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the water. The acceleration of the dolphin is 2.6 m/s2 [up]. 8. Two boxes, A and B, are touching each other and are at rest on a horizontal, frictionless surface. Box A has a mass of 25 kg and box B a mass of 15 kg. A person applies a force of 30 N [right] on box A which, in turn, pushes on box B. Calculate the acceleration of the boxes. 9. A 4.0-kg oak block on a horizontal, rough oak surface is attached by a light string that passes over a light, frictionless pulley to a hanging 2.0-kg object. The magnitude of the force of friction on the 4.0-kg block is 11.8 N. 4.0 kg 2.0 kg (a) Calculate the acceleration of the system. (b) Calculate the tension in the string. (a) Calculate the mass of the dolphin. Extension (b) What would be the acceleration of the dolphin if it had the same strength but half the mass? 6. An ice hut used for winter fishing is resting on a level patch of snow. The mass of the hut is 80 kg. A wind exerts a horizontal force of 205 N on the hut, and causes it to accelerate. While in motion, the magnitude of the force of friction acting on the hut is 196 N. What is the acceleration of the hut? 7. Suppose the only horizontal forces acting on a 20-N object on a smooth table are 36 N [45] and 60 N [125]. (a) What is the net force acting on the object? (b) Calculate the acceleration of the object. 10. Summarize concepts and ideas associated with Newton’s second law using a graphic organizer of your choice. See Student References 4: Using Graphic Organizers on pp. 869–871 for examples of different graphic organizers. Make sure that the concepts and ideas are clearly presented and are linked appropriately. e TEST To check your understanding of Newton’s second law, follow the eTest links at www.pearsoned.ca/school/physicssource. 158 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 159 3.4 Newton’s Third Law Volleyball is a sport that involves teamwork and players knowing how to apply forces to the ball to redirect it. When the velocity of the ball
is large, a player will usually bump the ball to slow it down so that another player can redirect it over the net (Figure 3.51). At the instant the player bumps the ball, the ball exerts a large force on the player’s arms, often causing sore arms. Immediately after the interaction, the ball bounces upward. To explain the motion of each object during and after this interaction requires an understanding of Newton’s third law. Newton’s first two laws describe the motion of an object or a system of objects in isolation. But to describe the motion of objects that are interacting, it is important to examine how the force exerted by one object on another results in a change of motion for both objects. Find out what happens when two initially stationary carts interact by doing 3-7 QuickLab. info BIT In order to walk, you must apply a force backward on the ground with one foot. The ground then pushes forward on that foot. Figure 3.51 Conrad Leinemann of Kelowna, British Columbia, bumps the ball while teammate Jody Holden of Shelburne, Nova Scotia, watches during the beach volleyball competition at the 1999 Pan Am Games in Winnipeg, Manitoba. Chapter 3 Forces can change velocity. 159 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 160 3-7 QuickLab 3-7 QuickLab Exploding Carts Problem If a stationary cart exerts a net force on another identical cart, what will be the motion of both carts after the interaction? Materials dynamics cart with spring dynamics cart without spring 500-g standard mass Procedure 1 Note the position of the spring on the one cart, and standard mass barrier C-clamp cocked spring spring release Figure 3.52 Questions 1. What did you observe when you released the spring when the cart was initially at rest and not touching the other cart? how to cock and release the spring. 2. (a) What did you observe when you released the 2 Cock the spring and place the cart on the table. Release the spring. CAUTION: Do not cock the spring unless it is safely attached to the cart. Do not point the spring at anyone when releasing it. 3 Repeat step 2, this time making the cart with the spring touch the second cart (Figure 3.52). Release the spring. 4 Repeat step 3 but add a 500-g standard mass to one of the carts before releasing the spring. spring when one cart was touching the other cart? (b
) What evidence do you have that two forces were present? (c) What evidence do you have that a force was exerted on each cart? (d) How do the magnitudes and directions of the two forces compare? 3. Compare and contrast the results from steps 3 and 4. action force: force initiated by object A on object B reaction force: force exerted by object B on object A Forces Always Exist in Pairs When two objects interact, two forces will always be involved. One force is the action force and the other is the reaction force. The important points to remember are that the reaction force always acts on a different object than the action force, and that the reaction force acts in the opposite direction. Concept Check Is it possible to have an action force without a reaction force? 160 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 161 Newton’s Third Law and Its Applications Newton found that the reaction force is equal in magnitude to the action force, but opposite in direction. This relationship is called Newton’s third law of motion. If object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction. F A on B F B on A Some people state Newton’s third law as “for every action force, there is an equal and opposite reaction force.” However, remembering Newton’s third law this way does not emphasize that the action and reaction forces are acting on different objects (Figure 3.53). PHYSICS INSIGHT In order to show action-reaction forces, you must draw two free-body diagrams, one for each object. Faction force exerted by student on ground Freaction force exerted by ground on student Figure 3.53 The action force is the backward force that the student exerts on the ground. The reaction force is the forward force that the ground exerts on the student. Only the action-reaction pair are shown here for simplicity. Concept Check If the action force is equal in magnitude to the reaction force, how can there ever be an acceleration? Explain using an example and free-body diagrams. Action-Reaction Forces Acting on Objects in Contact Let’s revisit the scenario of the volleyball player bumping the ball. At the instant that both the ball and the player’s arms are in contact, the action force is the upward force that the player exerts on the ball. The
reaction force is the downward force that the ball exerts on the player’s arms. During the collision, the ball accelerates upward and the player’s arms accelerate downward (Figure 3.54). e TECH Explore how a stranded astronaut can return to a spacecraft by throwing away tools. Follow the eTech links at www.pearsoned.ca/school/ physicssource. F action force exerted by player on ball force exerted by ball on player F reaction Figure 3.54 The action-reaction forces at collision time when a volleyball player bumps the ball Chapter 3 Forces can change velocity. 161 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 162 A similar reasoning applies when a baseball bat strikes a baseball. The action force is the forward force that the bat exerts on the ball. The reaction force is the backward force that the ball exerts on the bat. During the collision, the ball accelerates forward and the bat slows down as it accelerates backward (Figure 3.55). Freaction force exerted by ball on bat Faction force exerted by bat on ball e WEB Fire hoses and extinguishers are difficult to control because their contents exit at high speed as they are redirected when putting out a fire. Research the operation and safe use of fire extinguishers. How do Newton’s three laws apply to fire extinguishers? Interview an experienced firefighter. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/school/ physicssource. Figure 3.55 The action-reaction forces at collision time when a baseball bat strikes a baseball Action-Reaction Forces Acting on Objects Not in Contact Sometimes an object can exert a force on another without actually touching the other object. This situation occurs when an object falls toward Earth’s surface, or when a magnet is brought close to an iron nail. Action-reaction forces still exist in these interactions. When an apple falls toward the ground, the action force is the force of gravity that Earth exerts on the apple. The falling apple, in turn, exerts a reaction force upward on Earth. So while the apple is accelerating down, Earth is accelerating up (Figure 3.56). You see the acceleration of the apple but not of Earth because the inertial mass of the apple is far less than that of Earth. In fact, Earth does accelerate but at a negligible rate because the magnitude of the acceleration is inversely proportional to mass. force
exerted by Earth on apple Faction force exerted by apple on Earth Freaction Figure 3.56 The action-reaction forces when an apple falls toward Earth’s surface 162 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 163 When a magnet is brought close to an iron nail, the action force is the magnetic force that the magnet exerts on the nail. The reaction force is the force that the nail exerts on the magnet. So the nail accelerates toward the magnet, and at the same time the magnet is accelerating toward the nail (Figure 3.57). Investigate the validity of Newton’s third law by doing 3-8 QuickLab. Freaction Faction N force exerted by nail on magnet force exerted by magnet on nail S Figure 3.57 The action-reaction forces when a magnet is brought close to an iron nail 3-8 QuickLab 3-8 QuickLab Skateboard Interactions Problem How does Newton’s third law apply to interactions involving skateboards? Materials two skateboards CAUTION: Wear a helmet and knee pads when doing this activity. Procedure 1 Choose a partner with a mass about the same as yours. 2 Sit on skateboards on a hard, level surface with your feet toward one another and touching (Figure 3.58). Figure 3.58 3 Give your partner a gentle push with your feet. Observe what happens to both skateboards. 4 Repeat steps 2 and 3 but this time, give your partner a harder push. Observe what happens to both skateboards. 5 Repeat steps 2 and 3 but this time, have you and your partner push simultaneously. Observe what happens to both skateboards. 6 Choose a partner with a significantly different mass than yours. 7 Repeat steps 2 to 5 with your new partner. 8 Sit on a skateboard near a wall. Then push against the wall. Observe the motion of your skateboard. Questions 1. Describe the motion of each skateboard when (a) you pushed a partner of equal mass, and (b) you pushed a partner of significantly different mass. 2. Compare and contrast the results from steps 4 and 5. 3. What happened to your skateboard when you pushed against the wall? 4. Explain each interaction in this activity using Newton’s laws. Draw a sketch showing the action-reaction forces in each situation. Chapter 3 Forces can change velocity. 163 03-Phys20-Chap03.q
xd 7/24/08 10:37 AM Page 164 Applying Newton’s Third Law to Situations Involving Frictionless Surfaces In Example 3.12, an applied force acts on box A, causing all three boxes to accelerate. Newton’s third law is used to calculate the force that box C exerts on box B. Example 3.12 Three boxes, A, B, and C, are positioned next to each other on a horizontal, frictionless surface (Figure 3.59). An applied force acting on box A causes all the boxes to accelerate at 1.5 m/s2 [right]. Calculate the force exerted by box C on box B. 10 kg 8.0 kg A B 5.0 kg C Figure 3.59 Practice Problems 1. For the situation in Example 3.12, calculate the force that box B exerts on box A. 2. For the situation in Example 3.9 Practice Problem 1 on page 153, calculate the applied force needed to lift both buckets up. Answers 1. 23 N [left] 2. 1.2 102 N [up] Given mA 8.0 kg a 1.5 m/s2 [right] mB 10 kg mC 5.0 kg Required C on B) force exerted by box C on box B (F FN Analysis and Solution Draw a free-body diagram for box C (Figure 3.60). Box C is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on box C in both the horizontal and vertical directions. up left down right horizontal direction F neth Fneth F B on C FB on C F g vertical direction FF F netv N 0 Fnetv Calculations in the vertical direction are not required in this problem. FB on C mC Apply Newton’s second law. FB on C F B on C m s2 mCa (5.0 kg)1.5 m 7.5 kg 2 s 7.5 N 7.5 N [right] Apply Newton’s third law. F C on B F B on C 7.5 N [left] Fg Paraphrase The force exerted by box C on box B is 7.5 N [left]. Figure 3.60 164 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 165 Applying Newton’s Third Law to Situations
Involving Friction In Example 3.13, a rough surface exerts a force of friction on two boxes. Newton’s third law is used to calculate the force exerted by box B on box A in this situation. Example 3.13 Two boxes, A and B, of identical material but different mass are placed next to each other on a horizontal, rough surface (Figure 3.61). An applied force acting on box A causes both boxes to accelerate at 2.6 m/s2 [right]. If the magnitude of the force of friction on box B is 28.3 N, calculate the force exerted by box B on box A. Given mA F f on B 6.5 kg 28.3 N [left] mB 8.5 kg 6.5 kg 8.5 kg A B Figure 3.61 FN left up down right a 2.6 m/s2 [right] Ff on B mB FA on B Required B on A) force exerted by box B on box A (F Analysis and Solution Draw a free-body diagram for box B (Figure 3.62). Box B is not accelerating up or down. 0 N. So in the vertical direction, F net Write equations to find the net force on box B in both the horizontal and vertical directions. horizontal direction FF neth Fneth FF A on B FA on B FF f on B Ff on B F g vertical direction FF F netv N 0 Fnetv Calculations in the vertical direction are not required in this problem. Apply Newton’s second law. mBa FA on B Ff on B FA on B F A on B mBa Ff on B (8.5 kg)(2.6 m/s2) (28.3 N) (8.5 kg)(2.6 m/s2) 28.3 N 50 N 50 N [right] Apply Newton’s third law. F B on A F A on B 50 N [left] Paraphrase The force exerted by box B on box A is 50 N [left]. FA on B Fg Figure 3.62 Ff on B Fneth Practice Problem 1. To minimize the environmental impact of building a road through a forest, a logger uses a team of horses to drag two logs, A and B, from the cutting location to a nearby road. A light chain connects log A with a mass of 150 kg to the horses’ harness. Log B with a mass of 250
kg is connected to log A by another light chain. (a) The horses can pull with a combined force of 2600 N. The ground exerts a force of friction of magnitude 2400 N on the logs. Calculate the acceleration of the logs. (b) If the force of friction on log A is 900 N, calculate the force exerted by log B on log A. Answer 1. (a) 0.500 m/s2 [forward] (b) 1.63 103 N [backward] Chapter 3 Forces can change velocity. 165 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 166 Applying Newton’s Second and Third Laws to Propeller Aircraft The acceleration of many devices such as propeller aircraft can be controlled in midair. To explain how these machines accelerate involves applying Newton’s second and third laws. A propeller airplane can move through air because as the propeller rotates, it exerts an action force on the air, pushing the air backward. According to Newton’s third law, the air, in turn, exerts a reaction force on the propeller, pushing the airplane forward (Figure 3.63). Propeller blades are slanted so that they scoop new air molecules during each revolution. The faster a propeller turns, the greater is the mass of air accelerated backward and, by Newton’s second law, the force exerted by the air on the propeller increases. Faction force exerted by propeller on air Freaction force exerted by air on propeller Figure 3.63 The action-reaction forces when a propeller airplane is in flight THEN, NOW, AND FUTURE Wallace Rupert Turnbull (1870–1954) Wallace Rupert Turnbull was an aeronautical engineer interested in finding ways to make aircraft wings stable (Figure 3.64). In 1902, he built the first wind tunnel in Canada at Rothesay, New Brunswick, for his experiments on propeller design. In 1909, Turnbull was awarded a bronze medal from the Royal Aeronautical Society for his research on efficient propeller design. One of his major inventions was the variable-pitch propeller, which is still used on aircraft today. During takeoff, the angle of the blades is adjusted to scoop more air. Air moving at a high speed backward gives a plane thrust, which causes the plane to accelerate forward. Once a plane maintains a constant altitude, the blade angle, or pitch, is decreased, reducing fuel
consumption. This allows greater payloads to be carried efficiently and safely through the sky. By 1925, Turnbull had perfected a propeller that used an electric motor to change its pitch. In 1927, the Canadian Air Force successfully tested the propeller at Borden, Ontario. Turnbull was later inducted into the Canadian Aviation Hall of Fame in 1977. Questions 1. Research the forces that act on airplanes in flight. Define these forces and compare them to forces already discussed in this chapter. 2. Explain how and where the forces on an airplane act to cause changes in its horizontal motion. Use Newton’s laws and diagrams to support your explanations. Figure 3.64 Canadian inventor Wallace Rupert Turnbull 166 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 167 Applying Newton’s Third Law to Rockets The motion of rockets is a little different from that of propeller airplanes because a rocket does not have propellers that scoop air molecules. In fact, a rocket can accelerate in outer space where there is a vacuum. When a rocket engine is engaged, the highly combustible fuel burns at a tremendous rate. The action force of the exhaust gas leaving the rocket, according to Newton’s third law, causes a reaction force that pushes against the rocket. It is the action force of the exhaust gas being directed backward that accelerates the rocket forward (Figure 3.65). That is why a rocket can accelerate in outer space. Test out Newton’s third law with a toy rocket by doing 3-9 Design a Lab. Faction force exerted by rocket on exhaust gas Freaction force exerted by exhaust gas on rocket Figure 3.65 The action-reaction forces when a rocket is in flight 3-9 Design a Lab 3-9 Design a Lab Motion of a Toy Rocket Figure 3.66 shows a toy rocket partially filled with water about to be released from an air pump. The pump is used to add pressurized air into the rocket. air under pressure air pump water rocket release Figure 3.66 The Question What effect does increasing each of these quantities have on the motion of the rocket? • the amount of water inside the rocket • the air pressure inside the rocket Design and Conduct Your Investigation • State a hypothesis. Then design and conduct an experiment to test your hypothesis. Be sure to identify all variables and to control the appropriate ones. Caution: Never point the rocket at anyone. Perform this activity outside. • Compare the direction of motion of the water and the
rocket when the rocket is released. • Explain the motion of the rocket, water, and air in terms of Newton’s third law. Include sketches showing at least three action-reaction pairs of forces. • How well did your results agree with your hypothesis? Chapter 3 Forces can change velocity. 167 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 168 3.4 Check and Reflect 3.4 Check and Reflect Knowledge 1. In your own words, state Newton’s third law. 2. Explain why (a) a swimmer at the edge of a pool pushes backward on the wall in order to move forward, and (b) when a person in a canoe throws a package onto the shore, the canoe moves away from shore. 3. No matter how powerful a car engine is, a car cannot accelerate on an icy surface. Use Newton’s third law and Figure 3.53 on page 161 to explain why. 4. State and sketch the action-reaction forces in each situation. (a) Water pushes sideways with a force of 600 N on the centreboard of a sailboat. (b) An object hanging at the end of a spring exerts a force of 30 N [down] on the spring. Applications 5. An object is resting on a level table. Are the normal force and the gravitational force acting on the object action-reaction forces? Explain your reasoning. 6. A vehicle pushes a car of lesser mass from rest, causing the car to accelerate on a rough dirt road. Sketch all the actionreaction forces in this situation. 7. Suppose you apply a force of 10 N to one spring scale. What is the reading on the other spring scale? What is the force exerted by the anchored spring scale on the wall? 8. Blocks X and Y are attached to each other by a light rope and can slide along a horizontal, frictionless surface. Block X has a mass of 10 kg and block Y a mass of 5.0 kg. An applied force of 36 N [right] acts on block X. Y X Fapp (a) Calculate the action-reaction forces the blocks exert on each other. (b) Suppose the magnitudes of the force of friction on blocks X and Y are 8.0 N and 4.0 N respectively. Calculate the action-reaction forces the blocks exert on each other. 9. A rectangular juice box has two holes punched near the bottom corners
on opposite sides, and another hole at the top. The box is hung from a rigid support with a string. Predict what will happen if the box is filled with water through the top hole and the holes at the bottom are open. Use Newton’s third law to explain your answer. Test your prediction. Cover the holes at the bottom with tape before filling the box with water. Then remove the tape to let the water out and observe the motion of the box. string hole water juice box holes? F 10 N water stream e TEST To check your understanding of Newton’s third law, follow the eTest links at www.pearsoned.ca/school/physicssource. 168 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 169 3.5 Friction Affects Motion Throughout this chapter, you encountered friction in all the lab activities and when solving several problems. Friction is a force that is present in almost all real-life situations. In some cases, friction is desirable while in other cases, friction reduces the effectiveness of mechanical systems. Without friction, you would not be able to walk. The wheels on a vehicle would have no traction on a road surface and the vehicle would not be able to move forward or backward. Parachutists would not be able to land safely (Figure 3.67). On the other hand, friction causes mechanical parts to seize and wear out, and mechanical energy to be converted to heat. For example, snowmobiles cannot move for long distances over bare ice. Instead, snowmobilers must detour periodically through snow to cool the moving parts not in contact with the ice. To determine the direction of the force of friction acting on an object, you need to first imagine the direction in which the object would move if there were no friction. The force of friction, then, opposes motion in that direction. info BIT Olympic cyclists now wear slipperier-than-skin suits with seams sown out of the airflow to reduce friction and improve race times by as much as 3 s. friction: force that opposes either the motion of an object or the direction the object would be moving in if there were no friction Figure 3.67 When a person falls in midair, the air resistance that acts on a parachute slows the fall. In this case, friction allows a parachutist to land without injury. Chapter 3 Forces can change velocity. 169 03-Phys20-Chap03.q
