98×1024 kg (6.38×106 m)2 6.67×10−11N ⋅ m2 kg2 × = and we obtain a value for the acceleration of a falling body: = 9.80 m/s2., (6.43) (6.44) (6.45) Figure 6.22 The distance between the centers of mass of Earth and an object on its surface is very nearly the same as the radius of Earth, because Earth is so much larger than the object. This is the expected value and is independent of the body's mass. Newton's law of gravitation takes Galileo's observation that all masses fall with the same acceleration a step further, explaining the observation in terms of a force that causes objects to fall—in fact, in terms of a universally existing force of attraction between masses. Gravitational Mass and Inertial Mass Notice that, in Equation 6.40, the mass of the objects under consideration is directly proportional to the gravitational force. More mass means greater forces, and vice versa. However, we have already seen the concept of mass before in a different context. In Chapter 4, you read that mass is a measure of inertia. However, we normally measure the mass of an object by measuring the force of gravity (F) on it. How do we know that inertial mass is
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identical to gravitational mass? Assume that we compare the mass of two objects. The objects have inertial masses m1 and m2. If the objects balance each other on a pan balance, we can conclude that they have the same gravitational mass, that is, that they experience the same force due to gravity, F. Using Newton's second law of motion, F = ma, we can write m1 a1 = m2 a2. If we can show that the two objects experience the same acceleration due to gravity, we can conclude that m1 = m2, that is, that the objects' inertial masses are equal. In fact, Galileo and others conducted experiments to show that, when factors such as wind resistance are kept constant, all objects, regardless of their mass, experience the same acceleration due to gravity. Galileo is famously said to have dropped two balls of different masses off the leaning tower of Pisa to demonstrate this. The balls accelerated at the same rate. Since acceleration due to gravity is constant for all objects on Earth, regardless of their mass or composition, i.e., a1 = a2, then m1 = m2. Thus, we can conclude that inertial mass is identical to gravitational mass. This allows us to calculate the acceleration of free fall due to gravity, such as in the orbits of planets. Take-Home Experiment Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a piece of paper as well, does it behave like the other objects? Explain your observations. Making Connections: Gravitation, Other Forces, and General Relativity Attempts are still being made to understand the gravitational force. As we shall see in Particle Physics, modern physics is exploring the connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading us to think of gravitation as bending space and time. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 239 Applying the Science Practices: All Objects Have Gravitational Fields We can use the formula developed above, = 2, to calculate the gravitational fields of other objects. For example, the Moon has a radius of 1.7 × 106 m and a mass of 7.3 × 1022 kg. The gravitational field on the surface of the Moon can be expressed as = 2 =
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6.67×10−11 N·m2 kg2 × = 1.685 m/s2 7.3×1022 kg 2 1.7×106 m This is about 1/6 of the gravity on Earth, which seems reasonable, since the Moon has a much smaller mass than Earth does. A person has a mass of 50 kg. The gravitational field 1.0 m from the person's center of mass can be expressed as = 2 = 6.67×10−11 N·m2 kg2 × 50 kg (1 m)2 = 3.34×10−9 m/s2 This is less than one millionth of the gravitational field at the surface of Earth. In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force caused the Moon to orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its orbit. Newton found that the two accelerations agreed “pretty nearly.” Example 6.6 Earth's Gravitational Force Is the Centripetal Force Making the Moon Move in a Curved Path (a) Find the acceleration due to Earth's gravity at the distance of the Moon. (b) Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth), and compare it with the value of the acceleration due to Earth's gravity that you have just found. Strategy for (a) This calculation is the same as the one finding the acceleration due to gravity at Earth's surface, except that is the distance from the center of Earth to the center of the Moon. The radius of the Moon's nearly circular orbit is 3.84×108 m. Solution for (a) Substituting known values into the expression for found above, remembering that is the mass of Earth not the Moon, yields = = 2 6.67×10−11N ⋅ m2 kg2 = 2.70×10−3 m/s.2 Strategy for (b) Centripetal acceleration can be calculated using either form of We choose to use the second form: = 2 = 2. × 5.98×1024 kg (3.84×108 m)2 (6.46) (6.47) 240 Chapter 6 | Gravitation and Uniform Circular Motion where is the angular velocity of the Moon about Earth. Solution for (b) = 2, Given that the period