xd 7/24/08 10:38 AM Page 170 In a sport such as curling, friction affects how far the stone will travel along the ice. Sweeping the ice in front of a moving stone reduces the force of friction acting on the stone (Figure 3.68). The result is that the stone slides farther. To better understand how the nature of a contact surface affects the force of friction acting on an object, do 3-10 QuickLab. Figure 3.68 Brad Gushue, from St. John’s, Newfoundland, and his team won the gold medal in men’s curling at the 2006 Winter Olympics in Turin, Italy. 3-10 QuickLab 3-10 QuickLab Friction Acting on a Loonie Problem What factors affect the ability of a loonie to start sliding? 4 Use a piece of tape to fasten the fine sandpaper on the textbook, sandy-side facing up. Repeat step 3. Materials two loonies textbook tape protractor coarse, medium, and fine sandpaper: a 10 cm 25 cm piece of each Procedure 1 Read the procedure and design a chart to record your results. 2 Place your textbook flat on a lab bench and place a loonie at one end of the book. 5 Repeat step 4 for the medium sandpaper and then for the coarse sandpaper. Carefully remove and save the sandpaper. 6 Repeat steps 2 and 3 but this time increase the mass (not the surface area) by stacking one loonie on top of the other. Use a piece of tape between the two loonies to fasten them together. Questions 1. How consistent were your results for each trial? 2. Explain how the angle needed to start the loonie sliding down the incline was affected by 3 Slowly raise this end of the textbook until the loonie starts to slide down the incline (Figure 3.69). • • the roughness of the contact surface the mass of the coins (number of stacked coins) in contact with the contact surface Use the protractor to measure the angle the textbook makes with the lab bench when the loonie first starts to slide. Repeat this step several times, and find the average of the angles. 170 Unit II Dynamics 3. Identify the controlled, manipulated, and responding variables in this activity. Figure 3.69 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 171 Static Friction In 3-10 QuickLab
, you discovered that the force of friction depends on the nature of the two surfaces in contact. If you drag an object on a smooth surface, the force of friction acting on the object is less than if you drag it on a rough or bumpy surface. If you drag a smooth block and a rough block on the same surface, the force of friction acting on each block will be different. Although there are different types of friction, the force of friction that acts on objects sliding across another surface is the main focus in this section. e SIM Learn how friction is created and how it affects the net force on an object. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Suppose an object A (the desk) is in contact with another object B (the floor) as in Figure 3.70. The contact surface would be the horizontal surface at the bottom of each leg of the desk. app, such that F Now suppose that a force acts on the desk, say F app has a vertical component as well as a horizontal component. If the desk remains at rest, even though F app acts on it, then the net force on the desk is zero, F net 0 N. y x vd 0 Fappx θ Fapp Fappy Ffstatic Figure 3.70 An applied force F floor exerts a force of static friction on the bottom of each leg of the desk. app is acting on the desk at a downward angle. The In the x direction, Fnetx 0 N, which means that F appx must be balanced by another force. This balancing force is the force of static friction, F fstatic. The equation for the net force acting on the desk in the x direction would then be F F F netx fstatic appx Ffstatic Fappx Fnetx 0 Fappcos Ffstatic Fappcos Ffstatic So the direction of FF fstatic opposes the x component of the applied force acting on the desk. static friction: force exerted on an object at rest that prevents the object from sliding Chapter 3 Forces can change velocity. 171 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 172 The Magnitude of Static Friction An important point about static friction is that its magnitude does not have a fixed value. Instead, it varies from zero to some maximum value. This maximum value is reached at the instant the object starts to move. If you push on a table with
a force of ever-increasing magnitude, you will notice that the table remains at rest until you exceed a critical value. Because of Newton’s second law, the magnitude of the force of static friction must increase as the applied force on the table increases, if the forces are to remain balanced. Static Friction on a Horizontal Surface Suppose the applied force acting on the desk in Figure 3.70 on page 171 is given. Example 3.14 demonstrates how to calculate the force of static friction by using a free-body diagram to help write the equation for the net force on the desk. Since F app acts at an angle to the surface of the desk, it is convenient to use Cartesian axes to solve this problem. Example 3.14 The magnitude of the applied force in Figure 3.71 is 165 N and 30.0. If the desk remains stationary, calculate the force of static friction acting on the desk. Given magnitude of F app 30.0 165 N Required fstatic) force of static friction (F y x x θ 30.0° Fapp Figure 3.71 Practice Problem 1. A mountain climber stops during the ascent of a mountain (Figure 3.73). Sketch all the forces acting on the climber, and where those forces are acting. Figure 3.73 172 Unit II Dynamics Analysis and Solution Draw a free-body diagram for the desk (Figure 3.72). Fappx Ffstatic 0 Fnetx Fappy 0 Fnety FN Fg y x FN Ffstatic Fappx θ 30.0° Fappy Fapp Fg Figure 3.72 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 173 0 N in both Since the desk is not accelerating, F net the x and y directions. Write equations for the net force on the desk in both directions. x direction F FF netx appx Fappx Fnetx 0 Fappx y direction FF F nety N 0 Fnety Calculations in the y direction are not required in this problem. FF fstatic Ffstatic Ffstatic FF appy F g Ffstatic Fappx (165 N)(cos ) (165 N)(cos 30.0) 143 N F fstatic prevents the desk from sliding in the x direction. The negative value for Ffstatic indicates that the direction of FF fstatic is along the negative x-axis or [180]. F fstatic 143 N [180] Par
aphrase The force of static friction acting on the desk is 143 N [180]. Answer 1. FT Ffstatic FN FN Ffstatic Fg Static Friction on an Incline If an object is at rest on an incline, the net force acting on the object 0 N. Let’s first examine the forces acting on the object is zero, F net (Figure 3.74). FN Ffstatic Figure 3.74 (left) Free-body diagram for an object at rest on an incline; (below) vector addition diagrams for the and forces Fg θ Fg Fg θ Fg Forces Forces Fnet Fg Ffstatic Fnet 0 Ffstatic Fg Fg Fnet FN Fnet 0 FN Chapter 3 Forces can change velocity. 173 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 174 When working with inclines, it is easier to rotate the reference coordinates so that motion along the incline is described as either uphill or downhill. This means that only the gravitational force needs to be resolved into components, one parallel to the incline F g and one perpendicular to the incline F g. Usually, uphill is chosen to be positive unless the object is accelerating downhill. In Figure 3.74 on page 173, if there were no friction, the component g would cause the object to accelerate down the incline. So for the object F fstatic) must be acting up the incline. to remain at rest, a balancing force (FF The equation for the net force acting on the object parallel to the incline would then be g F net F F fstatic Fnet Fg Ffstatic 0 Fg Ffstatic Ffstatic Fg g requires using the geometry of To determine the expression for F g is 90.0. Since a triangle. In Figure 3.75, the angle between F g and FF g is. g is 90.0, the angle between FF g and F g and F the angle between F 90° θ θ Fg Fg θ Fg Figure 3.75 Diagram for an object at rest on an incline showing only the force of gravity vector resolved into components Since the object is not accelerating perpendicular to the incline, the equation for the net force acting on the object in this direction is net FF F N Fnet FN 0 FN FN Fg F g Fg Fg In 3-10 QuickLab, the sandpaper
exerted a force of static friction on the loonie, preventing the coin from sliding down the incline. Example 3.15 demonstrates how to calculate the force of static friction acting on a loonie at rest on an incline of 25.0. 174 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 175 Example 3.15 A loonie with a mass of 7.0 g is at rest on an incline of 25.0 (Figure 3.76). Calculate the force of static friction acting on the loonie. Given m 7.0 g 7.0 103 kg g 9.81 m/s2 Required fstatic) force of static friction (F 25.0 loonie θ 25.0° Figure 3.76 Analysis and Solution Draw a free-body diagram for the loonie (Figure 3.77). p u FN p u h ill h ill o w n d o w n d Ffstatic Fg Fg θ 25.0° θ 25.0° Fg Fg Ffstatic Fnet 0 Figure 3.77 Practice Problems 1. A loonie of mass 7.0 g is taped on top of a toonie of mass 7.3 g and the combination stays at rest on an incline of 30.0. Calculate the force of static friction acting on the face of the toonie in contact with the incline. 2. A loonie of mass 7.00 g is placed on the surface of a rough book. A force of static friction of magnitude 4.40 102 N acts on the coin. Calculate the maximum angle at which the book can be inclined before the loonie begins to slide. Answers 1. 7.0 102 N [uphill] 2. 39.8 0 N both Since the loonie is not accelerating, F net parallel and perpendicular to the incline. Write equations to find the net force on the loonie in both directions. direction net FF FF N Fnet 0 Calculations in the direction are not required in this problem. direction g FF net F FF fstatic Fnet Fg Ffstatic Fnet Fg Ffstatic Now, Fg mg sin So, Ffstatic F g 0 (mg sin ) mg sin (7.0 103 kg)(9.81 m/s2)(sin 25.0) 2.9 102 N F f
static prevents the loonie from sliding downhill. The positive value for Ffstatic indicates that the direction of F fstatic is uphill. 2.9 102 N [uphill] F fstatic Paraphrase The force of static friction acting on the loonie is 2.9 102 N [uphill]. Chapter 3 Forces can change velocity. 175 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 176 kinetic friction: force exerted on an object in motion that opposes the motion of the object as it slides on another object Kinetic Friction Suppose you apply a force to the desk in Figure 3.78 and the desk starts to slide across the floor at constant velocity. In this situation, the force of static friction is not able to balance the applied force, so motion occurs. Now the floor will exert a force of friction on the desk that opposes the direction of motion of the desk. This force is the force f kinetic. of kinetic friction, F Kinetic friction is present any time an object is sliding on another, whether or not another force acts on the sliding object. If you stop pushing the desk once it is in motion, the desk will coast and eventually stop. While the desk is sliding, the floor exerts a force of kinetic friction on the desk. This frictional force is directed backward, and causes the desk to eventually come to a stop. y x vd constant Fappx θ Fapp Fappy Ffkinetic Figure 3.78 The applied force F desk, causing the desk to slide. While the desk is in motion, the floor exerts a force of kinetic friction that opposes the motion of the desk. app overcomes the force of static friction acting on the The Direction of Kinetic Friction on an Incline If an object is on an incline and the object begins to slide, the surface of the incline exerts a force of kinetic friction on the object that opposes its motion. Whether the object is accelerating uphill or downhill, F net 0 N parallel to the incline. 176 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 177 Accelerating Down an Incline Let’s first consider the case where an object accelerates downhill g causes the object to accelerate (Figure 3.79). In this situation, F downhill. The force of kinetic friction acts to oppose the motion of the fkinetic is uphill as shown below. object
. So F a FN Ffkinetic Fg Fg θ Fg θ Figure 3.79 (left) Free-body diagram for an object accelerating downhill; (below) vector addition diagrams for the and forces Forces Forces Fg Fnet Ffkinetic Fg Ffkinetic Fnet Fnet 0 Fg FN Fg Fnet Fnet FN 0 The equation for the net force acting on the object parallel to the incline is g F net F F fkinetic net becomes If you apply Newton’s second law, the equation for F ma F g F fkinetic ma Fg Ffkinetic fkinetic acts uphill. For the object to In Figure 3.79, F g acts downhill and F accelerate downhill, the net force on the object, F net, is directed downhill. So the magnitude of F fkinetic. g must be greater than the magnitude of F Since the object is not accelerating perpendicular to the incline, the net force acting on the object in this direction is zero. The equation for the net force on the object in the perpendicular direction is net F F N Fnet FN 0 FN FN Fg F g Fg Fg Chapter 3 Forces can change velocity. 177 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 178 Accelerating Up an Incline If an object is accelerating uphill, the force of kinetic friction acts app, g also acts downhill. A force, F downhill to oppose the motion. F must act uphill on the object that is great enough to overcome both F fkinetic g (Figure 3.80). and F Fapp Forces a FN Ffkinetic Fg Fg θ Fg θ Fapp Fg Ffkinetic Fg Ffkinetic Fapp Fnet Fnet ≠ 0 Fnet Fg FN Forces Fg Fnet FN Fnet 0 Figure 3.80 (left) Free-body diagram for an object accelerating uphill; (right) vector addition diagrams for the and forces The equation for the net force acting on the object parallel to the incline is net F F app g F F fkinetic net becomes If you apply Newton’s second law, the equation for F ma F app ma Fapp g F F fkinetic Fg Ffkinetic app acts uphill. For fkinetic act downhill and F g and F In Figure 3.80, both F net is directed uphill.
So the magnitude the object to accelerate uphill, F app must be greater than the sum of the magnitudes of F fkinetic. g and F of F Since the object is not accelerating perpendicular to the incline, the net force acting on the object in this direction is zero. The equation for the net force on the object in the perpendicular direction is net F F N Fnet FN 0 FN FF g Fg Fg FN Fg Concept Check What is the angle between the normal force and the force of friction? Is this angle always the same size? Explain your reasoning. 178 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 179 Example 3.16 demonstrates how to calculate the acceleration of a block sliding down an incline. Since the direction of motion of the block is downhill, it is convenient to choose downhill to be positive. Example 3.16 A 3.5-kg block is sliding down an incline of 15.0 (Figure 3.81). The surface of the incline exerts a force of kinetic friction of magnitude 3.9 N on the block. Calculate the acceleration of the block. 3.5 kg 15.0° Figure 3.81 Given m 3.5 kg magnitude of F fkinetic 3.9 N Required acceleration of block (a ) 15.0 g 9.81 m/s2 Analysis and Solution Draw a free-body diagram for the block (Figure 3.82). Since the block is accelerating downhill, F net to the incline, but F net 0 N perpendicular to the incline. 0 N parallel F Write equations to find the net force on the block in both directions. direction net F F N Fnet 0 Calculations in the direction are not required in this problem. direction g F net F F fkinetic Fnet Fg Ffkinetic ma Fg Ffkinetic Now, Fg mg sin So, ma mg sin (3.9 N) mg sin 3.9 N g FN Fg Ff kinetic θ 15.0° θ 15.0° Fg Fg Fg Ff kinetic Fnet 9 N 3. a g sin m 9 N. 3 (9.81 m/s2)(sin 15.0) g k 5. 3 1.4 m/s2 Figure 3.82 The positive value for a indicates that the direction of a is downhill. a 1.4 m/s2 [down
hill] Paraphrase The acceleration of the block is 1.4 m/s2 [downhill]. Practice Problems 1. Determine the acceleration of the block in Example 3.16 if friction is not present. 2. A 55.0-kg skier is accelerating down a 35.0 slope. The magnitude of the skier’s acceleration is 4.41 m/s2. Calculate the force of kinetic friction that the snowy surface exerts on the skis. Answers 1. 2.5 m/s2 [downhill] 2. 66.9 N [uphill] Chapter 3 Forces can change velocity. 179 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 180 Comparing the Magnitudes of Static and Kinetic Friction The magnitude of the force of kinetic friction is never greater than the maximum magnitude of the force of static friction. Often, the magni- fstatic. fkinetic is less than the magnitude of F tude of F Figure 3.83 shows a graph of a situation where a person is applying very little force to an object during the first 2 s. Then the person begins to push harder, and at t 4 s, the object starts to move. The graph does not provide any information about the applied force after 4 s ( Ff Magnitude of the Force of Friction vs. Time static friction maximum value of static friction kinetic friction 0 4 8 12 16 20 Time t (s) Figure 3.83 The force of static friction increases up to a maximum value. Concept Check Explain why it makes sense that the magnitude of the force of kinetic friction does not exceed the maximum magnitude of the force of static friction. e WEB Leonardo da Vinci was as creative in science as he was in art. Research some of da Vinci’s scientific ideas. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/ school/physicssource. Determining the Magnitude of Frictional Forces Leonardo da Vinci (1452–1519) was one of the first people to experimentally determine two important relationships about friction. He discovered that for hard contact surfaces, the force of friction does not depend on the contact surface area. If you push a heavy box across the floor, the force of friction acting on the box is the same whether you push it on its bottom or on its side [Figure 3.84 (a)
and (b)]. Da Vinci also discovered that the force of friction acting on an object depends on the normal force acting on that object. Find out what this relationship is by doing 3-11 Inquiry Lab. (a) (b) Figure 3.84 The force of friction acting on the box in each of these pictures is the same. For hard contact surfaces, the force of friction does not depend on contact surface area. 180 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 181 3-11 Inquiry Lab 3-11 Inquiry Lab Relating Static Friction and the Normal Force Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question What is the relationship between the maximum magnitude of the force of static friction and the magnitude of the normal force acting on an object? Hypothesis State a hypothesis relating the magnitude of F and the magnitude of F N. Write an “if/then” statement. fstatic Variables Read the procedure and identify the controlled, manipulated, and responding variable(s). Materials and Equipment balance wooden block with different face areas and a hook horizontal board spring scale, calibrated in newtons set of standard masses Procedure 1 Read the steps of the procedure and design a chart to record your results. 2 Measure the mass of the block using the balance. 3 Place the largest face of the block on the horizontal board. Attach the spring scale to the block. Pull with an ever-increasing horizontal force until the block just starts to move. Record this force, which is the maximum magnitude of the force of static friction. 4 Increase the mass of the block system by placing a standard mass on the upper surface. Record the total mass of the block with the standard mass. Use the spring scale to determine the maximum magnitude of the force of static friction for this system (Figure 3.85). Figure 3.85 5 Repeat step 4 three more times, increasing the added mass each time until you have five different masses and five corresponding maximum magnitudes of static friction. 6 Calculate the magnitude of the weight corresponding to each mass system. Record the magnitude of the normal force. 7 (a) Graph the maximum magnitude of the force of static friction as a function of the magnitude of the normal force. (b) Draw the line of best fit and calculate the slope of the graph. Analysis 1. Describe the graph you drew in step 7. 2. As the magnitude of the normal force acting on
the mass system increased, what happened to the maximum magnitude of the force of static friction? 3. What is the relationship between the maximum magnitude of the force of static friction and the magnitude of the normal force? Write this as a proportionality statement. Does this relationship agree with your hypothesis? 4. On a level surface, how does the magnitude of the weight of an object affect the magnitude of the normal force and the maximum magnitude of the force of static friction? 5. Explain why adding a bag of sand to the trunk of a rear-wheel-drive car increases its traction. 6. Design and conduct an experiment to verify that contact surface area does not affect the maximum magnitude of the force of static friction for a sliding object. Identify the controlled, manipulated, and responding variables. Analyze your data and form conclusions. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 3 Forces can change velocity. 181 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 182 Project LINK How will the force of static friction acting on each vehicle in the Unit II Project on page 232 affect the stopping distance? How will the types of treads of the tires affect the force of static friction? coefficient of static friction: proportionality constant relating (Ffstatic)max and FN Coefficient of Static Friction In 3-11 Inquiry Lab, you found that the maximum magnitude of the force of static friction is directly proportional to the magnitude of the normal force. This proportionality can be written mathematically: (Ffstatic)max FN As an equation, the relationship is (Ffstatic)max sFN where s is a proportionality constant called the coefficient of static friction. Since the magnitude of the force of static friction can be anywhere from zero to some maximum value just before motion occurs, the general equation for the magnitude of the force of static friction must have an inequality sign. Ffstatic sFN for static friction Coefficient of Kinetic Friction Find out how the force of kinetic friction acting on an object is related to the normal force on that object by doing 3-12 Design a Lab. 3-12 Design a Lab 3-12 Design a Lab Relating Kinetic Friction and the Normal Force In this lab, you will investigate the relationship between the force of kinetic friction acting on an object and the normal force acting on that object. The Question What is the relationship between the magnitude of the force
of kinetic friction and the magnitude of the normal force acting on an object? Design and Conduct Your Investigation • State a hypothesis relating the magnitudes of F • Then use the set-up in Figure 3.85 on page 181 to design an experiment. List the materials you will use as well as a detailed procedure. You will need to place objects of different mass on the block for each trial. and F N. fkinetic • For each trial, measure the force that must be applied to keep the block system moving at constant velocity. Then calculate the magnitude of the normal force. • Plot a graph of Ffkinetic • Analyze your data and form conclusions. as a function of FN. How well did your results agree with your hypothesis? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. From 3-12 Design a Lab, just as with static friction, the magnitude of kinetic friction is directly proportional to the magnitude of the normal force. This proportionality can be written mathematically: Ffkinetic FN 182 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 183 As an equation, the relationship is kFN for kinetic friction Ffkinetic where k is a proportionality constant called the coefficient of kinetic friction. The force of kinetic friction has only one value, unlike the force of static friction which varies from zero to some maximum value. So the equation for the force of kinetic friction has an equal sign, not an inequality as does the equation for the force of static friction. Characteristics of Frictional Forces and Coefficients of Friction There are a few important points to keep in mind about the force of friction and the variables that affect its magnitude: • The equations for static friction and kinetic friction are not fundamental laws. Instead, they are approximations of experimental results. coefficient of kinetic friction: proportionality constant relating Ffkinetic and FN • The equations (Ffstatic)max kFN cannot be written sFN and Ffkinetic N are perpendicular f and F as vector equations because the vectors F to each other. s and • Both • For a given pair of surfaces, the coefficient of static friction is usually k are proportionality constants that have no units. greater than the coefficient of kinetic friction. • The coefficients of friction depend on the materials forming the contact surface, how smooth or rough a surface is, whether the surface is wet or dry, the