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(the time it takes to make one complete rotation) of the Moon's orbit is 27.3 days, (d) and using we see that The centripetal acceleration is 1 d×24hr d ×60min hr ×60 s min = 86,400 s = Δ Δ = 2π rad (27.3 d)(86,400 s/d) = 2.66×10−6rad s. = 2 = (3.84×108 m)(2.66×10−6 rad/s)2 = 2.72×10−3 m/s.2 (6.48) (6.49) (6.50) (6.51) The direction of the acceleration is toward the center of the Earth. Discussion The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth's gravity found in (a). This agreement is approximate because the Moon's orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth's surface). The clear implication is that Earth's gravitational force causes the Moon to orbit Earth. Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton's third law, if Earth exerts a force on the Moon, then the Moon should exert an equal and opposite force on Earth (see Figure 6.23). We do not sense the Moon's effect on Earth's motion, because the Moon's gravity moves our bodies right along with Earth but there are other signs on Earth that clearly show the effect of the Moon's gravitational force as discussed in Satellites and Kepler's Laws: An Argument for Simplicity. Figure 6.23 (a) Earth and the Moon rotate approximately once a month around their common center of mass. (b) Their center of mass orbits the Sun in an elliptical orbit, but Earth's path around the Sun has “wiggles” in it. Similar wiggles in the paths of stars have been observed and are considered direct evidence of planets orbiting those stars. This is important because the planets' reflected light is often too dim to be observed. Tides Ocean tides are one very observable result of the Moon's gravity acting on Earth. Figure 6.24 is a simplified drawing of the Moon's position relative to the tides. Because water easily flows on Earth's surface, a high tide is created
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on the side of Earth nearest to the Moon, where the Moon's gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth? The answer is that Earth is pulled toward the Moon more than the water on the far side, because Earth is closer to the Moon. So the water on the side of Earth closest to the Moon is pulled away from Earth, and Earth is pulled away from water on the far side. As Earth rotates, the tidal bulge (an effect of the tidal forces between an orbiting natural satellite and the primary planet that it orbits) keeps its orientation with the Moon. Thus there are two tides per day (the actual tidal period is about 12 hours and 25.2 minutes), because the Moon moves in its orbit each day as well). This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 241 Figure 6.24 The Moon causes ocean tides by attracting the water on the near side more than Earth, and by attracting Earth more than the water on the far side. The distances and sizes are not to scale. For this simplified representation of the Earth-Moon system, there are two high and two low tides per day at any location, because Earth rotates under the tidal bulge. The Sun also affects tides, although it has about half the effect of the Moon. However, the largest tides, called spring tides, occur when Earth, the Moon, and the Sun are aligned. The smallest tides, called neap tides, occur when the Sun is at a 90º angle to the Earth-Moon alignment. Figure 6.25 (a, b) Spring tides: The highest tides occur when Earth, the Moon, and the Sun are aligned. (c) Neap tide: The lowest tides occur when the Sun lies at 90º to the Earth-Moon alignment. Note that this figure is not drawn to scale. Tides are not unique to Earth but occur in many astronomical systems. The most extreme tides occur where the gravitational force is the strongest and varies most rapidly, such as near black holes (see Figure 6.26). A few likely candidates for black holes have been observed in our galaxy. These have masses greater than the Sun but have diameters only a few kilometers across. The tidal forces near them are so great that they can actually tear matter from a companion star. 242 Chapter 6 | Gravitation and Uniform Circular Motion Figure 6.26 A black
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hole is an object with such strong gravity that not even light can escape it. This black hole was created by the supernova of one star in a two-star system. The tidal forces created by the black hole are so great that it tears matter from the companion star. This matter is compressed and heated as it is sucked into the black hole, creating light and X-rays observable from Earth. ”Weightlessness” and Microgravity In contrast to the tremendous gravitational force near black holes is the apparent gravitational field experienced by astronauts orbiting Earth. What is the effect of “weightlessness” upon an astronaut who is in orbit for months? Or what about the effect of weightlessness upon plant growth? Weightlessness doesn't mean that an astronaut is not being acted upon by the gravitational force. There is no “zero gravity” in an astronaut's orbit. The term just means that the astronaut is in free-fall, accelerating with the acceleration due to gravity. If an elevator cable breaks, the passengers inside will be in free fall and will experience weightlessness. You can experience short periods of weightlessness in some rides in amusement parks. Figure 6.27 Astronauts experiencing weightlessness on board the International Space Station. (credit: NASA) Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface. Many interesting biology and physics topics have been studied over the past three decades in the presence of microgravity. Of immediate concern is the effect on astronauts of extended times in outer space, such as at the International Space Station. Researchers have observed that muscles will atrophy (waste away) in this environment. There is also a corresponding loss of bone mass. Study continues on cardiovascular adaptation to space flight. On Earth, blood pressure is usually higher in the feet than in the head, because the higher column of blood exerts a downward force on it, due to gravity. When standing, 70% of your blood is below the level of the heart, while in a horizontal position, just the opposite occurs. What difference does the absence of this pressure differential have upon the heart? Some findings in human physiology in space can be clinically important to the management of diseases back on Earth. On a somewhat negative note, spaceflight is known to affect the human immune system, possibly making the crew members more vulnerable to infectious diseases. Experiments flown in space also have shown that some bacteria grow faster in microgravity than they do on Earth. However, on a positive note
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, studies indicate that microbial antibiotic production can increase by a factor of two in space-grown cultures. One hopes to be able to understand these mechanisms so that similar successes can be achieved on the ground. In another area of physics space research, inorganic crystals and protein crystals have been grown in outer space that have much higher quality than any grown on Earth, so crystallography studies on their structure can yield much better results. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 243 Plants have evolved with the stimulus of gravity and with gravity sensors. Roots grow downward and shoots grow upward. Plants might be able to provide a life support system for long duration space missions by regenerating the atmosphere, purifying water, and producing food. Some studies have indicated that plant growth and development are not affected by gravity, but there is still uncertainty about structural changes in plants grown in a microgravity environment. The Cavendish Experiment: Then and Now As previously noted, the universal gravitational constant is determined experimentally. This definition was first done accurately by Henry Cavendish (1731–1810), an English scientist, in 1798, more than 100 years after Newton published his universal law of gravitation. The measurement of is very basic and important because it determines the strength of one of the four forces in nature. Cavendish's experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses (tens of kilograms at most), using apparatus like that in Figure 6.28. Remarkably, his value for differs by less than 1% from the best modern value. One important consequence of knowing was that an accurate value for Earth's mass could finally be obtained. This was done by measuring the acceleration due to gravity as accurately as possible and then calculating the mass of Earth from the relationship Newton's universal law of gravitation gives = 2 where is the mass of the object, is the mass of Earth, and is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See Figure 6.21. The mass of the object cancels, leaving an equation for :, (6.52) Rearranging to solve for yields = 2. = 2. (6.53) (6.54) So can be calculated because all quantities on the right, including the radius of Earth, are known from direct measurements. We shall see in Satellites and Kepler's
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Laws: An Argument for Simplicity that knowing also allows for the determination of astronomical masses. Interestingly, of all the fundamental constants in physics, is by far the least well determined. The Cavendish experiment is also used to explore other aspects of gravity. One of the most interesting questions is whether the gravitational force depends on substance as well as mass—for example, whether one kilogram of lead exerts the same gravitational pull as one kilogram of water. A Hungarian scientist named Roland von Eötvös pioneered this inquiry early in the 20th century. He found, with an accuracy of five parts per billion, that the gravitational force does not depend on the substance. Such experiments continue today, and have improved upon Eötvös' measurements. Cavendish-type experiments such as those of Eric Adelberger and others at the University of Washington, have also put severe limits on the possibility of a fifth force and have verified a major prediction of general relativity—that gravitational energy contributes to rest mass. Ongoing measurements there use a torsion balance and a parallel plate (not spheres, as Cavendish used) to examine how Newton's law of gravitation works over sub-millimeter distances. On this small-scale, do gravitational effects depart from the inverse square law? So far, no deviation has been observed. Figure 6.28 Cavendish used an apparatus like this to measure the gravitational attraction between the two suspended spheres ( ) and the two on the stand ( ) by observing the amount of torsion (twisting) created in the fiber. Distance between the masses can be varied to check the dependence of the force on distance. Modern experiments of this type continue to explore gravity. Chapter 6 | Gravitation and Uniform Circular Motion 249 Figure 6.31(b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all motions in the solar system, but all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. As our knowledge of nature has grown, the basic simplicity of its laws has become ever more evident. Figure 6.31 (a) The Ptolemaic model of the universe has Earth at the center with the Moon, the planets, the Sun, and the stars revolving about it in complex superpositions of circular paths. This geocentric model, which can be made progressively more accurate by adding more circles, is purely descriptive, containing no hints as to what are the causes