temperature of the two contact surfaces, and other factors. Table 3.4 lists coefficients of friction between pairs of materials. Table 3.4 Approximate Coefficients of Friction for Some Materials Material Coefficient of Static Friction s Coefficient of Kinetic Friction k Copper on copper Steel on dry steel Steel on greased steel Dry oak on dry oak Rubber tire on dry asphalt Rubber tire on wet asphalt Rubber tire on dry concrete Rubber tire on wet concrete Rubber tire on ice Curling stone on ice Teflon™ on Teflon™ Waxed hickory skis on dry snow Waxed hickory skis on wet snow Synovial fluid on joint 1.6 0.41 0.15 0.5 1.2 0.6 1.0 0.7 0.006 0.003 0.04 0.06 0.20 0.01 1.0 0.38 0.09 0.3 0.8 0.5 0.7 0.5 0.005 0.002 0.04 0.04 0.14 0.01 Chapter 3 Forces can change velocity. 183 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 184 How Friction Affects Motion Movable joints in the human body, such as elbows, knees, and hips, have membranes that produce a lubricating fluid called synovial fluid. Among other factors, the amount of synovial fluid and the smoothness of adjacent bone surfaces affect the coefficients of friction in synovial joints (Figure 3.86). The movement of synovial joints is very complicated because various biological processes are involved. In diseases such as arthritis, physical changes in joints and/or the presence of too much or too little synovial fluid affect the coefficients of friction. This, in turn, results in limited and painful movement. Figure 3.86 The amount of synovial fluid present depends on the need for a joint to move in a particular direction. info BIT Cars with wide tires experience no more friction than if the cars had narrow tires. Wider tires simply spread the weight of a vehicle over a greater surface area. This reduced pressure on the road reduces heating and tire wear. The effect of temperature on the coefficients of friction plays a role in drag racing. Drag racers often warm the tires on their cars by driving for a while. Tires that are warm stick to a racing track better than cooler tires. This increased coefficient of static friction increases traction and
improves the acceleration of the car. The amount of moisture on a road surface, the temperature of the road surface and tires, and the type of tire treads are some factors that determine if a vehicle will skid. For a given tire, the coefficients of static and kinetic friction are greater on a dry road than if the same road is wet. The result is that vehicles are less likely to skid on a dry road than on a wet road. Tire treads and road surfaces also affect the force of friction acting on a vehicle (Figure 3.87). A ribbed tire increases friction acting sideways which helps a driver steer better. A lug tread provides more traction than a ribbed tire. Slicks, the tires on drag racing cars, have no treads at all to increase the surface area of the tire in contact with the racing track to better dissipate heat. (a) (b) (c) Figure 3.87 Different types of tires: (a) a ribbed tire with chains on it for better traction on snowy and icy surfaces, (b) a lug tread, and (c) slicks on a racing car Example 3.17 demonstrates how to use the coefficients of friction in Table 3.4 on page 183 to calculate the mass of a sled. Since the sled is at rest, the snowy surface exerts a force of static friction on the sled. 184 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 185 Example 3.17 A sled with waxed hickory runners rests on a horizontal, dry snowy surface (Figure 3.88). Calculate the mass of the sled if the maximum force that can be applied to the sled before it starts moving is 46 N [forward]. Refer to Table 3.4 on page 183. up down forward backward Figure 3.88 Given F app s 46 N [forward] 0.06 from Table 3.4 (waxed hickory skis on dry snow) g 9.81 m/s2 [down] up Required mass of sled (m) backward down Analysis and Solution Draw a free-body diagram for the sled (Figure 3.89). Since the sled is not accelerating, F net in both the horizontal and vertical directions. Write equations to find the net force on the sled in both directions. 0 N FF neth Fneth horizontal direction F app Fapp 0 Fapp Fapp Fapp sFN FF fstatic Ffstatic Ffstatic (
Fapp sFN sFN) vertical direction F F netv N FN Fnetv 0 FN FN mg FF g Fg (mg) mg FN mg into the equation for Fapp. Fapp Substitute FN smg Fapp sg m 46 N m (0.06)9.81 s2 46 kgm s2 (0.06)9.81 m s2 8 101 kg Paraphrase The mass of the sled is 8 101 kg. Fapp Ffstatic 0 Fneth FN forward Ffstatic Fapp Fg Figure 3.89 Practice Problems 1. An applied force of 24 N [forward] causes a steel block to start moving across a horizontal, greased steel surface. Calculate the mass of the block. Refer to Table 3.4 on page 183. 2. Suppose the sled in Example 3.17 is resting on a horizontal, wet snowy surface. Would the sled move if the applied force is 125 N? Explain. Refer to Table 3.4 on page 183. Answers 1. 16 kg 2. no, F fstatic > F app In Example 3.18, a toboggan is initially at rest on a snowy hill. By knowing only the angle of the incline, it is possible to determine the coefficient of static friction for the toboggan on the hill. Chapter 3 Forces can change velocity. 185 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 186 Example 3.18 A 50-kg toboggan is on a snowy hill. If the hill forms an angle of at least 20.0 with the horizontal, the toboggan just begins to slide downhill (Figure 3.90). Calculate the coefficient of static friction for the toboggan on the snow. Practice Problems 1. Calculate the coefficient of static friction if the toboggan in Example 3.18 is 20 kg and the hill forms an angle of 30.0 with the horizontal. 2. An 80-kg skier on a slushy surface starts moving down a hill forming an angle of at least 25.0 with the horizontal. (a) Determine the coefficient of static friction. (b) Calculate the maximum force of static friction on the skier. Given m 50 kg 20.0 Required coefficient of static friction ( s) Analysis and Solution Draw a free-body diagram for the toboggan [Figure 3.91 (a)]. When the angle of the incline is just enough for the
toboggan to start moving, the surface of the incline is exerting the maximum magnitude of the force of static friction on the toboggan. u p p u h ill h ill o w n d o w n d θ 20.0° Figure 3.90 g 9.81 m/s2 u p p u h ill h ill o w n d o w n d Ffstatic FN Fg θ 20.0° θ 20.0° Fg Fg Answers 1. 0.58 2. (a) 0.47 (b) 3.3 102 N [uphill] info BIT The trigonometric function tan can be expressed in terms of sin and cos. in s tan s o c net 0 N in both the parallel and perpendicular Just before the toboggan begins to slide, F directions to the incline. Write equations to find the net force on the toboggan in both directions [Figure 3.91 (b)]. Figure 3.91 (a) direction net F F N Fnet FN 0 FN FN Fg F g Fg Fg direction g F net F F fstatic Fnet Fg Ffstatic 0 Fg Ffstatic Fg Ffstatic Fg mg sin Ffstatic Now, Fg mg cos So, FN (mg cos ) mg cos (mg sin ) mg sin mg sin mg cos into the last equation for the direction. sFN Figure 3.91 (b) Substitute FN Fg Ffstatic Fnet 0 s(mgcos ) mgsin s cos sin in s c s o tan tan 20.0 0.36 s Paraphrase The coefficient of static friction for the toboggan on the snow is 0.36. Note that angle of the hill. s does not depend on the mass of the toboggan, only on the 186 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 187 Kinetic Friction Applies to Skidding Tires When the tires of a vehicle lock or if the tires skid on a road surface, the tires no longer rotate. Instead, the tires slide along the road surface. At the area where the tire and the road are in contact, the road surface exerts a force of kinetic friction directed backward on the tire (Figure 3.92). e TECH Explore how the initial velocity of a skidding car and its mass affect the braking distance. Follow the eTech
links at www.pearsoned.ca/school/ physicssource. Ffkinetic Ffkinetic Figure 3.92 Diagram showing the force of kinetic friction acting on the tires of a skidding car Safety features on vehicles such as anti-lock braking systems are designed to prevent the wheels of a vehicle from locking when a driver steps on the brakes. If the wheels lock, the tires no longer rotate on the road surface and the vehicle ends up skidding. As long as the wheels continue to turn, the road surface exerts a force of static friction on the tires. Anti-lock braking systems maximize the force of static friction acting on the tires, allowing the driver of a vehicle to come to a more controlled stop. In Example 3.19, a lift truck is skidding on a concrete surface. Since the wheels are not rotating, the concrete surface is exerting a force of kinetic friction on the tires. e WEB Research how anti-lock braking systems work, and identify the strengths and weaknesses. Interview a car salesperson and/or an owner. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/school/ physicssource. Example 3.19 A 1640-kg lift truck with rubber tires is skidding on wet concrete with all four wheels locked (Figure 3.93). Calculate the acceleration of the truck. Refer to Table 3.4 on page 183. backward up down forward Given m 1640 kg k g 9.81 m/s2 [down] 0.5 from Table 3.4 (rubber on wet concrete) Required acceleration of lift truck (a) Analysis and Solution Draw a free-body diagram for the lift truck (Figure 3.94). Since the lift truck is accelerating forward, F net direction, but F net vertical direction. Write equations to find the net force on the lift truck in both directions. 0 N in the horizontal 0 N in the Figure 3.93 up backward FN down forward Ffkinetic Fg FN Fg 0 Fnetv Figure 3.94 Chapter 3 Forces can change velocity. 187 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 188 Practice Problems 1. An applied force of 450 N [forward] is needed to drag a 1000-kg crate at constant speed across a horizontal, rough floor. Calculate the coefficient of kinetic friction for the crate on the floor. 2. Calculate
the force of kinetic friction if the truck in Example 3.19 is skidding downhill at constant speed on a hill forming an angle of 15.0 with the horizontal. Answers 1. 4.59 102 2. 4.16 103 N [uphill] horizontal direction F F fkinetic neth Ffkinetic Fneth ma Ffkinetic kFN vertical direction FF F netv N FN Fnetv 0 FN FN mg F g Fg (mg) mg FN mg into the equation for Ffkinetic. Substitute FN kmg ma a kg (0.5)9.81 5 m/s2 m s2 The negative value for a indicates that the direction of a backward. is a 5 m/s2 [backward] Paraphrase The acceleration of the truck is 5 m/s2 [backward]. Example 3.20 involves a snowmobile accelerating uphill while towing a sled. Since the motion of the sled is uphill, it is convenient to choose uphill to be positive. Example 3.20 A person wants to drag a 40-kg sled with a snowmobile up a snowy hill forming an angle of 25.0 (Figure 3.95). The coefficient of kinetic friction for the sled on the snow is 0.04. Calculate the force of the snowmobile on the sled if the sled accelerates at 2.5 m/s2 [uphill]. Figure 3.95 p u p u h ill h ill o w n d o w n d θ 25.0° Given m 40 kg g 9.81 m/s2 25.0 a 2.5 m/s2 [uphill] k 0.04 Required app) applied force on sled (F 188 Unit II Dynamics 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 189 Analysis and Solution Draw a free-body diagram for the sled (Figure 3.96). FN Ffkinetic Figure 3.96 p u p u h ill h ill o w n d o w n d Fg θ 25.0° Fapp θ 25.0° Fg Fapp Fg Ffkinetic Fg Fnet Since the sled is accelerating uphill, 0 N parallel to the incline, but F 0 N perpendicular to the incline. F net net Write equations to find the net force on the sled in both directions. direction net FF F N Fnet FN 0 FN FN direction net FF F
app Fnet Fapp ma Fapp Fapp F g Fg Fg Fg FF g F fkinetic Fg Ffkinetic Fg Ffkinetic ma Fg Ffkinetic Now, Fg mg cos So, FN (mg cos ) mg cos Also, Fg mg sin and kFN Ffkinetic Fapp ma (mg sin ) kFN) ( ma mg sin kFN Substitute FN mg cos into the equation for Fapp. Fapp ma mg sin ma mg(sin kmg cos kcos ) Practice Problems 1. A roofer is shingling a roof that rises 1.0 m vertically for every 2.0 m horizontally. The roofer is pulling one bundle of shingles (A) with a rope up the roof. Another rope connects bundle A to bundle B farther down the roof (Figure 3.97). mA 15 kg mB 15 kg 10 m 20 m Figure 3.97 Each of the two bundles of shingles has a mass of 15 kg. The coefficient of kinetic friction for the bundles on plywood sheeting is 0.50. (a) What force must the roofer exert up the roof to drag the bundles at constant speed? (b) Calculate the force exerted by bundle A on bundle B. (c) What total force would the roofer have to exert to accelerate both bundles at 2.0 m/s2 [up roof]? Answers 1. (a) 2.6 102 N [up roof] (b) 1.3 102 N [up roof] (c) 3.2 102 N [up roof] (40 kg)(2.5 m/s2) (40 kg)(9.81 m/s2) [(sin 25.0) (0.04)(cos 25.0)] 3 102 N app is uphill. The positive value for Fapp indicates that the direction of F F app 3 102 N [uphill] Paraphrase The snowmobile must apply a force of 3 102 N [uphill]. Chapter 3 Forces can change velocity. 189 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 190 3.5 Check and Reflect 3.5 Check and Reflect Knowledge 1. In your own words, define friction. 2. What are some situations where friction is so small that it could be neglected? 3. Distinguish between static friction and kinetic friction. Applications 4. A pair of sk
is weigh 15 N [down]. Calculate the difference in the maximum force of static friction for the skis on a wet and dry snowy, horizontal surface. Refer to Table 3.4 on page 183. 5. A force of 31 N [forward] is needed to start an 8.0-kg steel slider moving along a horizontal steel rail. What is the coefficient of static friction? 6. A biker and his motorcycle have a weight of 2350 N [down]. Calculate the force of kinetic friction for the rubber tires and dry concrete if the motorcycle skids. Refer to Table 3.4 on page 183. 7. A 15-kg box is resting on a hill forming an angle with the horizontal. The coefficient of static friction for the box on the surface is 0.45. Calculate the maximum angle of the incline just before the box starts to move. 8. The coefficient of static friction for a wheelchair with its brakes engaged on a conveyor-type ramp is 0.10. The average mass of a person including the wheelchair is 85 kg. Determine if a ramp of 8.0° with the horizontal will prevent motion. 9. A truck loaded with a crate of mass m is at rest on an incline forming an angle of 10.0 with the horizontal. The coefficient of static friction for the crate on the truck bed is 0.30. Find the maximum possible acceleration uphill for the truck before the crate begins to slip backward. 190 Unit II Dynamics 10. A loaded dogsled has a mass of 400 kg and is being pulled across a horizontal, packed snow surface at a velocity of 4.0 m/s [N]. Suddenly, the harness separates from the sled. If the coefficient of kinetic friction for the sled on the snow is 0.0500, how far will the sled coast before stopping? Extensions 11. A warehouse employee applies a force of 120 N [12.0] to accelerate a 35-kg wooden crate from rest across a wooden floor. The coefficient of kinetic friction for the crate on the floor is 0.30. How much time elapses from the time the employee starts to move the crate until it is moving at 1.2 m/s [0°]? 12. Make a Venn diagram to summarize the similarities and differences between static and kinetic friction. See Student References 4: Using Graphic Organizers on page 869 for an example. 13. Research how the type of tread on a tire affects the coefficients of static friction and kinetic friction given the
same road surface. Find out what hydroplaning is and how tires are designed to minimize this problem. Write a brief report of your findings, including diagrams where appropriate. Begin your search at www.pearsoned.ca/school/physicssource. 14. Design an experiment to determine the coefficients of static and kinetic friction for a curling stone on an icy surface. Perform the experiment at a local arena or club. Ask the icemaker to change the temperature of the ice, and repeat the experiment to determine if there is a difference in your values. Write a brief report of your findings. e TEST To check your understanding of friction and inclines, follow the eTest links at www.pearsoned.ca/school/physicssource. 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 191 CHAPTER 3 SUMMARY Key Terms and Concepts dynamics force free-body diagram normal force net force inertia inertial mass action force Key Equations Newton’s first law: F net Newton’s second law: F net F A on B Newton’s third law: 0 when v 0 ma F B on A reaction force friction static friction kinetic friction coefficient of static friction coefficient of kinetic friction Static friction: Ffstatic sFN Kinetic friction: Ffkinetic kFN Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. Dynamics involves forces free-body diagrams vector sums Newton’s first law Newton’s second law Newton’s third law are are measured in show all the forces is adding the involves inertia states that states that states that sliding friction two kinds kinetic equations so they have acting on determines the when magnitude and direction on Figure 3.98 Chapter 3 Forces can change velocity. 191 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 192 CHAPTER 3 REVIEW Knowledge 1. (3.1, 3.3) Two people, A and B, are pushing a stalled 2000-kg truck along a level road. Person A exerts a force of 300 N [E]. Person B exerts a force of 350 N [E]. The magnitude of the force of friction on the truck is 550 N. Calculate the acceleration of the truck. 2. (3.2) Use a free-body diagram and Newton’s first law to
explain the motion of (a) a figure skater during a glide, and (b) a hockey puck during a cross-ice pass. Assume ice is frictionless. 3. (3.4) A transport truck pulls a trailer with a force of 1850 N [E]. What force does the trailer exert on the transport truck? 4. (3.5) An inexperienced driver, stuck in snow, tends to spin the car tires to increase the force of friction exerted by the snow on the tires. What advice would you give to the driver? Why? Applications 5. A device used to treat a leg injury is shown below. The pulley is attached to the foot, and the weight of the 3.0-kg object provides a tension force to each side of the pulley. The pulley is at rest to the pulley, because the foot applies a force F T2 in T1 and FF which is balanced by the forces FF the rope. The weight of the leg and foot is supported by the pillow. (a) Using a free-body diagram for the pulley, determine the force FF. (b) What will happen to the magnitude of F T2 decreases? Why? T1 and FF the angle between FF if support leg harness foot pulley F T1 F 45.0º 45.0º F T2 pillow 3.0 kg 192 Unit II Dynamics 6. Refer to Example 3.6 Practice Problem 1 on page 150. In a second practice run, the initial acceleration of the bobsled, pilot, and brakeman is 4.4 m/s2 [forward]. Rider A exerts an average force of magnitude 1200 N on the bobsled, and the force of friction decreases to 400 N. What average force does rider B exert? 7. During its ascent, a loaded jet of mass 4.0 105 kg is flying at constant velocity 20.0 above the horizontal. The engines of the plane provide a of 4.60 106 N [forward] to provide the thrust T [perpendicular to wings]. The air lift force L opposes the motion of the jet. resistance R Determine the magnitudes of L and R. L T 20.0° Fg R 8. Suppose the force of kinetic friction on a sliding block of mass m is 2.5 N [backward]. What is the force of kinetic friction on the block if another block of mass 2m is placed on its upper surface? 9. A 1385-kg pickup
truck hitched to a 453-kg trailer accelerates along a level road from a stoplight at 0.75 m/s2 [forward]. Ignore friction and air resistance. Calculate (a) the tension in the hitch, (b) the force of friction exerted by the road on the pickup truck to propel it forward, and (c) the force the trailer exerts on the pickup truck. 10. Two curlers, A and B, have masses of 50 kg and 80 kg respectively. Both players are standing on a carpet with shoes having Teflon™ sliders. The carpet exerts a force of friction of 24.5 N [E] on player A and a force of friction of 39.2 N [W] on player B. Player A pushes player B with a force of 60 N [E]. (a) Calculate the net force acting on each player. (b) Calculate the acceleration of each player. 03-Phys20-Chap03.qxd 7/24/08 10:38 AM Page 193 11. A force of 15 N [S] moves a case of soft drinks weighing 40 N [down] across a level counter at constant velocity. Calculate the coefficient of kinetic friction for the case on the counter. 12. A 1450-kg car is towing a trailer of mass 454 kg. The force of air resistance on both vehicles is 7471 N [backward]. If the acceleration of both vehicles is 0.225 m/s2, what is the coefficient of static friction for the wheels on the ground? 13. Two bags of potatoes, m1 60 kg and m2 40 kg, are connected by a light rope that passes over a light, frictionless pulley. The pulley is suspended from the ceiling using a light spring scale. (a) What is the reading on the scale if the pulley is prevented from turning? (b) Draw a free-body diagram for each bag when the pulley is released. (i) Calculate the acceleration of the system. (ii) Calculate the tension in the rope. (c) What is the reading on the scale when the bags are accelerating? (d) Explain the difference between your answers in parts (a) and (c). spring scale 10 pulley m1 60 kg POTATOES POTATOES m2 40 kg POTATOES m1 60 kg 14. A drag racing car initially at rest can reach a
speed of 320 km/h in 6.50 s. The wheels of the car can exert an average horizontal force of 1.52 104 N [backward] on the pavement. If the force of air resistance on the car is 5.2 103 N [backward], what is the mass of the car? 15. A tractor and tow truck have rubber tires on wet concrete. The tow truck drags the tractor at constant velocity while its brakes are locked. If the tow truck exerts a horizontal force of 1.0 104 N on the tractor, determine the mass of the tractor. Refer to Table 3.4 on page 183. 16. Create a problem involving an object of mass m on an incline of angle. Write a complete solution, including an explanation of how to resolve the gravitational force vector into components. 17. The table below shows some coefficients of static and kinetic friction ( tires in contact with various road surfaces. s and k) for rubber Coefficient Dry Concrete Wet Concrete Dry Asphalt Wet Asphalt s k 1.0 0.7 0.7 0.5 1.2 0.6 0.6 0.5 (a) Which road surface exerts more static friction on a rubber tire, dry concrete or dry asphalt? Explain. (b) On which surface does a car slide more easily, on wet concrete or on wet asphalt? Why? (c) On which surface will a moving car begin to slide more easily, on dry concrete or on dry asphalt? Why? (d) On which surface will a car with locked brakes slide a shorter distance, on dry concrete or on dry asphalt? Explain. Extensions 18. An 80-kg baseball player slides onto third base. The coefficient of kinetic friction for the player on the ground is 0.70. His speed at the start of the slide is 8.23 m/s. (a) Calculate his acceleration during the slide. (b) For how long does he slide until he stops? (c) Show that the time it takes the player to come to a stop is given by the equation t vi kg. Consolidate Your Understanding 19. Write a paragraph explaining the similarities and differences among Newton’s three laws. Include an example that involves all three laws and explain how each law applies. Use the example to teach the laws to a student who has not studied dynamics. 20. Write a paragraph describing the differences between static and kinetic friction, and between the coefficients of static and kinetic friction. Include an example with
a free-body diagram for each type of friction. Think About It Review your answers to the Think About It questions on page 125. How would you answer each question now? e TEST To check your understanding of forces and Newton’s laws of motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 3 Forces can change velocity. 193 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 194 Gravity extends throughout the universe. Skydiving, hang-gliding, bungee jumping, and hot-air ballooning are just a few activities that take advantage of gravitational forces for a thrill (Figure 4.1). Gravitational force attracts all objects in the universe. It holds you to Earth, and Earth in its orbit around the Sun. In 1665, Isaac Newton began his study of gravity when he attempted to understand why the Moon orbits Earth. His theories led him to an understanding of the motion of planets and their moons in the solar system. Several centuries later, these theories led to the launch of satellites and the success of various space missions such as Mariner and Voyager. Gravity is one of the four basic forces of nature, called fundamental forces, that physicists think underlie all interactions in the universe. These forces are the gravitational force, the electromagnetic force, the weak nuclear force, and the strong nuclear force. In this chapter, you will investigate how gravity affects the motion of objects on Earth and on other planets, and how it affects the motion of satellites orbiting Earth. Figure 4.1 Understanding how forces and gravity affect motion determines how successful the design of a hot-air balloon will be and the best way to navigate it. C H A P T E R 4 Key Concepts In this chapter, you will learn about: gravitational force Newton’s law of universal gravitation gravitational field Learning Outcomes When you have completed this chapter, you will be able to: Knowledge identify gravity as a fundamental force in nature describe Newton’s law of universal gravitation explain the Cavendish experiment define and apply the concept of a gravitational field compare gravitational field strength and acceleration due to gravity predict the weight of objects on different planets Science, Technology, and Society explain that concepts, models, and theories help interpret observations and make predictions 194 Unit II 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 195 4-1 QuickLab 4-1 QuickLab Falling Coins Problem Suppose you
drop two coins of different shapes and sizes from the same height. How do the rates of both coins falling compare? Materials variety of coins (penny, nickel, dime, quarter, loonie, and toonie) ruler Styrofoam™ disk (size of a loonie) Procedure 1 Choose any two different coins and place them at the edge of a table above an uncarpeted floor. 2 Using a ruler, push the coins off the table so they leave at the same time (Figure 4.2). 3 Listen carefully for the sounds of the coins as they hit the floor. 4 Repeat this activity with different combinations of two coins, including the loonie with the Styrofoam™ disk. Record your observations. Figure 4.2 Questions 1. When the coins landed, how many sounds did you hear? Did all combinations of two coins give the same result? Explain. 2. If all the coins fall at the same rate, how many sounds would you expect to hear when they land? 3. How would the results compare if two coins were released at the same time from a greater height, such as 10 m? 4. How did the average acceleration of the loonie differ from that of the Styrofoam™ disk? Explain why. Think About It 1. (a) What factors affect the weight of an astronaut during a rocket flight? (b) How does the astronaut’s weight change? 2. What would be the motion of Earth if the Sun’s gravity were zero? Assume that no other celestial bodies affect Earth. Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 4 Gravity extends throughout the universe. 195 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 196 info BIT Gravitational force is an attraction force only. There is no such thing as a repulsive gravitational force. gravitational force: attractive force between any two objects due to their masses 4.1 Gravitational Forces due to Earth One of Newton’s great achievements was to identify the force that causes objects to fall near Earth’s surface as the same force that causes the Moon to orbit Earth. He called this force “gravity,” and he reasoned that this force is present throughout the universe. g, is the force that attracts any two objects Gravitational force, F
together. Although this force is the weakest fundamental force, you can feel its effect when you interact with an object of very large mass such as Earth. When you slide down a waterslide, you can feel the gravitational force exerted by Earth pulling you downward toward the bottom of the slide (Figure 4.3). But if you want to feel the gravitational force exerted by the person sitting next to you, you will not be able to sense anything because the magnitude of the force is so small. Figure 4.3 The attractive force between Earth and you is far greater than that between you and another person coming down a waterslide. 196 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 197 Gravitational force is a force that always exists in pairs. This is another example of Newton’s third law. If Earth exerts a gravitational force of magnitude 700 N on you, then you exert a gravitational force of magnitude 700 N on Earth. Earth attracts you and you attract Earth. The force you exert on Earth has a negligible effect because Earth’s mass (5.97 1024 kg) is huge in comparison to yours. However, the gravitational force that Earth exerts on you causes a noticeable acceleration because of your relatively small mass. Concept Check Which diagram best represents the gravitational force acting on you and on Earth (Figure 4.4)? Explain your reasoning. (a) (b) (c) (d) Figure 4.4 The Concept of Weight In a vacuum, all objects near Earth’s surface will fall with the same acceleration, no matter what the objects consist of or what their masses are. The only force acting on a falling object in a vacuum is the gravitational force exerted by Earth on the object (Figure 4.5). Suppose you analyze this situation using a free-body diagram and Newton’s second law. Fg Figure 4.5 Free-body diagram for a falling object in a vacuum Chapter 4 Gravity extends throughout the universe. 197 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 198 Fs 2 6 10 14 18 0 4 8 12 16 20 The equation for the net force acting on the falling object is F net F g ma F g Since the object is accelerating due to gravity, a g. So the equation for the net force becomes mg F g or F g mg The equation F mg is valid in general, because the gravitational force acting on an object
is the same, whether or not the object is at rest or is moving. This equation relates the gravitational force acting on an object, the so-called weight of the object, to its mass. g Fg Figure 4.6 Diagram showing the forces acting on an object that is suspended from a spring scale weight: gravitational force exerted on an object by a celestial body 4-2 QuickLab 4-2 QuickLab One way to measure the magnitude of the weight of an object directly involves using a spring scale (Figure 4.6). When the object stops moving at the end of the spring, Earth exerts a downward gravitational force on the object while the spring exerts an upward elastic force of equal magnitude on the object. Find out what the relationship is between mass and gravitational force in the vicinity of your school by doing 4-2 QuickLab. Relating Mass and Weight Problem What is the relationship between the mass of an object and the local value of the gravitational force exerted on that object? Materials set of standard masses with hooks spring scale (010 N) graph paper Procedure 1 Design a procedure to determine the gravitational force acting on a set of standard masses (Figure 4.7). 2 Use a table to record the magnitude of the gravitational force and mass. Add another column in the table for the ratio of Fg to m. 3 Calculate the ratio of Fg to m for each standard mass. Calculate the average ratio for all masses. Include units 10 Figure 4.7 198 Unit II Dynamics 4 Plot a graph of Fg vs. m. Draw a line of best fit through the data points. Calculate the slope of the graph. Questions 1. What does the ratio of Fg to m represent? How constant is this value? 2. Describe the graph of Fg vs. m. How does the slope compare to the average ratio calculated in step 3? 3. Write an equation relating Fg and m. Use the symbol g for the proportionality constant. 4. (a) The Moon exerts a gravitational force that is 1 that exerted by Earth. If you did this about 6 activity on the Moon using an appropriate spring scale and the same standard masses, would • the graph be a straight line? • the slope be the same as before? • the line go through the origin? • the proportionality constant g be the same as before? (b) Why would a 05 N spring scale be more ideal to use on the Moon, rather than a 010 N spring scale? 04
-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 199 Gravitational Mass mg is determined by finding the ratio of The mass in the equation F g the gravitational force acting on an object to the acceleration due to g and g are in the same direction, the scalar form of the gravity. Since F equation may be used: Fg mg m Fg g A practical way to measure this mass involves using a balance. In Figure 4.8, an object of unknown mass (A) is placed on one pan and standard masses (B) are added to the other pan until both pans balance. This method involves comparing the weights of two objects: one unknown and the other known. Mass measured using the concept of weight is called gravitational mass. gravitational force exerted on standard masses gravitational force exerted on an unknown mass gravitational mass: mass measurement based on comparing the known weight of one object to the unknown weight of another object Figure 4.8 When both pans are balanced, the gravitational force acting on the unknown mass is equal to the gravitational force acting on the standard masses. If the balance in Figure 4.8 were moved to the Moon, the process of determining the gravitational mass of object A would be the same. However, the weight of A and B would be different from that at Earth’s surface because the acceleration due to gravity at the Moon’s surface, gMoon, is 1.62 m/s2 compared to 9.81 m/s2, the average value at Earth’s surface. e SIM Explore how mass and weight are measured. Follow the eSim links at www.pearsoned.ca/school/ physicssource. FgA FgB mAgMoon mBgMoon mA mB But since both objects A and B experience the same value of gMoon, mB. So the gravitational mass of an object is the same whether the mA object is on Earth, the Moon, or anywhere else in the universe. Both gravitational mass and inertial mass are properties of an object that do not depend on the location of the object. Is Inertial Mass the Same as Gravitational Mass? You can determine the mass of an object by using either the concept of inertia or weight. Experiments since Newton’s day have shown that for any object, the numerical value of its inertial mass is equal to its gravitational mass. Later, Albert Einstein (1879–1955) showed that inertial mass is actually
equivalent to gravitational mass. So it does not matter whether you determine the mass of an object using inertia or weight, because the numerical value of the mass is the same. Chapter 4 Gravity extends throughout the universe. 199 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 200 action-at-a-distance force: force that acts even if the objects involved are not touching field: three-dimensional region of influence gravitational field: region of influence surrounding any object that has mass Describing Gravitational Force as a Field Gravitational force is an example of a force that acts on objects whether or not they actually touch each other, even if the objects are in a vacuum. These forces are referred to as action-at-a-distance forces. In the 1800s, physicists introduced the concept of a field to explain action-at-a-distance forces. You encountered some fields in previous science courses when you worked with magnets. Imagine you are moving the north pole of a magnet close to the north pole of a fixed magnet. As you move the magnet closer to the fixed magnet, you can feel an increasing resistance. Somehow, the fixed magnet has created a region of influence in the space surrounding it. Physicists refer to a three-dimensional region where there is some type of an influence, whether it is an attraction or a repulsion, on a suitable object as a field. Since every object exerts a gravitational force in three dimensions, it influences the space around it (Figure 4.9). This region of influence is a gravitational field, and it is through this region that two objects interact. gravitational field of Earth gravitational field of the Moon Figure 4.9 This figure shows a two-dimensional representation of Earth’s and the Moon’s gravitational field. A gravitational field is three-dimensional and is directed toward the centre of the object. 200 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 201 To determine the magnitude and direction of a gravitational field created by an object, you could use a test mass mtest. At different locations around the object, this test mass will experience a gravitational force that has a certain magnitude and direction. The direction of the gravitational force will be directed toward the centre of the object. Gravitational field strength is defined as the gravitational force per F unit mass, g g. If you release the test mass, it will accelerate toward m te st the object with an acceleration equal
to g. Figure 4.10 shows how the magnitude of Earth’s gravitational field strength changes as a test mass is moved farther away from Earth’s centre. The farther the test mass is moved, the more significant is the decrease in g. In fact, the graph in Figure 4.10 shows an inverse square relationship: g 1 r 2 Magnitude of Gravitational Field Strength vs. Distance from Earth’s Centre gravitational field strength: gravitational force per unit mass at a specific location 10.0 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 Distance from Earth’s centre r (Earth radii) Figure 4.10 The magnitude of the gravitational field strength as a function of distance from Earth’s centre Since force is measured in newtons and mass in kilograms, the units of gravitational field strength are newtons per kilogram, or N/kg. The ratio you determined in 4-2 QuickLab was the gravitational field strength at the vicinity of your school. Concept Check What happens to the magnitude of the gravitational field strength if (a) r decreases by a factor of four? (b) r increases by a factor of two? (c) mtest doubles? (d) mtest is halved? Chapter 4 Gravity extends throughout the universe. 201 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 202 4.1 Check and Reflect 4.1 Check and Reflect Knowledge 1. Distinguish between mass and weight. Explain using an example and a diagram. 10. How could you distinguish between a 5.0-kg medicine ball and a basketball in outer space without looking at both objects? 2. Distinguish between inertial mass and Extensions gravitational mass. 3. In your own words, define gravitational acceleration and gravitational field strength. State the units and symbol for each quantity. 11. Visit a local fitness gymnasium. Find out how athletes use gravitational and elastic forces to improve their fitness. Is friction a help or a hindrance? Write a brief report of your findings. 4. In your own words, explain the concept of a gravitational field. Include an example of how a gravitational field affects another object. 12. List some occupations that might require a knowledge of gravitational field strength. Briefly explain how gravitational field strength applies to these occupations. 5. Why do physicists use the concept of 13. Complete the gathering grid below to summarize the similarities
and differences among gravitational mass, inertial mass, and gravitational force. Gravitational Mass Inertial Mass Gravitational Force Definition SI Unit Measuring Instrument(s) How the Quantity Is Measured Factors It Depends On Variability with Location e TEST To check your understanding of gravitational force, weight, mass, and gravitational field strength, follow the eTest links at www.pearsoned.ca/school/physicssource. a field to describe gravity? 6. In a vacuum, a feather and a bowling ball are released from rest at the same time from the same height. Compare the time it takes for each object to fall. Explain your answer. Applications 7. The Moon exerts a gravitational force that is about 1 6 that exerted by Earth. Explain why the mass of an object measured on the Moon using a balance is the same as if the object were on Earth’s surface. 8. Describe a situation where measuring the inertial mass of an object is easier than measuring its gravitational mass. 9. The table below shows the magnitude of the gravitational force on objects of different mass in Banff, Alberta. Mass (kg) 0 1.50 3.00 4.50 6.00 7.50 10.0 Magnitude of 0 14.7 29.4 44.1 58.9 73.6 98.1 Gravitational Force (N) (a) Graph the data. (b) Calculate the slope of the line. (c) What does the slope represent? 202 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 203 info BIT Gravity is the dominant force throughout the universe. This force is attractive and has an infinite range. 4.2 Newton’s Law of Universal Gravitation Gravity affects all masses in the universe. No matter where you are on Earth or in outer space, you exert a gravitational force on an object and an object exerts a gravitational force, of equal magnitude but opposite direction, on you. Because gravitational force acts over any distance, the range of its effect is infinite. Near Earth’s surface, the magnitude of the gravitational force exerted mBg. by Earth (object A) on object B is given by the equation FA on B But object B also exerts a gravitational force of equal magnitude on mAg. Newton hypothesized that, given two objects A Earth, FB on A and B, the magnitude of the gravitational force exerted by one object on the other is directly
proportional to the product of both masses: Fg mAmB Figure 4.11 shows the magnitude of the gravitational force acting on an object at Earth’s surface (rEarth), one Earth radius above Earth (2rEarth), and two Earth radii above Earth (3rEarth). If the separation distance from Earth’s centre to the centre of the object doubles, Fg decreases 1 of its original value. If the separation distance from Earth’s centre to to 4 1 of its original value. the centre of the object triples, Fg decreases to 9 1, Fg is inversely proportional to the square of the separation So, just as g 2 r distance (Figure 4.12): Fg 1 2 r Magnitude of Gravitational Force vs. Distance from Earth’s Centre Earth Fg rEarth 1 4 Fg 1 9 Fg 3rEarth 2rEarth 10.0 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 Distance from Earth’s centre r (Earth radii) Figure 4.11 The magnitude of the gravitational force acting on an object some distance from Earth varies inversely with the square of the separation distance. Figure 4.12 The magnitude of the gravitational force acting on a 1.00-kg object as a function of distance from Earth’s centre Chapter 4 Gravity extends throughout the universe. 203 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 204 If you combine both proportionalities into one statement, you get mAmB r 2 GmAmB r 2 ship is Newton’s law of universal gravitation. (Figure 4.13). This mathematical relation- or Fg Fg Any two objects, A and B, in the universe exert gravitational forces of equal magnitude but opposite direction on each other. The forces are directed along the line joining the centres of both objects. The magnitude of the gravitational force is given by Fg GmAmB r 2, where mA and mB are the masses of the two objects, r is the separation distance between the centres of both objects, and G is a constant called the universal gravitational constant. First mass Separation distance Second mass Magnitude of gravitational force Fg 2 Fg 6 Fg 4 Fg 1 4 Fg m 3 m 6 4 Fg Figure 4.13 The magnitude of the gravitational force is directly proportional to the
product of the two masses, and inversely proportional to the square of the separation distance. Experiments have shown that the magnitude of the gravitational force acting on any pair of objects does not depend on the medium in which the objects are located (Figure 4.14). In other words, given two fish, the gravitational force acting on either fish will have the same magnitude if both fish are underwater or in midair. (a) (b) 1 m 1 m Figure 4.14 The magnitude of the gravitational force acting on either fish is the same whether both fish are (a) underwater or (b) above water. e SIM Explore the relationship among mA, mB, r, and Fg in Newton’s law of gravitation. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 204 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 205 Concept Check Two identical stationary baseballs are separated by a distance r. What will happen to the magnitude of the gravitational force acting on either ball if (a) the mass of each ball doubles? (b) r is halved? (c) the mass of each ball is halved and r doubles? e WEB Edmund Halley, an associate of Newton, paid for the publication of some of Newton’s famous work. Find out about Edmund Halley and his contributions to Newton’s work. Begin your search at www.pearsoned.ca/school/ physicssource. Determining the Value of the Universal Gravitational Constant Although Newton found a mathematical relationship for gravitational force, he was unable to determine the value of G. In 1798, scientist Henry Cavendish (1731–1810) confirmed experimentally that Newton’s law of gravitation is valid, and determined the density of Earth. Cavendish’s experimental set-up was later used to determine the value of G. The magnitude of the gravitational force acting on most pairs of objects is very weak and the magnitude decreases significantly as the separation distance between the objects increases. However, if you use two light spheres (each of mass m) and two heavy spheres (each of mass M) that are very close to each other, it is possible to determine the magnitude of the gravitational force exerted by M on m. The trick is to use a device that can accurately measure the very small gravitational force. A modern torsion balance is a device
that uses a sensitive fibre and a beam of light to measure very minute forces due to gravity, magnetic fields, or electric charges. Cavendish used a modified torsion balance invented by John Michell (1724–1793) to verify Newton’s law of gravitation. A modern torsion balance consists of a small, light, rigid rod with two identical, light, spheres (m) attached to each end (Figure 4.15). The rod is suspended horizontally by a thin fibre connected to the centre of the rod. A mirror is also attached to the fibre and rod so that when the rod turns, the mirror also turns by the same amount. The entire assembly is supported in an airtight chamber. The torsion balance initially experiences no net force and the spheres m are stationary. fibre support torsion fibre mirror m zero deflection screen laser M m rod M Figure 4.15 A modern torsion balance uses a laser beam to measure the amount of twist in the fibre. The most accurate value of G has been determined using such a device. torsion balance: device used to measure very small forces info BIT Charles de Coulomb invented the original torsion balance in 1777 to measure small magnetic forces and forces in fluids. However, John Michell independently invented the same type of device in 1784. Chapter 4 Gravity extends throughout the universe. 205 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 206 m Fg M on m Fg m on M M fibre direction of motion M Fg m on M Fg M on m m Figure 4.16 This figure shows the top view of spheres m and M from Figure 4.15. The original position of spheres m is shown with dashed lines. The gravitational force exerted by M on m causes m to rotate toward M. The greater the magnitude of the gravitational force, the greater the angle of rotation. When two identical, heavy, spheres (M) are moved close to spheres m, the gravitational force exerted by M on m causes m to rotate horizontally toward M. This rotation causes the fibre to twist slightly (Figure 4.16). As the fibre twists, the mirror attached to both the fibre and the rod turns through an angle in the horizontal plane. A beam of light reflected from the mirror becomes deflected as spheres m rotate. The amount of deflection is an indication of how much the spheres rotate. The greater the magnitude of the gravitational force, the more the fibre