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of these motions. (b) The Copernican model has the Sun at the center of the solar system. It is fully explained by a small number of laws of physics, including Newton's universal law of gravitation. Glossary angular velocity:, the rate of change of the angle with which an object moves on a circular path arc length: Δ, the distance traveled by an object along a circular path banked curve: the curve in a road that is sloping in a manner that helps a vehicle negotiate the curve center of mass: the point where the entire mass of an object can be thought to be concentrated centrifugal force: a fictitious force that tends to throw an object off when the object is rotating in a non-inertial frame of reference centripetal acceleration: the acceleration of an object moving in a circle, directed toward the center centripetal force: any net force causing uniform circular motion Coriolis force: reference the fictitious force causing the apparent deflection of moving objects when viewed in a rotating frame of fictitious force: a force having no physical origin gravitational constant, G: a proportionality factor used in the equation for Newton's universal law of gravitation; it is a universal constant—that is, it is thought to be the same everywhere in the universe ideal angle: the angle at which a car can turn safely on a steep curve, which is in proportion to the ideal speed ideal banking: the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a certain speed without the aid of friction between the tires and the road; the net external force on the vehicle equals the horizontal centripetal force in the absence of friction ideal speed: the maximum safe speed at which a vehicle can turn on a curve without the aid of friction between the tire and the road microgravity: an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface Newton's universal law of gravitation: every particle in the universe attracts every other particle with a force along a line joining them; the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them non-inertial frame of reference: an accelerated frame of reference pit: a tiny indentation on the spiral track moulded into the top of the polycarbonate layer of CD 250 Chapter 6 | Gravitation and Uniform Circular Motion radians: a unit of angle measurement radius of curvature: radius
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of a circular path the ratio of the arc length to the radius of curvature on a circular path: rotation angle: Δ = Δ ultracentrifuge: a centrifuge optimized for spinning a rotor at very high speeds uniform circular motion: the motion of an object in a circular path at constant speed Section Summary 6.1 Rotation Angle and Angular Velocity • Uniform circular motion is motion in a circle at constant speed. The rotation angle Δ is defined as the ratio of the arc length to the radius of curvature: Δ = Δ, where arc length Δ is distance traveled along a circular path and is the radius of curvature of the circular path. The quantity Δ is measured in units of radians (rad), for which • The conversion between radians and degrees is 1 rad = 57.3º. • Angular velocity is the rate of change of an angle, 2π rad = 360º= 1 revolution. = Δ Δ, where a rotation Δ takes place in a time Δ. The units of angular velocity are radians per second (rad/s). Linear velocity and angular velocity are related by = or =. 6.2 Centripetal Acceleration • Centripetal acceleration c is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity and has the magnitude • The unit of centripetal acceleration is m / s2. 6.3 Centripetal Force c = 2 c = 2. • Centripetal force Fc is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity and has magnitude which can also be expressed as c = c, c = 2 or c = 2 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force • Rotating and accelerated frames of reference are non-inertial. • Fictitious forces, such as the Coriolis force, are needed to explain motion in such frames. 6.5 Newton's Universal Law of Gravitation • Newton's universal law of gravitation: Every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In equation form, this is = 2, This content is available for free at http://cnx.org/content/col
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11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 251 where F is the magnitude of the gravitational force. is the gravitational constant, given by = 6.673×10–11 N ⋅ m2/kg2. • Newton's law of gravitation applies universally. 6.6 Satellites and Kepler's Laws: An Argument for Simplicity • Kepler's laws are stated for a small mass orbiting a larger mass in near-isolation. Kepler's laws of planetary motion are then as follows: Kepler's first law The orbit of each planet about the Sun is an ellipse with the Sun at one focus. Kepler's second law Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times. Kepler's third law The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun: where is the period (time for one orbit) and is the average radius of the orbit. • The period and radius of a satellite's orbit about a larger body are related by 2 1 2 2 = 3 1 3 2, or Conceptual Questions 6.1 Rotation Angle and Angular Velocity 2 = 4π2 3 3 2 = 4π2. 1. There is an analogy between rotational and linear physical quantities. What rotational quantities are analogous to distance and velocity? 6.2 Centripetal Acceleration 2. Can centripetal acceleration change the speed of circular motion? Explain. 6.3 Centripetal Force 3. If you wish to reduce the stress (which is related to centripetal force) on high-speed tires, would you use large- or smalldiameter tires? Explain. 4. Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force? 5. If centripetal force is directed toward the center, why do you feel that you are ‘thrown' away from the center as a car goes around a curve? Explain. 6. Race car drivers routinely cut corners as shown in Figure 6.32. Explain how this allows the curve to be taken at the greatest speed. 252 Chapter 6 | Gravitation and Uniform Circular Motion Figure 6.32 Two paths around a race track curve are