twists, and the greater the angle of rotation. By measuring the amount of deflection, the gravitational force exerted by M on m can be determined. Spheres M are then moved to a symmetrical position on the opposite side of m, and the procedure is repeated. Since the separation distance between m and M, the values of m and M, and the gravitational force can all be measured, it is possible to calculate G using Newton’s law of gravitation. Fg GmM r 2 Fgr 2 mM G G Fgr 2 mM info BIT Scientists use Newton’s law of universal gravitation to calculate the masses of planets and stars. The current accepted value of G to three significant digits is 6.67 1011 Nm2/kg2. In Example 4.1, Newton’s law of gravitation is used to show that a person weighs slightly less at the top of the mountain than at its base. Example 4.1 Mount Logan in the Yukon is 5959 m above sea level, and is the highest peak in Canada. Earth’s mass is 5.97 1024 kg and Earth’s equatorial radius is 6.38 106 m. What would be the difference in the magnitude of the weight of a 55.0-kg person at the top of the mountain as compared to at its base (Figure 4.17)? Assume that Earth’s equatorial radius is equal to the distance from Earth’s centre to sea level. Given mp 55.0 kg h 5959 m mEarth rEarth 5.97 1024 kg 6.38 106 m Required difference in magnitude of weight (Fg) Mt. Logan 5959 m Figure 4.17 206 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 207 Practice Problems 1. Two people, A and B, are sitting on a bench 0.60 m apart. Person A has a mass of 55 kg and person B a mass of 80 kg. Calculate the magnitude of the gravitational force exerted by B on A. 2. The mass of the Titanic was 4.6 107 kg. Suppose the magnitude of the gravitational force exerted by the Titanic on the fatal iceberg was 61 N when the separation distance was 100 m. What was the mass of the iceberg? Answers 1. 8.2 107 N 2. 2.0 108 kg Analysis and Solution Assume that the separation distance between the person at the base of the mountain and
Earth is equal to Earth’s equatorial radius. Base of mountain: rB rEarth 6.38 106 m Top of mountain: rT 6.38 106 m 5959 m The person’s weight is equal to the gravitational force exerted by Earth on the person, and is directed toward Earth’s centre both at the base and at the top of the mountain. Calculate Fg at the base of the mountain using Newton’s law of gravitation. (Fg)B GmpmEarth (rB)2 2 m N (55.0 kg)(5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 538.049 N Calculate Fg at the top of the mountain using Newton’s law of gravitation. (Fg)T GmpmEarth (rT)2 2 m N (55.0 kg)(5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m 5959 m)2 537.045 N The difference in the magnitude of the weight is equal to the difference in magnitude of both gravitational forces. Fg (Fg)T (Fg)B 538.049 N 537.045 N 1.00 N Paraphrase The difference in the magnitude of the person’s weight is 1.00 N. Using Proportionalities to Solve Gravitation Problems Example 4.2 demonstrates how to solve gravitation problems using proportionalities. This technique is useful if you are given how the separation distance and masses change from one situation to another. Chapter 4 Gravity extends throughout the universe. 207 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 208 Example 4.2 Object A exerts a gravitational force of magnitude 1.3 1010 N on object B. Determine the magnitude of the gravitational force if the separation distance is doubled, mA increases by 6 times, and mB is halved. Explain your reasoning. Practice Problem 1. Object A exerts a gravitational force of magnitude 5.9 1011 N on object B. For each situation, determine the magnitude of the gravitational force. Explain your reasoning. (a) the separation distance 4 increases to of its original 3 3 of its value, mA increases to 2 original value, and mB is halved (b) the separation distance decreases 1 of its original value, m
A is to 6 5 halved, and mB increases to 4 of its original value Answer 1. (a) 2.5 1011 N (b) 1.3 109 N Analysis and Solution mAmB and Fg From Newton’s law of gravitation, Fg Figure 4.18 represents the situation of the problem. 1 r 2. before after mA mA Fg 1.3 1010 N mB r Fg? 2r Figure 4.18 Fg (6mA) mB 1 2 and Fg 1 (2r)2 (6) 1 2 mAmB 3mAmB 1 1 22 r 2 1 1 2 r 4 mB Calculate the factor change of Fg. 3 1 4 3 4 Calculate Fg. 3 4 3 4 Fg (1.3 1010 N) 9.8 1011 N The new magnitude of the gravitational force will be 9.8 1011 N. Using Superposition to Find the Net Gravitational Force on an Object Example 4.3 demonstrates how to calculate the gravitational force exerted by both the Moon and the Sun on Earth. A free-body diagram is used to determine the gravitational forces acting on Earth. The technique of adding the gravitational force due to each pair of objects (Earth and the Moon, and Earth and the Sun) to find the net gravitational force is called superposition. 208 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 209 Example 4.3 During a lunar eclipse, Earth, the Moon, and the Sun are aligned on the same plane as shown in Figure 4.19. Using the data in the chart below, calculate the net gravitational force exerted by both the Moon and the Sun on Earth. Celestial Body Mass* (kg) Mean Separation Distance from Earth* (m) Earth Earth’s Moon Sun 5.97 1024 7.35 1022 1.99 1030 — 3.84 108 1.50 1011 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) 3.84 108 m 1.50 1011 m Moon Earth Sun Figure 4.19 Given mEarth mMoon mSun 5.97 1024 kg 7.35 1022 kg 1.99 1030 kg rE to M rE to S 3.84 108 m 1.50 1011 m Required net gravitational
force on Earth (F g) Analysis and Solution Draw a free-body diagram for Earth (Figure 4.20). Practice Problems 1. During a solar eclipse, Earth, the Moon, and the Sun are aligned on the same plane as shown in Figure 4.21. Calculate the net gravitational force exerted by both the Moon and the Sun on Earth. 1.50 1011 m 3.84 108 m Earth Moon Sun Figure 4.21 2. During the first quarter phase of the Moon, Earth, the Moon, and the Sun are positioned as shown in Figure 4.22. Calculate the net gravitational force exerted by both the Moon and the Sun on Earth. Top View of Earth, the Moon, and the Sun Moon y 3.84 108 m Earth x Sun toward Sun toward Moon Fg1 Figure 4.20 Fg1 Fg2 Fnet 1.50 1011 m Figure 4.22 Answers 1. 3.54 1022 N [toward Sun’s centre] 2. 3.52 1022 N [0.3] Fg2 Diagram is not to scale. Calculate Fg exerted by the Moon on Earth using Newton’s law of gravitation. (Fg)1 GmEarthmMoon (rE to M)2 m N (5.97 1024 kg)(7.35 1022 kg) 6.67 1011 2 g k (3.84 108 m)2 2 1.985 1020 N 1.985 1020 N [toward Moon’s centre] F g1 Chapter 4 Gravity extends throughout the universe. 209 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 210 Calculate Fg exerted by the Sun on Earth using Newton’s law of gravitation. (Fg)2 GmEarthmSun (rE to S)2 m N (5.97 1024 kg)(1.99 1030 kg) 6.67 1011 2 g k (1.50 1011 m)2 2 3.522 1022 N 3.522 1022 N [toward Sun’s centre] F g2 gnet Fgnet F g2 Fg2 Find the net gravitational force on Earth. F F g1 Fg1 1.985 1020 N 3.522 1022 N 3.50 1022 N 3.50 1022 N [toward Sun’s centre] F gnet Paraph
rase The net gravitational force on Earth due to the Sun and Moon during a lunar eclipse is 3.50 1022 N [toward Sun’s centre]. The Role of Gravitational Force on Earth’s Tides Newton used the concept of gravitational force to account for Earth’s tides. Although he correctly identified the gravitational force exerted by the Moon and the Sun on Earth as the major cause, a complete understanding of tides must take into account other factors as well. The height of the tides varies depending on the location on Earth (Figure 4.23). In the middle of the Pacific Ocean, the difference between high and low tides is about 0.5 m. But along the coastline of the continents, the difference may be considerably greater. Figure 4.23 The tides in the Bay of Fundy are the highest in the world. In some locations, the water level rises up to 18 m between low and high tides. 210 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 211 Some factors that affect tides are • the shape of the coastline • the topography of the ocean floor near the coastline • friction between Earth and the ocean water • Earth’s position in its orbit around the Sun • Earth’s rotation about its axis • the tilt of Earth’s axis • the alignment of Earth, the Moon, and the Sun First consider only the Moon’s influence on Earth. Since the Moon exerts a gravitational force on Earth, the Moon is in a sense pulling Earth closer to it. So the land mass and ocean water on Earth are all “falling” toward the Moon. In Figure 4.24, this gravitational force is greatest at side A, then decreases at the midpoints of A and B, and is least at side B, because the magnitude of the gravitational force varies inversely with the square of the separation distance. e WEB Newspapers in cities near an ocean, such as Halifax, often publish tidal charts listing the times of local high and low tides. Find an example of a tidal chart and suggest how different people would find this information useful. Begin your search at www.pearsoned.ca/school/ physicssource. 23.5° B Earth A Moon Figure 4.24 Earth experiences high tides on sides A and B at the same time. The vectors show the relative gravitational force exerted by the Moon on a test mass at various locations near Earth’s surface.
The bulges at A and B are the high tides. The low tides occur at the midpoints of A and B. The bulge at B occurs because the land mass of Earth at B is pulled toward the Moon, leaving the ocean water behind. Next consider the fact that Earth rotates on its axis once every 24 h. As the bulges remain fixed relative to the Moon, Earth rotates underneath those bulges. So at a given location on Earth’s surface, a high tide is replaced by a low tide about 6 h later, followed again by a high tide about 6 h later, and so on. These time intervals are actually a bit longer than 6 h because the Moon is orbiting Earth every 27 1 3 days with respect to distant stars, and the Moon is taking the bulges along with it. Now consider that Earth is tilted on its axis. When the northern hemisphere is under the bulge at A, the southern hemisphere is under the bulge at B. So the high tides that are 12 h apart are not equally high, and low tides that are 12 h apart are not equally low. Example 4.4 demonstrates how to calculate the gravitational force exerted by the Moon on 1.0000 kg of water at A. info BIT Io, one of Jupiter’s moons, has tidal bulges of up to 100 m compared to typical tidal bulges of 1 m on Earth. Chapter 4 Gravity extends throughout the universe. 211 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 212 Example 4.4 Calculate the gravitational force exerted by the Moon on 1.0000 kg of water at A (Figure 4.25). Use G 6.672 59 1011 Nm2/kg2. Celestial Body Mass* (kg) Equatorial Radius* (m) Mean Separation Distance from Earth* (m) Earth 5.9742 1024 6.3781 106 — Earth’s Moon 7.3483 1022 1.7374 106 3.8440 108 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) Earth A water B rEarth 3.8440 108 m Moon Figure 4.25 Given mw mEarth mMoon rE to M 1.0000 kg 5.9742 1024 kg 7.3483 1022 kg 3.8440 108 m rEarth rMoon 6.3781 106 m
1.7374 106 m Required g) gravitational force exerted by Moon on water (F Analysis and Solution Draw a free-body diagram for the water showing only F g due to the Moon (Figure 4.26). away from Moon toward Moon Fg Figure 4.26 Practice Problem 1. Using the value of G given in Example 4.4, calculate the gravitational force exerted by the Moon on 1.0000 kg of water (a) at the midpoints of A and B, and (b) at B. Answer 1. (a) 3.3183 105 N [toward Moon’s centre] (b) 3.2109 105 N [toward Moon’s centre] 212 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 213 Find the separation distance between the water and the Moon. r rE to M rEarth 3.8440 108 m 6.3781 106 m 3.780 22 108 m Calculate Fg exerted by the Moon on the water using Newton’s law of gravitation. GmwmMoon r 2 Fg m N (1.0000 kg)(7.3483 1022 kg) 6.672 59 1011 2 g k (3.780 22 108 m)2 2 3.4312 105 N 3.4312 105 N [toward Moon’s centre] F g Paraphrase The gravitational force exerted by the Moon on the water at A is 3.4312 105 N [toward Moon’s centre]. The Role of Gravitational Force on Interplanetary Travel Scientists who plan space missions take advantage of the gravitational force exerted by planets and other celestial bodies to change the speed and direction of spacecraft. Distances between celestial bodies are huge compared to distances on Earth. So a space probe leaving Earth to study Jupiter and Saturn and their moons would take many years to arrive there. Scientists have to calculate the position and velocity of all the celestial bodies that will affect the motion of the probe many years in advance. If several planets are moving in the same direction and their positions are aligned, a space probe launched from Earth can arrive at its destination many years sooner, provided the probe moves near as many of those planets as possible (Figure 4.27). info BIT The Voyager mission was intended to take advantage of a geometric alignment of Jupiter, Saturn, Uranus, and Neptune. This arrangement occurs approximately every 175 years. By using the concept of gravity
assist, the flight time to Neptune was reduced from 30 to 12 years, and a minimum of onboard propellant on the spacecraft was required. Voyager 1 Voyager 1 Launch Launch Sept. 5, 1977 Sept. 5, 1977 Voyager 2 Voyager 2 Launch Launch Aug. 20, 1977 Aug. 20, 1977 Jupiter Mar. 5, 1979 Jupiter Mar. 5, 1979 Jupiter Jul. 9, 1979 Jupiter Jul. 9, 1979 Saturn Aug. 25, 1981 Saturn Aug. 25, 1981 Saturn Nov. 12, 1980 Saturn Nov. 12, 1980 Voyager 2 Neptune Aug. 25, 1989 Neptune Aug. 25, 1989 Uranus Jan. 24, 1986 Uranus Jan. 24, 1986 Voyager 1 Figure 4.27 Both Voyager spacecraft were launched from Cape Canaveral, Florida, in 1977. Voyager 1 had close encounters with Jupiter and Saturn, while Voyager 2 flew by all four of the gaseous planets in the solar system. Chapter 4 Gravity extends throughout the universe. 213 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 214 Each time the probe gets near enough to one of these planets, the gravitational field of the planet causes the path of the probe to curve (Figure 4.28). The planet deflects the space probe and, if the planet is moving in the same direction as the probe, the speed of the probe after its planetary encounter will increase. The use of the gravitational force exerted by celestial bodies to reduce interplanetary travel times is called gravity assist. info BIT Voyager 1 identified nine active volcanoes on Io, one of Jupiter’s moons. Up until that point, scientists knew of no other celestial body in the solar system, other than Earth, that has active volcanoes. According to Voyager 1’s instruments, the debris being ejected from Io’s volcanoes had a speed of 1.05 103 m/s compared to speeds of 50 m/s at Mount Etna on Earth. Figure 4.28 Voyager 1 passed close to Io, Ganymede, and Callisto, three of Jupiter’s moons. Jupiter has a total of 62 moons. Europa Ganymede Voyager 1 Jupiter Io Callisto THEN, NOW, AND FUTURE Small Steps Lead to New Models When Newton began his study of gravity in 1665, he was not the first person to tackle the challenge of explaining planetary motion. Ancient Greek astronomer Ptolemy (Claudius Ptolemaeus, 2nd century A.D.) and
eventually Nicolaus Copernicus (1473–1543) proposed two different models of the solar system: one Earth-centred and the other Sun-centred. Later Johannes Kepler (1571–1630) developed three empirical laws describing planetary motion using astronomical data compiled by astronomer Tycho Brahe (1546–1601). Kepler’s laws confirmed that a Sun-centred system is the correct model, because it was possible to predict the correct position of plan- ets. However, Kepler was unable to explain why planets move. Newton was the first scientist to explain the motion of planets in terms of forces. By using his three laws of motion and his law of gravitation, Newton derived Kepler’s laws, providing further evidence of the validity of his force laws and of the model of a Sun-centred solar system. Newton was able to develop his law of gravitation because many scientists before him developed theories and made observations about planetary motion. Newton’s laws and the concept of gravity could describe the motion of objects on Earth and throughout the universe. This was a tremendous breakthrough because up until that time, scientists were unable to predict or explain motion. While Newton’s model can still be used today for most everyday situations, scientists have further modified it. The process of developing new models and theories has helped scientists tackle questions about the universe in ways Newton could never have imagined. Questions 1. Research the scientific developments that led to Newton’s law of gravitation. 2. What are some benefits of developing new scientific models and theories? 214 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 215 4.2 Check and Reflect 4.2 Check and Reflect Knowledge 1. Why is G called a “universal” constant? 2. Describe how a torsion balance can be used to measure the constant G. 3. Suppose Fg is the magnitude of the gravitational force between two people with a separation distance of 1.0 m. How would Fg change if (a) the separation distance became 2.0 m? (b) one person was joined by an equally massive friend while at this 2.0-m separation distance? Applications 4. The Moon has a mass of 7.35 1022 kg and its equatorial radius is 1.74 106 m. Earth’s mass is 5.97 1024 kg and its equatorial radius is 6.38 106 m. (a)
Calculate the magnitude of the gravitational force exerted by (i) the Moon on a 100-kg astronaut standing on the Moon’s surface, and (ii) Earth on a 100-kg astronaut standing on Earth’s surface. (b) Explain why the values of Fg in part (a) are different. 5. Mars has two moons, Deimos and Phobos, each named after an attendant of the Roman war god Mars. Deimos has a mass of 2.38 1015 kg and its mean distance from Mars is 2.3 107 m. Phobos has a mass of 1.1 1016 kg and its mean distance from Mars is 9.4 106 m. (a) Without doing any calculations, predict which moon will exert a greater gravitational force on Mars. Explain your reasoning. (b) Check your prediction in part (a) by calculating the magnitude of the gravitational force exerted by each moon on Mars. Mars’ mass is 6.42 1023 kg. Show complete solutions. 6. Suppose the equatorial radius of Earth was the same as the Moon, but Earth’s mass remained the same. The Moon has an equatorial radius of 1.74 106 m. Earth’s mass is 5.97 1024 kg and its equatorial radius is 6.38 106 m. (a) Calculate the gravitational force that this hypothetical Earth would exert on a 1.00-kg object at its surface. (b) How does the answer in part (a) compare to the actual gravitational force exerted by Earth on this object? Extensions 7. Prepare a problem involving Newton’s law of gravitation for each situation. Work with a partner to solve each problem, and discuss the steps you use. (a) Choose the values of the two masses and the separation distance. (b) Use values of mA, mB, and r that are multiples of those in part (a). Use proportionalities to solve the problem. 8. During Newton’s time, scientists often • worked alone and contact with other scientists working on similar problems was difficult. • were knowledgeable in many different fields. Newton, for example, spent many years doing alchemy. (a) In paragraph form, assess the impacts that these factors might have had on science in Newton’s day. (b) In a paragraph, describe in what ways these factors are relevant to scientists today. e TEST To check your understanding of Newton’s law of grav
itation, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 215 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 216 info BIT A g force is a force that causes an acceleration with a magnitude of some multiple of g. A force of 2g means the magnitude of the acceleration is 2 9.81 m/s2 19.6 m/s2. 4.3 Relating Gravitational Field Strength to Gravitational Force The acceleration due to gravity g near or on Earth’s surface is about 9.81 m/s2 [down]. But where does the value of 9.81 m/s2 come from? Consider the forces acting on a test mass mtest some distance above Earth’s surface, where Earth has a mass of Msource. The only force acting on mtest is the gravitational force exerted by Earth on the test mass (Figure 4.29). r Msource mtest Fg Figure 4.29 The gravitational force exerted by Earth on test mass mtest info BIT During a roller coaster ride, riders may experience a 4g change in acceleration between the top and bottom of a loop. This dramatic change in acceleration causes the thrill and occasional dizziness experienced by riders. The magnitude of F weight or using Newton’s law of gravitation. g can be evaluated two ways: using the concept of Weight Newton’s law of gravitation Fg mtestg Fg GmtestMsource r 2 Since the value of Fg is the same no matter which equation you use, set both equations equal to each other. mtest g Gmtest Msource r 2 g GMsource r 2 So no matter where the test mass is located in the universe, you can calculate the magnitude of the gravitational field strength (or gravitational acceleration) at any distance from a celestial body if you know the mass of the celestial body Msource and the separation distance between the centre of the test mass and the celestial body r. Find out the relationship between gravitational field strength and the acceleration due to gravity by doing 4-3 Design a Lab. 216 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 217 4-3 Design a Lab 4-3 Design a Lab Comparing Gravitational Field Strength to Gravitational Acceleration The Question What is the relationship between gravitational field strength and the local
value of the gravitational acceleration? Design and Conduct Your Investigation State a hypothesis. Then design an experiment. Identify the controlled, manipulated, and responding variables. Review the procedure in 4-2 QuickLab on page 198. List the materials you will use, as well as a detailed procedure. Check the procedure with your teacher and then do the investigation. Analyze your data and form a conclusion. How well did your results agree with your hypothesis? How Is Gravitational Field Strength Related to Gravitational Acceleration? To determine how gravitational field strength is related to gravitational acceleration, use the definition of a newton, 1 N 1 kgm/s2. Then substitute kilogram-metres per second squared for newtons in the equation for gravitational field strength: 1 N kg 1 m kg s2 kg m 1 s2 Metres per second squared are the units of acceleration. So in terms of units, gravitational field strength and gravitational acceleration are equivalent (Figure 4.30). Figure 4.30 Jennifer Heil of Spruce Grove, Alberta, won the gold medal in the women’s freestyle skiing moguls in the 2006 Winter Olympics in Turin, Italy. The gravitational field strength at the surface of the Moon is about 1 6 that at Earth’s surface. How would Jennifer’s jump on Earth compare with one on the Moon? Chapter 4 Gravity extends throughout the universe. 217 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 218 Calculating the Gravitational Acceleration of an Object on Two Celestial Bodies Example 4.5 demonstrates how to calculate the gravitational acceleration at the equator on Earth’s surface and that on the surface of the Moon. These two values are then compared to find the ratio of gEarth to gMoon. To solve the problem requires using data from Table 4.1, which shows the mass and equatorial radius of the Sun, the Moon, and each planet in the solar system. Table 4.1 Masses and Radii for Celestial Bodies in the Solar System* Celestial Body Sun Mercury Venus Earth Earth’s Moon Mars Jupiter Saturn Uranus Neptune Mass (kg) 1.99 1030 3.30 1023 4.87 1024 5.97 1024 7.35 1022 6.42 1023 1.90 1027 5.69 1026 8.68 1025 1.02 1026 Equatorial Radius (m) 6.96 108 2.44 106 6.05 106 6
.38 106 1.74 106 3.40 106 7.15 107 6.03 107 2.56 107 2.48 107 *Source: Jet Propulsion Laboratory, California Institute of Technology (See JPL link at www.pearsoned.ca/school/physicssource.) Example 4.5 (a) Calculate the magnitude of the gravitational acceleration of an object at the equator on the surface of Earth and the Moon (Figure 4.31). Refer to Table 4.1 above. (b) Determine the ratio of gEarth to gMoon. How different would your weight be on the Moon? gEarth? gMoon? Moon Earth Figure 4.31 Given mEarth mMoon 5.97 1024 kg 7.35 1022 kg rEarth rMoon 6.38 106 m 1.74 106 m Required (a) magnitude of gravitational acceleration at equator on Earth and the Moon (gEarth and gMoon) (b) ratio of gEarth to gMoon e WEB Globular clusters are groups of about 1 000 000 stars that are bound together by gravity. Find out who discovered the first cluster and how many have been identified so far. Research the approximate size and shape of a globular cluster and the forces involved in its formation. Summarize your findings. Begin your search at www.pearsoned.ca/school/ physicssource. 218 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 219 Analysis and Solution (a) Use the equation g to calculate the GMsource r2 magnitude of the gravitational field strength on each celestial body. The magnitude of the gravitational acceleration is numerically equal to the magnitude of the gravitational field strength. Earth gEarth GmEarth (rEarth)2 m N (5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 2 9.783 N/kg 9.783 m/s2 The Moon gMoon GmMoon (rMoon)2 m N (7.35 1022 kg) 6.67 1011 2 g k (1.74 106 m)2 2 1.619 N/kg 1.619 m/s2 (b) Calculate the ratio of gEarth to gMoon. m 9.783 s2 m 1.619 s2 gEarth gMoon 6.04 Paraphrase (a) The magnitude of the gravitational acceleration at the equator on
the surface of Earth is 9.78 m/s2 and of the Moon is 1.62 m/s2. Practice Problems 1. A satellite orbits Earth at a distance of 3rEarth above Earth’s surface. Use the data from Table 4.1 on page 218. (a) How many Earth radii is the satellite from Earth’s centre? (b) What is the magnitude of the gravitational acceleration of the satellite? 2. An 80.0-kg astronaut is in orbit 3.20 104 km from Earth’s centre. Use the data from Table 4.1 on page 218. (a) Calculate the magnitude of the gravitational field strength at the location of the astronaut. (b) What would be the magnitude of the gravitational field strength if the astronaut is orbiting the Moon with the same separation distance? 3. The highest satellites orbit Earth at a distance of about 6.6rEarth from Earth’s centre. What would be the gravitational force on a 70-kg astronaut at this location? Answers 1. (a) 4rEarth (b) 6.11 101 m/s2 2. (a) 3.89 101 N/kg (b) 4.79 103 N/kg 3. 16 N [toward Earth’s centre] (b) The ratio of gEarth to gMoon is 6.04. So your weight would be about 6 times less on the surface of the Moon than on Earth. Calculating the Weight of an Object on Mars The equation g GMsource r 2 can be used with F mg to calculate the weight of an object on any celestial body. In Example 4.6, the weight of a student on Mars is calculated. This quantity is then compared with the student’s weight on Earth’s surface. An interesting application of the variation in the weight of an object involves the Mars rover (Figure 4.32). The rover had a mass of about 175 kg, but on the surface of Mars, the rover weighed about 2.5 times less than on Earth’s surface. The rover was designed to avoid inclines greater than 30 but gEarth, the rover could travel farther up an because gMars incline on Mars than on Earth using the same battery charge. Figure 4.32 In full sunlight, a 140-W battery enabled the Mars rover to travel about 100 m per day on level ground with an average speed of 1.0 cm/s between charges. Chapter 4 Gravity extends
throughout the universe. 219 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 220 Example 4.6 (a) What is the mass of a 60.0-kg student on Mars and on Earth? (b) What is the student’s weight at the equator on the surface of Mars and of Earth (Figure 4.33). Use the data from Table 4.1 on page 218. Fg? Mars Given (a) ms (b) mMars mEarth 60.0 kg 6.42 1023 kg 5.97 1024 kg rMars rEarth 3.40 106 m 6.38 106 m Fg? Earth Required (a) mass on Mars and on Earth (m) gEarth) gMars and F (b) weight on Mars and on Earth (F Figure 4.33 Analysis and Solution (a) Mass is a scalar quantity and does not depend on location. So the (b) Use the equation g student’s mass will be the same on Mars as on Earth. GMsource r 2 gravitational field strength on Mars and on Earth. to calculate the magnitude of the Practice Problems Use the data from Table 4.1, page 218, to answer the following questions. 1. What would be the weight of a 22.0-kg dog at the equator on Saturn’s surface? 2. (a) Do you think your skeleton could support your weight on Jupiter? (b) Compared to Earth, how much stronger would your bones need to be? 3. (a) What is the magnitude of the gravitational field strength at the equator on Uranus’ surface? (b) Compared to Earth, how would your weight change on Uranus? Answers 1. 230 N [toward Saturn’s centre] 2. (a) no (b) 2.53 times 3. (a) 8.83 N/kg, (b) 0.903 FgEarth Mars gMars GmMars (rMars)2 2 m N (6.42 1023 kg) 6.67 1011 2 g k (3.40 106 m)2 3.704 N/kg Earth gEarth GmEarth (rEarth)2 2 m N (5.97 1024 kg) 6.67 1011 2 g k (6.38 106 m)2 9.783 N/kg Since the direction of F celestial body, use the scalar equation Fg
magnitude of the weight. g will be toward the centre of each mg to find the Mars FgMars msgMars Earth FgEarth msgEarth N (60.0 kg)3.704 g k 222 N N (60.0 kg)9.783 g k 587 N 220 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 221 Paraphrase and Verify (a) The mass of the student would be 60.0 kg on both Mars and Earth. (b) The weight of the student on Mars would be 222 N [toward Mars’ centre] and on Earth 587 N [toward Earth’s centre]. So the student would weigh about 2.6 times more on Earth than Mars. Different Values of Gravitational Field Strength on Earth For a long time, people thought that the magnitude of the gravitational field strength was constant at any location on Earth’s surface. However, scientists discovered that the value of g depends on both latitude and altitude. Latitude is the angular distance north or south of the equator. Altitude is the elevation of the ground above sea level. Figure 4.34 shows how the magnitude of the gravitational field strength at sea level varies with latitude. Measured Magnitude of Gravitational Field Strength at Sea Level vs. Latitude info BIT The value 9.81 N/kg is an average of the magnitude of the gravitational field strength at different locations on Earth’s surface. 9.84 9.83 9.82 9.81 9.80 9.79 9.78 20° 60° 80° 40° Latitude Equator Figure 4.34 Gravitational field strength at sea level as a function of latitude. At what location on Earth’s surface would you weigh the least? The most? P ole s The value of g increases as you move toward either the North or South Pole, because Earth is not a perfect sphere. It is flatter at the poles and it bulges out slightly at the equator. In fact, Earth’s radius is 21 km greater at the equator than at the poles. So an object at the equator is farther away from Earth’s centre than if the object were at the North Pole. Since g, the farther an object is from Earth’s centre, the smaller 1 r 2 the value of g will be. Other factors affect the value of g at Earth’s surface. The materials
that make up Earth’s crust are not uniformly distributed. Some materials, such as gold, are more dense than materials such as zinc. Earth’s rotation about its axis also reduces the measured value of g, but the closer an object is to the North or South Pole, the less effect Earth’s rotation has on g. Concept Check Leo weighs 638 N [down] in Calgary, Alberta. What are some problems with Leo saying he weighs 638 N [down] anywhere on Earth? What property of matter would he be more accurate to state? Chapter 4 Gravity extends throughout the universe. 221 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 222 Gravity (mGal 20.5 20. 20.0 20.0 0. 5 0. 5 2 2 AA Applications of the Variation in g in Geology The variation in the value of g on Earth is used to detect the presence of minerals and oil. Geophysicists and geologists use sensitive instruments, called gravimeters, to detect small variations in g when they search for new deposits of ore or oil. Gold and silver deposits increase the value of g, while deposits of oil and natural gas decrease g. Figure 4.35 is an example of a map that shows different measured values of g as lines, where each line represents a specific value of g. 19.0 19.1 19.2 19.3 19.4 19.5 19.6 19.7 19.8 19.9 20.0 20.1 20.2 20.3 20.4 20.5 20.6 20.7 20.8 20.9 21.0 21.1 21.2 21.3 21.4 21.5 21.6 21.7 21.8 21.9 22.0 22.5 23.0 23.5 mGal Figure 4.35 A map showing the location of sulphide deposits in northern New Brunswick (shown in black) true weight: gravitational force acting on an object that has mass a 0 v constant FN Fg True Weight vs. Apparent Weight mg to calculate the weight of an object So far, you used the equation F g at any location in the universe. The gravitational force that you calculate g, of an object. with this equation is really called the true weight, F Suppose a student is standing on a scale calibrated in newtons in an elevator (Figure 4.36). If the elevator is at rest or is
moving at constant velocity, the scale reads 600 N. Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.36 The elevator and student are either at rest or moving at constant velocity. Using the free-body diagram for the student (Figure 4.37), the equation for the net force on the student is F N F F FN F g net 0 F g 0 Fg 0 mg FN N Figure 4.37 The free-body diagram for the student in Figure 4.36 FN mg 222 Unit II Dynamics a FN Fg Figure 4.39 The free-body diagram for the student in Figure 4.38 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 223 So when F is equal to the magnitude of the student’s weight. 0 N on the student, the magnitude of the normal force net Now suppose the elevator is accelerating up uniformly (Figure 4.38). In this situation, the scale reads 750 N. Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.38 The elevator and student are accelerating up uniformly. N net To understand why the reading on the scale is different, draw the freebody diagram for the student (Figure 4.39) and write the equation for the net force: F F F g ma F F g N ma F F N ma mg m(a g) g The equation for F N is valid whether the student is accelerating up or down. In Figure 4.38, the student feels heavier than usual because the scale is pushing up on him with a force greater than mg. If the elevator is accelerating down uniformly, the scale reads m(a g ) but this time a and g are 525 N (Figure 4.40). As before, F in the same direction. So FN is less than mg, and the student feels lighter than usual. N Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.40 The elevator and student are accelerating down uniformly. Chapter 4 Gravity extends throughout the universe. 223 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 224 apparent weight: negative of the normal force acting on an object The quantity F N is called the apparent weight, w, of an object. For the situations shown in Figures 4.38 and 4.40 on page 223, the equation for the apparent weight of the student is e SIM Calculate the true weight, normal force, and apparent weight of a
person during an elevator ride. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 224 Unit II Dynamics w F N m(a g) m(g a ) Example 4.7 demonstrates how to calculate the true weight and apparent weight of an astronaut in a rocket during liftoff on Earth’s surface. Example 4.7 A 100.0-kg astronaut in a spacesuit is standing on a scale in a rocket (Figure 4.41). The acceleration of the rocket is 19.6 m/s2 [up]. Calculate her true weight and apparent weight during liftoff on Earth. The acceleration due to gravity on Earth’s surface is 9.81 m/s2 [down]. Given m 100.0 kg a 19.6 m/s2 [up] g 9.81 m/s2 [down] up down Required true weight and apparent weight during liftoff (F g and w) 0 3000 1000 2000 Figure 4.41 Analysis and Solution Draw a free-body diagram and a vector addition diagram for the astronaut (Figure 4.42). up down FN Fg FN Fnet Fg Use the equation F g mg to find the astronaut’s true weight. Figure 4.42 F g mg (100.0 kg)9.81 9.81 102 N m s2 The astronaut is not accelerating left or right. So in the horizontal direction, F 0 N. net 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 225 For the vertical direction, write an equation to find the net force on the astronaut. F F F net N g g N N FN Apply Newton’s second law. ma F F g ma FF FF ma Fg m (9.81 102 N) (100.0 kg)19.6 s2 m 9.81 102 N (100.0 kg)19.6 s2 2.94 103 N 2.94 103 N [up] F Use the equation w F N Practice Problems 1. In Example 4.7, draw the free-body diagram for the scale during liftoff. 2. Suppose the rocket in Example 4.7 has an acceleration of 19.6 m/s2 [down] while it is near Earth’s surface. What will be the astronaut’s apparent weight and true weight? Answers 1. See page 898. 2. 9.79
102 N [up], 9.81 102 N [down] N to find the astronaut’s apparent weight. w F N (2.94 103 N) 2.94 103 N Paraphrase During liftoff, the astronaut’s true weight is 9.81 102 N [down] and her apparent weight is 2.94 103 N [down]. In Example 4.8, an astronaut is accelerating in deep space, a location in which the gravitational force acting on an object is not measurable. So in deep space, F 0. g Example 4.8 Refer to Example 4.7 on pages 224 and 225. What is the magnitude of the astronaut’s true weight and apparent weight if the rocket is in deep space? The magnitude of the acceleration of the rocket is 19.6 m/s2. Given m 100.0 kg magnitude of a 19.6 m/s2 g 0 m/s2 Required magnitude of true weight and apparent weight in deep space (F g and w) Analysis and Solution In deep space, the mass of the astronaut is still 100.0 kg, but g is negligible. F So g mg 0 N The astronaut is not accelerating left or right. So in the horizontal direction, F 0 N. net Practice Problems 1. An 80.0-kg astronaut is standing on a scale in a rocket leaving the surface of the Moon. The acceleration of the rocket is 12.8 m/s2 [up]. On the Moon, g 1.62 N/kg [down]. Calculate the magnitude of the true weight and apparent weight of the astronaut (a) during liftoff, and (b) if the rocket has the same acceleration in deep space. 2. A 60.0-kg astronaut is standing on a scale in a rocket about to land on the surface of Mars. The rocket slows down at 11.1 m/s2 while approaching Mars. Use the data from Table 4.1 on page 218. Calculate the true weight and apparent weight of the astronaut (a) as the rocket lands, and (b) if the rocket is accelerating at 7.38 m/s2 [up] when leaving Mars. Chapter 4 Gravity extends throughout the universe. 225 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 226 For the vertical direction, write an equation to find the net force on the astronaut. Refer to the free-body diagram in Figure 4.42 on page
224. Answers 1. (a) (F g) 1.30 102 N [down], (w) 1.15 103 N [down] F net b) (F 2. (a) (F (b) (F g) 0 N, (w) 1.02 103 N [down] g) 2.22 102 N [down], (w) 8.88 102 N [down] g) 2.22 102 N [down], (w) 6.65 102 N [down] Apply Newton’s second law. F N FN ma ma m (100.0 kg)19.6 s2 1.96 103 N Use the equation w F N to find the astronaut’s apparent weight. w F N (1.96 103 N) 1.96 103 N Paraphrase In deep space, the astronaut’s true weight is zero and the magnitude of her apparent weight is 1.96 103 N. Free Fall Let’s revisit the elevator scenario on pages 222 and 223. Suppose the elevator cable breaks (Figure 4.43). Assuming that frictional forces are negligible, the elevator, student, and scale all fall toward Earth with an acceleration of g. The student is now in free fall, the condition in which the only force acting on an object is F g. free fall: situation in which the only force acting on an object that has mass is the gravitational force a g 0 0 6 0 0 7 0 Fg Fg 600 N 300 200 100 400 0 500 600 700 Figure 4.43 The elevator, student, and scale are in free fall. Figure 4.44 The free-body diagram for the student in Figure 4.43 226 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 227 To understand free fall, draw the free-body diagram for the student (Figure 4.44) and write the equation for the net force: F net F ma F g g ma mg a g So in free fall, a g and both the student and the scale are accelerating at g downward. In Figure 4.43 on page 226, the scale reads zero because 0. Since it no longer exerts a normal force on the student, so F 0, the student’s apparent weight is also zero. Sometimes an F object in free fall is described as being “weightless.” However, this description is incorrect. In free fall, F w 0
but F 0. N N N g Observe the motion of water in a cup while in free fall by doing 4-4 QuickLab. 4-4 QuickLab 4-4 QuickLab Water in Free Fall Problem What is the motion of water in a cup when the cup is dropped from several metres above Earth’s surface? CAUTION: Do this activity outside. Have someone steady the ladder and be careful when climbing it. Materials paper cup pointed pen or pencil water dishpan stepladder Procedure 1 Make two holes on opposite sides of the cup near the bottom using the pen or pencil. Cover the holes with your thumb and forefinger. Then fill the cup with water. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. 2 Hold the cup at shoulder height above a dishpan, and uncover the holes. Observe what happens to the water (Figure 4.45). Have a partner sketch the path the water takes. 3 Repeat step 1 but climb the ladder and drop the cup toward the dishpan from a height of several metres. Observe the motion of the water during the fall. Questions 1. Describe the path and motion of the water (a) when the cup was held stationary, and cup with two holes and filled with water Figure 4.45 (b) when the cup was dropped from the ladder. Give a reason for your observations. Chapter 4 Gravity extends throughout the universe. 227 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 228 Weightlessness Videos transmitted from a space shuttle or the space station often show astronauts floating in their cabin (Figure 4.46). Are the astronauts weightless in space? The answer is no. Then why do they appear to be weightless? Since the shuttle is some distance above Earth, g is less than its 1 r 2 value at Earth’s surface, because g. While the shuttle orbits Earth at high speed in an almost circular path, Earth exerts a gravitational force on the shuttle and everything in it. So the shuttle is able to remain in orbit. If an astronaut were standing on a scale in the shuttle, the scale would read zero, because the shuttle and everything in it are in free fall. The astronaut would feel “weightless” because the gravitational force exerted by Earth pulls the shuttle and the astronaut toward Earth. Suppose an astronaut is in a rocket in deep space and the acceleration of the
rocket is zero. The astronaut would experience no measurable gravitational forces from any celestial bodies, and the astronaut’s acceleration would be zero. In this situation, the astronaut would have a true weight of zero and an apparent weight of zero, a condition called true weightlessness. true weightlessness: situation in which w 0 for an object and F 0 on the object g Figure 4.46 At the altitude of the shuttle, the value of g is about 90% of its value at Earth’s surface. 228 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 229 4.3 Check and Reflect 4.3 Check and Reflect Knowledge 1. (a) What is the difference between true weight and apparent weight? (b) Describe a situation in which the true weight of an object is zero but its apparent weight is not zero. 2. A person orbiting Earth in a spacecraft has an apparent weight of zero. Explain if the person still experiences a gravitational force. 8. An astronaut in a rocket has an apparent weight of 1.35 103 N [down]. If the acceleration of the rocket is 14.7 m/s2 [up] near Earth’s surface, what is the astronaut’s true weight? The acceleration due to gravity on Earth’s surface is about 9.81 m/s2 [down]. 9. A 50-kg astronaut experiences an acceleration of 5.0g [up] during liftoff. (a) Draw a free-body diagram for the astronaut during liftoff. 3. List two factors that affect the magnitude (b) What is the astronaut’s true weight of the gravitational field strength at Earth’s surface. Applications 4. Is there a place in the universe where true weightlessness actually exists? 5. Calculate the gravitational field strength at the location of a 70-kg astronaut 2.0rEarth from Earth’s centre. Use the data from Table 4.1 on page 218. 6. Graph the equation g GmMoon r 2 using technology. Refer to Student References 5: Graphing Data on pp. 872–874. Plot g on the y-axis (range of 02.0 N/kg) and r on the x-axis (range of 1–5rMoon). Toggle through to read values of g corresponding to specific values of rMoon to answer these questions: (a) Describe the graph of g
vs. rMoon. How is it similar to Figure 4.10 on page 201? (b) What is the value of g (i) on the surface? (ii) at 1 2 rMoon above the surface? (iii) at rMoon above the surface? the (c) At what distance is g 1 100 and apparent weight? 10. Calculate the acceleration of the elevator in Figures 4.38 and 4.40 on page 223. Extensions 11. Draw a flowchart to summarize the steps needed to find the apparent weight of an object. Refer to Student References 4: Using Graphic Organizers on page 869. 12. Research how geophysicists and geologists use gravitational field strength to locate minerals, oil, and natural gas in Canada. Prepare a half-page report on your findings. Begin your search at www.pearsoned.ca/school/physicssource. 13. Suppose you are wearing a spacesuit. Where could you walk faster, on Earth or on the Moon? Explain your answer. 14. Draw a concept map to identify and link the concepts needed to understand gravitational acceleration and gravitational field strength near a celestial body other than Earth. Refer to Student References 4: Using Graphic Organizers on page 869. Create and solve a problem to demonstrate your understanding of these concepts. gravitational field strength on the surface of the Moon? e TEST 7. At the top of Mount Robson in British Columbia, a 7.5-kg turkey weighs 73.6 N [down]. Calculate the magnitude of the gravitational field strength at this location. To check your understanding of gravitational field strength, true weight, apparent weight, and free fall, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 229 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 230 CHAPTER 4 SUMMARY Key Terms and Concepts gravitational force weight gravitational mass action-at-a-distance force field gravitational field gravitational field strength Key Equations True weight: F mg g Newton’s law of gravitation: Fg GmAmB r 2 torsion balance true weight apparent weight free fall true weightlessness Gravitational field strength (or gravitational acceleration): g GMsource r 2 Apparent weight: w F N Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of