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shown. Race car drivers will take the inside path (called cutting the corner) whenever possible because it allows them to take the curve at the highest speed. 7. A number of amusement parks have rides that make vertical loops like the one shown in Figure 6.33. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b)The car goes over the top at slower than this speed? Figure 6.33 Amusement rides with a vertical loop are an example of a form of curved motion. 8. What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 6.33 under the following circumstances: (a) The car goes over the top at such a speed that the gravitational force is the only force acting? (b) The car goes over the top faster than this speed? (c) The car goes over the top slower than this speed? 9. As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer. 10. Suppose a child is riding on a merry-go-round at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merry-go-round. Which path shown in Figure 6.34 will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer. This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 253 Figure 6.34 A child riding on a merry-go-round releases her lunch box at point P. This is a view from above the clockwise rotation. Assuming it slides with negligible friction, will it follow path A, B, or C, as viewed from Earth's frame of reference? What will be the shape of the path it leaves in the dust on the merry-go-round? 11. Do you feel yourself thrown to either side when
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you negotiate a curve that is ideally banked for your car's speed? What is the direction of the force exerted on you by the car seat? 12. Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth's frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton's third law, explain what force stretches the string, identifying its physical origin. Figure 6.35 A mass attached to a nail on a frictionless table moves in a circular path. The force stretching the string is real and not fictional. What is the physical origin of the force on the string? 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force 13. When a toilet is flushed or a sink is drained, the water (and other material) begins to rotate about the drain on the way down. Assuming no initial rotation and a flow initially directly straight toward the drain, explain what causes the rotation and which direction it has in the northern hemisphere. (Note that this is a small effect and in most toilets the rotation is caused by directional water jets.) Would the direction of rotation reverse if water were forced up the drain? 14. Is there a real force that throws water from clothes during the spin cycle of a washing machine? Explain how the water is removed. 15. In one amusement park ride, riders enter a large vertical barrel and stand against the wall on its horizontal floor. The barrel is spun up and the floor drops away. Riders feel as if they are pinned to the wall by a force something like the gravitational force. This is a fictitious force sensed and used by the riders to explain events in the rotating frame of reference of the barrel. Explain in an inertial frame of reference (Earth is nearly one) what pins the riders to the wall, and identify all of the real forces acting on them. 254 Chapter 6 | Gravitation and Uniform Circular Motion 16. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted? 17. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in free
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fall because the acceleration due to gravity is not 9.80 m/s2. Who do you agree with and why? 18. A non-rotating frame of reference placed at the center of the Sun is very nearly an inertial one. Why is it not exactly an inertial frame? 6.5 Newton's Universal Law of Gravitation 19. Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted? 20. Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not 9.80 m/s2. Who do you agree with and why? 21. Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body and decreases as it moves away. 22. Newton's laws of motion and gravity were among the first to convincingly demonstrate the underlying simplicity and unity in nature. Many other examples have since been discovered, and we now expect to find such underlying order in complex situations. Is there proof that such order will always be found in new explorations? 6.6 Satellites and Kepler's Laws: An Argument for Simplicity 23. In what frame(s) of reference are Kepler's laws valid? Are Kepler's laws purely descriptive, or do they contain causal information? This content is available for free at http://cnx.org/content/col11844/1.13 Chapter 6 | Gravitation and Uniform Circular Motion 255 Problems & Exercises 6.1 Rotation Angle and Angular Velocity 1. Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions—it then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read? 2. Microwave ovens rotate at a rate of about 6 rev/min. What is this in revolutions per second? What is the angular velocity in radians per second? 3. An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting
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