the chapter. gravitational force mass is one of two types Gravitation involves weight is gravitational field strength equation equation other examples inertial strong nuclear using exerted by a using celestial body where direction acceleration standard masses and a and a G is value 6.67 1011 N•m2/kg2 Figure 4.47 230 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 231 CHAPTER 4 REVIEW Knowledge 1. (4.2) What is the significance of the term “universal” in Newton’s law of universal gravitation? 2. (4.1, 4.3) A brick placed on an equal arm balance requires 5.0 kg to just balance it. When the brick is hung from a spring scale, the scale reads 48 N. The balance, standard masses, spring scale, and brick are moved to a planet where the gravitational field strength is 2.0 times that on Earth. What will be the reading on the balance and on the spring scale in this new location? 3. (4.1, 4.3) How does gravitational field strength vary with the mass of a celestial body? Assume that the radius is constant. 4. (4.3) The gravitational field strength at Earth’s surface is about 9.81 N/kg [down]. What is the gravitational field strength exactly 1.6rEarth from Earth’s centre? Applications 5. A 1.0-kg object, initially at rest, is dropped toward Earth’s surface. It takes 2.26 s for the object to fall 25 m. Determine how long it takes a 2.0-kg object to fall this distance from rest on Jupiter. Use the data from Table 4.1 on page 218. 6. Describe the steps you would use to determine the distance from Earth’s centre where the gravitational force exerted by Earth on a spacecraft is balanced by the gravitational force exerted by the Moon. Assume that you know the distance from Earth’s centre to the centre of the Moon. Do not do the calculations. 7. Use the data from Table 4.1 on page 218. Calculate the true weight of a 60.0-kg astronaut on (a) the surface of Mars, and (b) the surface of Saturn. 8. Objects A and B experience a gravitational force of magnitude 2.5 108 N. Determine the magnitude of the gravitational force if the separation distance is
halved, mA increases by 8 times, and mB is reduced to 1 4 of its original value. 9. Suppose a 65-kg astronaut on Mars is standing on a scale calibrated in newtons in an elevator. What will be the reading on the scale when the acceleration of the elevator is (a) zero? (b) 7.2 m/s2 [up]? (c) 3.6 m/s2 [down]? 10. A 50-kg rock in a Nahanni River canyon breaks loose from the edge of a cliff and falls 500 m into the water below. The average air resistance is 125 N. (a) What is the average acceleration of the rock? (b) How long does the rock take to reach the water? (c) What is the true weight of the rock? (d) Does the rock have an apparent weight? Explain. Extensions 11. Research why astronauts do exercises in space and why they have difficulty walking when they return to Earth. Write a short paragraph of your findings. Begin your search at www.pearsoned.ca/ school/physicssource. 12. Pilots in high-speed planes are subject to g forces. Healthy people can withstand up to 3–4 gs. Beyond that limit, the blood will pool in the lower half of the body and not reach the brain, causing the pilot to lose consciousness. Research how Dr. Wilbur Franks from Toronto found a solution to this problem. What connection does this problem have to human survival during a space flight? Begin your search at www.pearsoned.ca/school/physicssource. Consolidate Your Understanding 13. Write a paragraph summarizing Newton’s law of gravitation. Include a numerical example that illustrates the law. Show a detailed solution. Think About It Review your answers to the Think About It questions on page 195. How would you answer each question now? e TEST To check your understanding of gravitation concepts, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 4 Gravity extends throughout the universe. 231 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 232 UNIT II PROJECT Tire Design, Stopping Distance, and Vehicle Mass Scenario Imagine that you work for Alberta Infrastructure and Transportation. A level section of highway in the mountains has an abnormally large number of accidents. The surface of one lane of the highway is concrete
and the other is asphalt. You are a member of a research team formed to determine the stopping distances in the summer for both wet and dry days for different types of tires and for different masses of vehicles. Your team is to prepare a written report on your findings for the Traffic Branch of the Ministry. Assume that the vehicles are travelling at the posted speed limit of 90 km/h when the brakes are applied and that the vehicles are not equipped with anti-lock braking systems (ABS). Assume that the reaction time of drivers before they apply the brakes is 1.8 s. Planning Form a team of three to five members. Summarize the question your group is researching. Make hypotheses about how tire tread, surface condition, and vehicle mass might affect stopping distance. Assign roles to different team members. Some examples are team leader, materials manager, liaison officer, record keeper, and safety officer. Brainstorm strategies for researching and reporting on the question and create a timeline. Research tire designs that are designed to work well on both wet and dry road surfaces. Use the Internet and consult local tire suppliers. Materials • a digital camera and a computer • force-measuring equipment • mass-measuring equipment • new and used tires having different treads and designs • vehicle brochures Procedure 1 Research the Internet and interview different tire suppliers to identify tires designed to work well at above-freezing temperatures on both wet and dry pavement. Assessing Results After completing the project, assess its success based on a rubric designed in class* that considers research strategies experiment techniques clarity and thoroughness of the written report effectiveness of the team’s presentation quality and fairness of the teamwork 2 Visit a local tire supplier and borrow new and used tires of different designs. Photograph the tires to record the tread designs. 3 Design and conduct an experiment to determine the coefficients of static and kinetic friction for the tires on wet and dry asphalt, and wet and dry concrete. Recall that you will need the mass of the tires and the local value of gravitational field strength. CAUTION: Take all your measurements on parking lots or sidewalks, and beware of traffic. Consult the local police department and ask for supervision while doing your experiments. 4 Research the masses of at least four small, medium, and large vehicles travelling highways that would use these tires. Determine the average mass of each class of vehicle. Sales brochures from vehicle dealers have mass information. 5 Determine the average deceleration on both wet and dry roadways for the vehicles of different mass equipped with
these tires. Remember that some drivers may lock their brakes while braking. 6 Determine the stopping distance for the different vehicles under different conditions. Remember to consider driver reaction time. 7 Write a report of your findings. Use graphs and tables where appropriate. Thinking Further Write a three-paragraph addition to your team’s report hypothesizing how driver education, changing the posted speed limit, and requiring that vehicles be equipped with ABS brakes might affect the results. *Note: Your instructor will assess the project using a similar assessment rubric. 232 Unit II Dynamics 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 233 UNIT II SUMMARY Unit Concepts and Skills: Quick Reference Concepts CHAPTER 3 Force Net force Newton’s first law Newton’s second law Summary Forces can change velocity. Resources and Skill Building 3.1 The Nature of Force Force is a push or a pull on an object. Force is a vector quantity measured in newtons (1 N 1 kg•m/s2). Net force is the vector sum of two or more forces acting simultaneously on an object. A free-body diagram helps you write the net force acting on an object. 3-1 QuickLab 3-2 QuickLab Examples 3.1–3.4 Examples 3.2–3.4 3.2 Newton’s First Law of Motion Newton’s first law states that an object will continue being at rest or moving at constant speed in a straight line unless acted upon by a non-zero net force. 3-3 QuickLab 3.3 Newton’s Second Law of Motion Newton’s second law states that when a non-zero net force acts on an object, the object accelerates in the direction of the net force. The magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object. Newton’s third law 3.4 Newton’s Third Law of Motion Newton’s third law states that if object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction. Types of friction Factors affecting friction Coefficients of static and kinetic friction 3.5 Friction Affects Motion Friction is a force that opposes either the motion of an object or the direction the object would be moving in if there were no friction. Static friction is present when an object is stationary but experiences an applied force. Kin
etic friction is present when an object is moving. The magnitude of the force of friction acting on an object is directly proportional to the normal force on the object. The coefficients of friction are proportionality constants that relate the magnitude of the force of friction to the magnitude of the normal force. Temperature, moisture, and the smoothness or roughness of the contact surfaces, and the materials forming the contact surface are some factors that affect the value of the coefficients of friction. 3-5 Inquiry Lab Figures 3.33–3.35 3-6 Design a Lab Examples 3.5–3.11 3-7 QuickLab Figures 3.53–3.57, 3.63, 3.65 3-8 QuickLab Examples 3.12, 3.13 3-9 Design a Lab 3-10 QuickLab Figures 3.70, 3.74, 3.78–3.80 Examples 3.14–3.16 3-11 Inquiry Lab Table 3.4 Examples 3.17–3.20 CHAPTER 4 Gravity extends throughout the universe. Gravitational force 4.1 Gravitational Forces due to Earth Gravitational force is a fundamental force, and can be described as an action-at-a-distance force or as a field. Gravitational field strength Gravitational field strength is the ratio of gravitational force to mass at a specific location. The units of gravitational field strength are N/kg. 4-1 QuickLab 4-2 QuickLab Figure 4.10 Newton’s law of universal gravitation Calculating g anywhere in the universe Variations of g True weight, apparent weight, free fall, and weightlessness 4.2 Newton’s Law of Universal Gravitation Newton’s law of universal gravitation states that the gravitational force of attraction between any two masses is directly proportional to the product of the masses and inversely proportional to the square of the separation distance between the centres of both masses. 4.3 Relating Gravitational Field Strength to Gravitational Force Newton’s law of gravitation can be used to determine the magnitude of gravitational field strength anywhere in the universe. The magnitude of gravitational field strength at a location is numerically equal to the magnitude of gravitational acceleration. The value of g at Earth’s surface depends on latitude, altitude, the composition of Earth’s crust, and Earth’s rotation about its axis. The true weight of an object is equal to the gravitational force acting on the mass, and depends on location. Apparent weight is the negative
of the normal force acting on an object. Free fall is the condition where the only force acting on an object is the gravitational force. True weightlessness is the condition in which w 0 for an object and F 0 on the object. g Figures 4.11–4.13, 4.16, 4.24 Examples 4.1–4.4 4-3 Design a Lab Examples 4.5–4.6 Figures 4.34, 4.35 Figures 4.36–4.40, 4.43, 4.44, 4.46 4-4 QuickLab Examples 4.7, 4.8 Unit II Dynamics 233 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 234 UNIT II REVIEW Vocabulary 1. Using your own words, define these terms, concepts, principles, or laws. action-at-a-distance force action force apparent weight coefficient of friction field free-body diagram free fall gravitational field strength gravitational force gravitational mass inertia inertial mass kinetic friction net force Newton’s first law Newton’s law of gravitation Newton’s second law Newton’s third law normal force reaction force static friction tension true weight Knowledge CHAPTER 3 2. An object experiences three forces: F 1 is 60 N [22.0], F 3 is 83 N [300]. 2 is 36 N [110], and F Explain, using words and diagrams, how to calculate the net force on the object. What is the net force? 3. An object experiences zero net force. Work with a partner to describe the possibilities for its motion. 4. A person with a plaster cast on an arm or leg experiences extra fatigue. Use Newton’s laws to explain to a classmate the reason for this fatigue. 5. Use inertia and Newton’s first law to explain how the spin cycle in a washing machine removes water from wet clothes. 234 Unit II Dynamics 6. A load is placed on a 1.5-kg cart. A force of 6.0 N [left] causes the cart and its load to have an acceleration of 3.0 m/s2 [left]. What is the inertial mass of the load? 7. What happens to the acceleration of an object if the mass is constant and the net force (a) quadruples? (b) is divided by 4? (c) becomes zero? 8. Two people, A and B, are pulling a wagon on a horizontal, friction
less surface with two ropes. Person A applies a force [50] on one rope. Person B applies a force of 25 N [345] on the other rope. If the net force on the wagon is 55.4 N [26], calculate the magnitude of person A’s applied force. A θ 1 50º x θ 2 345º B 9. A book is at rest on a table. The table is exerting an upward force on the book that is equal in magnitude to the downward force exerted by the book on the table. What law does this example illustrate? 10. A pencil exerts a force of 15 N [down] on a notebook. What is the reaction force? What object is exerting the reaction force? 11. Explain why the coefficients of static and kinetic friction are numerals without units. 12. Draw a free-body diagram for a stationary 5.0-kg block resting on a rough incline forming an angle of 30.0 with the horizontal. (a) Explain why the block is stationary. (b) Explain why a free-body diagram is helpful to describe the situation. 13. How does the ability of a car slowing down on wet asphalt compare to it slowing down on wet concrete? Use the data from Table 3.4 on page 183. 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 235 CHAPTER 4 26. A car is stopped at a stoplight facing due east. 14. Compare the acceleration due to gravity and the gravitational field strength at the top of a tall skyscraper on Earth. 15. Consider the quantities gravitational force, mass, and gravitational field strength. Which of these quantities affects the inertia of an object? 16. Suppose an athlete were competing in the 2010 Winter Olympics in Vancouver and Whistler, British Columbia. Whistler has an elevation of 2182 m at the top and 652 m at the base. If the ski jumping and bobsled events are held near the top of the mountain rather than at the base, how might the results of these events be affected? 17. Two bags, each containing 10 oranges of equal mass, are hung 4 m apart. In a small group, describe two situations, one involving mass and the other involving separation distance, that would double the gravitational force exerted by one bag on the other. Explain your answer. 18. A student working on a satellite problem got an Ns2 m answer of 57.3. What physical
quantity was the student solving for? 19. Use an example to explain the meaning of the statement: “The gravitational force exerted by Mars on a space probe varies inversely as the square of the separation distance between the centre of Mars and the centre of the probe.” 20. Is an object in free fall weightless? Explain your reasoning. 21. Compare the gravitational force exerted by Earth (mass M) on two satellites (masses m and 2m) in orbit the same distance from Earth. 22. Compare the magnitude of Earth’s gravitational field strength at the equator and at the North Pole. Explain your answer to a classmate. 23. On Earth, how does the mass of an object affect the values of the quantities below? Explain your answers. (a) acceleration due to gravity (b) gravitational field strength Applications 24. Two horizontal forces act on a soccer player: 150 N [40.0] and 220 N [330]. Calculate the net force on the player. 25. Calculate the acceleration of a 1478-kg car if it experiences a net force of 3100 N [W]. When the light turns green, the car gradually speeds up from rest to the city speed limit, and cruises at the speed limit for a while. It then enters a highway on-ramp and gradually speeds up to the highway speed limit all the while heading due east. Sketch a free-body diagram for the car during each stage of its motion (five diagrams in total). 27. A net force of magnitude 8.0 N acts on a 4.0-kg object, causing the velocity of the object to change from 10 m/s [right] to 18 m/s [right]. For how long was the force applied? 28. Two people, on opposite banks, are towing a boat down a narrow river. Each person exerts a force of 65.0 N at an angle of 30.0 to the bank. A force of friction of magnitude 104 N acts on the boat. (a) Draw a free-body diagram showing the horizontal forces acting on the boat. (b) Calculate the net force on the boat. 29. A force acting on train A causes it to have an acceleration of magnitude 0.40 m/s2. Train A has six cars with a total mass of 3.0 105 kg, and a locomotive of mass 5.0 104 kg. Train B has a locomotive of the same mass as train A, and four
cars with a total mass of 2.0 105 kg. If the same force acts on train B, what will be its acceleration? Ignore friction. 30. A submarine rescue chamber has a mass of 8.2 t and safely descends at a constant velocity of 10 cm/s [down]. If g 9.81 m/s2, what is the upward force exerted by the cable and water on the chamber? 31. A 240-kg motorcycle and 70-kg rider are travelling on a horizontal road. The air resistance acting on the rider-bike system is 1280 N [backward]. The road exerts a force of static friction on the bike of 1950 N [forward]. What is the acceleration of the system? 32. The velocity of a 0.25-kg model rocket changes from 15 m/s [up] to 40 m/s [up] in 0.60 s. Calculate the force that the escaping gas exerts on the rocket. 33. Two boxes, A and B, are in contact and initially stationary on a horizontal, frictionless surface. Box A has a mass of 60 kg and box B a mass of 90 kg. A force of 800 N [right] acts on box A causing it to push on box B. (a) What is the acceleration of both boxes? (b) What is the magnitude of the action-reaction forces between the boxes? Unit II Dynamics 235 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 236 34. A person exerts a force of 1.5 N [right] to pull a 2.0-kg block of glass at constant velocity along a horizontal surface on the Moon (gMoon kinetic friction for the glass on the surface? 1.62 m/s2). What is the coefficient of 35. Three oak blocks, mA 4.0 kg, mB 3.0 kg, are positioned next to each other on mC a dry, horizontal oak surface. Use the data from Table 3.4 on page 183. 6.0 kg, and (a) What horizontal force must be applied to accelerate the blocks at 1.4 m/s2 [forward], assuming the blocks are moving at a constant velocity? (b) Calculate the force exerted by mB on mC. (c) Calculate the force exerted by mB on mA. 38. Three objects, A, B, and C, are connected together by light strings that
pass over light, frictionless pulleys. The coefficient of kinetic friction for object B on the surface is 0.200. (a) What is the acceleration of object B? (b) What is the tension in each string? Explain why the tensions are different. (c) Draw a free-body diagram for object B. (d) Identify four action-reaction pairs associated with object B. mB 2.0 kg B Fapp B A C A mA 6.0 kg C 4.0 kg mC 36. A 10.0-kg block is placed on an incline forming an angle of 30.0 with the horizontal. Calculate the acceleration of the block if the coefficient of kinetic friction for the block on the incline is 0.20. m 30.0° 37. A rehabilitation clinic has a device consisting of a light pulley attached to a wall support. A patient pulls with a force of magnitude 416 N. The rope exerts a force of friction of magnitude 20 N on the pulley. Calculate the acceleration of mA and mB. wall support shaft mA 15 kg mB 20 kg A B 236 Unit II Dynamics 39. The gravitational force on an object located 2rEarth from Earth’s centre is 200 N [toward Earth’s centre]. What is the gravitational force if the object is 10rEarth from Earth’s centre? 40. A 50-kg diver steps off a 9.0-m diving tower at the same time as a 100-kg diver. Work with a partner to compare the times taken for the two divers to reach the water. Ignore air resistance. 41. Skylab 1, the first American space station, had a mass of about 68 t. It was launched into orbit 435 km above Earth’s surface. Calculate the gravitational field strength at the location of Skylab 1 at this altitude. Use the data from Table 4.1 on page 218. 42. A spring scale is used to measure the gravitational force acting on a 4.00-kg silver block on Earth’s surface. If the block and spring scale are taken to the surface of Mars, by how much does the reading on the spring scale change? Use the data from Table 4.1 on page 218. 43. A 60-kg student is standing on a scale in an elevator on Earth. What will be the reading on the scale when the elevator is (a) (i) moving down at constant speed
? (ii) at rest at a floor? (iii) accelerating at 4.9 m/s2 [up]? (iv) accelerating at 3.3 m/s2 [down]? (b) What is the student’s apparent weight and true weight in all the situations in part (a)? 04-Phys20-Chap04.qxd 7/24/08 10:59 AM Page 237 44. A 60-kg skydiver falls toward Earth with an unopened parachute. The air resistance acting on the skydiver is 200 N [up]. What is the true weight and acceleration of the skydiver? 45. A group of tourists on a ledge overlooking Pulpit Rock, Northwest Territories, dislodge a 25-kg boulder. The rock takes 8.0 s to fall 300 m into the water below. At this location, the gravitational field strength is 9.81 N/kg [down]. (a) Calculate the average acceleration of the boulder. (b) Calculate the average air resistance acting on the boulder. (c) What was the average apparent weight of the boulder during its fall? Extensions 46. The value of g on the Moon is less than that on Earth. So a pendulum on the Moon swings slower than it would on Earth. Suppose a pendulum is 36 cm long. Use the equation T 2 l g to calculate the period, T, at the equator on Earth and on the Moon. Use the data from Table 4.1 on page 218. 47. During the last seconds of a hockey game, the losing team replaces their goalie with a good shooter. The other team shoots the 150-g puck with an initial speed of 7.0 m/s directly toward the unguarded net from a distance of 32 m. The coefficient of kinetic friction for the puck on the ice is 0.08. (a) What is the force of kinetic friction acting on the puck? (b) What is the acceleration of the puck? (c) How long does it take the puck to stop? (d) Will the puck reach the net if no other player touches it? 48. Construct a gathering grid to distinguish among Newton’s three laws. In the left column, identify the criteria you will use to compare and contrast the laws. Add three additional columns, one for each law. Then place checkmarks in the appropriate columns to compare the laws. 49. During the 2000 Sydney Olympics, some swimmers wore special swim
suits designed to reduce water resistance. Compare the arguments that people might make to defend or oppose the standardization of athletic equipment in the interests of fair play. 50. List two different stakeholders in the airbag debate and describe how their positions on the issue compare. 51. The G rocket of the former Soviet Union has a mass of about 3.8 106 kg and its first-stage engines exert a thrust of about 5.0 107 N [up]. (a) What is the true weight of the rocket on Earth’s surface? (b) Calculate the net force acting on the rocket at liftoff. (c) Calculate the initial acceleration of the rocket. (d) What should happen to the acceleration if the force exerted by the engines remains constant as the fuel burns? (e) Why is the first stage jettisoned after the fuel is consumed? 52. Suppose the mass of the person sitting next to you is 70 kg and the separation distance between you and that person is 1.0 m. The mass of Mars is 6.42 1023 kg and the separation distance between Mars and Earth is 2.3 1011 m. Compare the gravitational force exerted by Mars on you and the gravitational force exerted by the person sitting next to you on you. 53. In a small group, research the materials being used to make artificial joints such as hips and knees. Find out how they are designed to provide enough friction for stability but not so much friction that the joints cannot move. Begin your search at www.pearsoned.ca/school/physicssource. 54. Manufacturers of skis recommend different waxes for different snow temperatures. Design and carry out an experiment to test the recommendations for three different waxes. 55. Research gait analysis is the study of how humans walk and run. This topic is central to physiotherapy, orthopedics, the design of artificial joints and sports footwear, and the manufacture of orthotics. How do Newton’s three laws apply to gait analysis? Interview people associated with rehabilitation and sports. Write a brief summary of your findings. Begin your search at www.pearsoned.ca/school/physicssource. e TEST To check your understanding of dynamics, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit II Dynamics 237 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 238 U N I T III
Circular Motion, Circular Motion, Work, and Work, and Energy Energy The International Space Station is a silent companion to Earth, placed into an orbit that is a precise balance of kinetic and gravitational potential energies. The International Space Station stays in orbit because physicists and engineers applied the laws of physics for circular motion and conservation of energy to determine the satellite’s speed and height above Earth. e WEB To learn more about the types of satellites placed in orbit, and the paths that they take, follow the links at www.pearsoned.ca/school/physicssource. 238 Unit III 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 239 Unit at a Glance C H A P T E R 5 Newton’s laws can explain circular motion. 5.1 Defining Circular Motion 5.2 Circular Motion and Newton’s Laws 5.3 Satellites and Celestial Bodies in Circular Motion C H A P T E R 6 In an isolated system, energy is transferred from one object to another whenever work is done. 6.1 Work and Energy 6.2 Mechanical Energy 6.3 Mechanical Energy in Isolated and Non-isolated Systems 6.4 Work and Power Unit Themes and Emphases • Energy, Equilibrium, and Systems • Nature of Science • Scientific Inquiry Focussing Questions This unit focusses on circular motion, work, and energy. You will investigate the conditions necessary to produce circular motion and examine some natural and human examples. You will consider energy, its transfer, and how it interacts with objects. As you study this unit, consider these questions: • What is necessary to maintain circular motion? • How does an understanding of conservation laws contribute to an understanding of the Universe? • How can mechanical energy be transferred and transformed? Unit Project Building a Persuader Apparatus • When you have finished this unit, you will understand how energy is transferred when objects interact. You will be able to use this understanding in the design and construction of a persuader apparatus that is able to protect its passenger from injury in different types of collisions. Unit III Circular Motion, Work, and Energy 239 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 240 Newton’s laws can explain circular motion. If humans hadn’t invented the wheel — the first circular motion machine — it’s hard to imagine what life would be like
. The number of devices that make use of the properties of circular motion is almost too large to count. Bicycles, gears, drills, transmissions, clutches, cranes, watches, and electric motors are just a few examples. The wheel and the many technologies derived from it are a uniquely human creation (Figure 5.1). But the principles of circular motion have always existed in nature. In fact, as you read this, you are spinning around in a large circle as Earth rotates on its axis. In Alberta, your speed is approximately 1000 km/h because of this rotation. At the same time, you are flying through space at an amazing speed of approximately 107 000 km/h as Earth revolves around the Sun. And you thought your car was fast! How is circular motion unique? What are the properties of objects moving with circular motion? In this chapter, you will explore the physics of circular motion that define and control these many technologies, as well as the motion of the planets in the solar system. C H A P T E R 5 Key Concepts In this chapter, you will learn about: uniform circular motion planetary and satellite motion Kepler’s laws Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe uniform circular motion as a special case of two-dimensional motion explain centripetal acceleration and force explain, quantitatively, the relationships among speed, frequency, period, and radius for circular motion explain, qualitatively, uniform circular motion using Newton’s laws of motion explain, quantitatively, the motion of orbiting bodies by using circular motion to approximate elliptical orbits predict the mass of a celestial body from orbital data explain, qualitatively, how Kepler’s laws were used to develop Newton’s universal law of gravitation Science, Technology, and Society explain the process of scientific inquiry illustrate how science and technology are developed to meet society’s needs and expand human capabilities analyze circular motion in daily situations Figure 5.1 240 Unit III 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 241 5-1 QuickLab 5-1 QuickLab Characteristics of Circular Motion Problem In what direction will an object moving in a circular path go when released? Materials a marble a circular barrier (e.g., rubber tubing, embroidery hoop, flexible toy car tracks) Procedure 1 Place the circular barrier on an unobstructed section of a table or the floor, then place the marble against the inside rim of
the barrier (Figure 5.2). 2 You will be rolling a marble around inside the barrier. Before you do, predict where you think the marble will go when you remove the barrier. Now roll the marble around the inside rim of the barrier. 3 As the marble is rolling around the rim, lift the barrier and make a note of the direction the marble rolls. Also pay attention to the position of the marble when you lift the barrier. 4 Sketch the circular path that the marble took inside the barrier, the position of the marble when you lifted the barrier, and the path that it rolled after the barrier was lifted. 5 Repeat steps 3 and 4 several times. Each time release the marble when it is in a different position. Figure 5.2 Questions 1. Was your prediction correct? Why or why not? 2. What similarities, if any, exist between your sketches? 3. What conclusions can you draw about the motion of the marble when it was released? 4. In each sketch that you made, draw a line from the centre of the circular motion to the position of the marble when it was released. What is the angle between this line and the path of the marble when it was released? Think About It 1. Can you predict the position the marble would have to be in for it to move away from your body when it was released? 2. What can you say about the direction of the velocity of the marble at any moment in its circular path? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 5 Newton’s laws can explain circular motion. 241 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 242 info BIT A boomerang has two axes of rotation: one for the spin given to the boomerang as it leaves the hand, and the other for the large circle in which it moves. axle axis of rotation Figure 5.3 The axle of a wheel is part of the axis of rotation. axle: shaft on which a wheel rotates axis of rotation: imaginary line that passes through the centre of rotation perpendicular to the circular motion uniform circular motion: motion in a circular path at a constant speed 5.1 Defining Circular Motion The bicycle is not the most sophisticated machine ever created, but it does have a lot going for it. It is easy to maintain; it does not require any fuel;
it is a very efficient form of transportation; and it is full of parts moving in circular motion. Perhaps most importantly for this lesson, it is easy for us to examine. When you pedal a bike, the force is transferred through the chain and the wheel turns (Figure 5.3). The wheel transmits the force to the road, which, according to Newton’s third law, pushes back, and the bicycle moves forward. The wheel’s axle is referred to as the axis of rotation because the entire wheel rotates around this shaft. If the wheels of the bike are moving at a constant speed, they are moving with uniform circular motion. That means their rotational speed is uniform. Because the wheel is circular, it can also provide a constant and uniform force to the road. This is important if you want the bike to move forward at a uniform speed. In the sections that follow, we will restrict our study of circular motion to two dimensions. In other words, the circular motion described in this chapter has only one axis of rotation and the motion is in a plane. Speed and Velocity in Circular Motion If you ride through puddles after a rainfall, you’ll come home with muddy water splashed on your back. Why is this? It has to do with the properties of an object moving in circular motion. As the wheel passes through the puddle, some of the water adheres to it. As the wheel rotates upward out of the puddle, excess water flies off. When it flies off, it moves along a path that is tangential to the wheel. Recall from Unit I that a tangent is a line that touches the circle at only one point and is perpendicular to the radius of the circle. In the case of the bicycle, that is the point where the water drops break free of the wheel. Unless the splashguard is big enough, it will not protect you from all the water that is flying off the wheel (Figure 5.4). path of the water drops when they leave the wheel at point A radial line A A tangential line Figure 5.4 Water that flies off the wheel at point A hits the rider in the back. The direction that the water drops fly is determined by the place where they leave the wheel. Figure 5.5 Water leaves the wheel at point A and flies off at a tangent. 242 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:
50 PM Page 243 PHYSICS INSIGHT A line tangent to the circle represents the velocity vector. It is perpendicular to the radial line at that point. v A r Figure 5.6 The velocity vector v shows the velocity of a particle at any instant as it moves on a circular path. A line drawn from point A in Figure 5.5 to the centre of the circle is called a radial line. The radial line and the tangential line are perpendicular to each other when they intersect at the same point on the circle. The tangential line represents the direction that the water drops are moving at any instant. The speed of the water drops is determined by their speed at the wheel. It is worth taking some time to review the difference between speed and velocity. Speed is a magnitude only (a scalar quantity) that does not have a direction. If a wheel is rotating with uniform circular motion, then the speed is constant, even though the wheel is continually turning and the water on the wheel is continually changing direction. Velocity, on the other hand, has a magnitude and a direction: it is a vector quantity. Because of this, even though the wheel is spinning at constant speed, the velocity of the water on the wheel is continually changing as the wheel’s direction of motion changes. That is, the water on the wheel is continually accelerating. It is correct to say that the speed of the water drops on the wheel represents the magnitude of the velocity. Knowing that the object moves in a circular path is often sufficient. However, we must specify the object’s velocity if we need to know its speed and direction at any instant in its circular path. Since we can assume the speed and direction of the water drop are known at any instant, we know its velocity. A velocity vector can be drawn at position A, and the drawing can be simplified as shown in Figure 5.6. Concept Check 1. 2. Imagine a Frisbee in level flight. Where is its axis of rotation, and what is its orientation? Identify all the axes of rotation in a typical bicycle. Figure 5.7 A mountain bike has many axes of rotation. Centripetal Acceleration and Force Now imagine that you drive over a small pebble that gets lodged in the treads of your bicycle wheel. As you ride, the pebble circles around with the wheel, as shown in Figure 5.8. At one moment it is in position A, and a fraction of a second later, it is in position
B. A short time has passed as the pebble moved from point A to point B. This small change in time is written as t. Chapter 5 Newton’s laws can explain circular motion. 243 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 244 The pebble has experienced a very small displacement from point A to point B, which is written as d. The velocities of the pebble at point A and point B are v B, respectively (Figure 5.9). C represents the centre of the circle (the axis of rotation). A and v vA B vB A d r r C vB –vA v C t B vB vA A Figure 5.8 A pebble caught in the wheel of a bike moves from position A to B in a small time t, and experiences a change in velocity. Figure 5.9 As the pebble moves from point A to point B, it moves through angle and experiences a change in velocity. Figure 5.10 The change in velocity, v, points inward. As t decreases, the angle between B will become smaller and vwill point A and v v toward the centre of the circle. The math to show this is beyond the scope of this book. The speed of the pebble does not change, but its direction does. Therefore, its velocity also changes. The change in velocity (vv) can best be shown by subtracting the velocity at A from the velocity at B (using the rules of graphical vector addition) as shown in Figure 5.10. Angle is the same in both Figures 5.9 and 5.10, and the two triangles are similar. Something subtle but significant happens when we subtract the two velocity vectors. The change in velocity, vv, points inward. As the interval of time, t, becomes smaller, vv begins to point toward the centre of the circle! This can be shown using calculus, but this is beyond the scope of this book. Velocity and Acceleration Toward the Centre The changing velocity (v ) represents the change in direction of the object, not its speed. If an object has a changing velocity, then it must be accelerating. Since the changing velocity is pointing inward toward the centre of the circle, the acceleration must also be in that direction (Figure 5.11). It is called centripetal acceleration (a c ). For an object to move with circular motion, it must experience
centripetal acceleration. If the circular motion is uniform, then so is the centripetal acceleration. According to Newton’s second law, if a mass is accelerating, it must also experience a non-zero net force. This non-zero net force is c ). In our example, the pebble stuck in the called the centripetal force (F wheel is experiencing a force, exerted by the rubber treads, that attempts to pull it toward the centre of the circle. This is the centripetal force. Why doesn’t the pebble actually move toward the centre of the wheel? It does, in a way. Remember: if the pebble were to break free of the tire’s grip, it would fly off at a tangent to its circular motion. While it remains stuck in the tire, it is forced to follow the circular path because the centripetal force attempts to pull it toward the centre. In other words, centripetal force is pulling the pebble toward the centre of the circle while at the same time, the pebble is moving off in a direction at a tangent to the circle. The result is the circular path in which it actually moves. v ac Figure 5.11 Although the velocity of an object moving with uniform circular motion is tangential, the centripetal acceleration (and centripetal force) is acting toward the centre of the circle. centripetal acceleration: acceleration acting toward the centre of a circle; centre-seeking acceleration centripetal force: force acting toward the centre of a circle causing an object to move in a circular path 244 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 245 5-2 Inquiry Lab 5-2 Inquiry Lab Speed and Radius Question For an object moving with uniform circular motion, what relationship exists between the radius of its path and its speed? (Assume the object is experiencing a constant centripetal force.) Hypothesis State a hypothesis relating the radius and speed. Remember to use an “if/then” statement. Variables The variables in this lab are the radius of the circular path, mass of the rubber stopper, mass of the hanging weight, number of revolutions, elapsed time, period, and speed. Read the procedure and identify the controlled, manipulated, and responding variables. Materials and Equipment 1-
hole rubber stopper (mass 25 g) 1.5 m of string or fishing line small-diameter plastic tube 100-g mass metre-stick felt marker safety goggles stopwatch CAUTION: Remember to swing the rubber stopper over your head and in a place clear of obstructions. Table 5.1 Data for 5-2 Inquiry Lab Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Procedure 1 Copy Table 5.1, shown at the bottom of this page, into your notebook. 2 Secure the rubber stopper to one end of the string. 3 Run the other end of the string through the plastic tube and attach the 100-g mass to it. 4 Hold the end of the string attached to the stopper at the zero mark of a metre-stick laid on a table. The zero mark of the ruler should line up with the centre of the stopper. While holding the stopper in position, pull the string taut along the ruler. 5 With the felt marker, mark the string at 20, 30, 40, 50, and 60 cm. 6 Adjust the string’s position so that the 20-cm mark is positioned on the lip of the plastic tube. Record 20 cm in the “Radius” column of the table. 7 Grasp the plastic tube in one hand and pinch the string to the lip of the tube using your thumb or forefinger. 8 Put on your safety goggles. Begin spinning the rubber stopper in a horizontal circle above your head as you release the string. Make sure the 100-g mass is hanging freely (Figure 5.12). At first, you may have to pull the string up or down using your other hand to position the mark as the stopper is spinning. Radius (m) Time for 20 Revolutions (s) Time 1 Time 2 Average Time (s) Period (s) Speed (m/s) Figure 5.12 Chapter 5 Newton’s laws can explain circular motion. 245 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 246 9 Adjust the rate at which you spin the rubber stopper so that the string does not slip up or down and the mark stays in position at the lip of the tube. Once you have reached a steady rate, your partner can begin timing. Do not pull the string up or down with the other hand. 10 Your partner should time 20 complete revolutions of the rubber stopper using a
stopwatch. While your partner does this, be sure to monitor the speed of the stopper so that the mark does not move off the lip. Record the time in the “Time 1” column of the table. 2. For each trial, divide the average time by the number of revolutions the stopper made. Record the values for the time it takes to make one revolution in the “Period” column. 3. For each trial, determine the speed of the stopper. The distance travelled is the circumference of the circle and the time is the period. 2 r Use the equation v, and record the value in T the “Speed” column. 4. Identify the controlled, responding, and manipulated 11 Repeat step 10 and record the time in the “Time 2” variables. column of the table. 12 Increase the radius by 10 cm, and record this radius in the “Radius” column of the table. Repeat steps 7 to 12 until all radii are used. Analysis 1. For each trial, average the two times and place the result in the “Average Time” column. 5. Identify the force that acted as the centripetal force, and determine its value. 6. Plot a graph of velocity versus radius. Remember to plot the manipulated variable on the horizontal axis and the responding variable on the vertical axis. 7. Complete the statement, “The speed varies with …” 8. Was your hypothesis accurate? If not, how can you modify it to reflect your observations from this lab? motion of ball if released force exerted by hand on rope (action force) Figure 5.13 The hand exerts a centripetal force on the rope and ball, but feels the reaction force exerted by the rope. This leads to a false impression that the centripetal force is acting outward. force exerted by rope on hand (reaction force) Misconceptions About Centripetal Force A common misconception is that centripetal force acts radially outward from the centre of a circle. This is not what happens. The change in the velocity of the object is inward, and therefore, so is the centripetal acceleration and force. What causes confusion is that when you spin an object in a circle at the end of a rope, you feel the force pulling outward on your hand (Figure 5.13). This outward pull is mistakenly thought to be the centripetal force. Newton’s
third law states that for every action there is an equal and opposite reaction. If we apply this law to our example, then the action force is the hand pulling on the rope to make the object move in a circle. This is the centripetal force. The reaction force is the force the rope exerts on your hand. It is outward. This is the force that people often believe is the centripetal force. The rope would not exert a force on your hand unless your hand exerted a force on the rope first. Another misconception is that centripetal force is a unique and separate force responsible for circular motion. This is not true. It is best to think of centripetal force as a generic term given to any force that acts toward the centre of the circular path. In fact, the translation from Latin of the word centripetal is “centre seeking.” 246 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:50 PM Page 247 Many different forces could actually be the centripetal force. For example, when a car turns a corner, the frictional force of the tires on the road acts as the centripetal force. If you spin an object around in a horizontal circle on a rope, the tension of the rope is the centripetal force. The force of gravity the Sun exerts on the planets is another example of a centripetal force. Sometimes several forces working together act as a centripetal force. For an object spinning in a vertical circle on a rope, two forces, gravity and tension, work together to act as the centripetal force at the top of the circle. This is because the centripetal force is a net force. It is often convenient to use it in place of the actual force or forces acting toward the centre. Table 5.2 summarizes circular motion quantities and their directions. Concept Check A pebble caught in the tread of a tire experiences a centripetal force as the tire turns. What force is responsible for it? Table 5.2 Circular Motion Quantities and Their Direction Quantity Velocity (v) Centripetal acceleration (a) Centripetal force (F net) c or F Change in velocity (v) Direction tangential to the circle toward the centre toward the centre toward the centre 5.1 Check and Reflect 5.1 Check and Reflect PHYSICS
INSIGHT Some texts refer to centrifugal force. This refers to the reaction force that exists as a result of centripetal force. If the centripetal force is removed, there is no centrifugal force. e SIM For an interactive demonstration that explores the relationship among centripetal acceleration, force, and velocity, visit www.pearsoned.ca/school/ physicssource. Knowledge 1. Give an example of an object that moves with uniform circular motion and one that moves with non-uniform circular motion. 2. Identify the force acting as the centripetal force in each of the following situations: (a) A car makes a turn without skidding. (b) A ball is tied to the end of a rope and spun in a horizontal circle. (c) The Moon moves in a circular orbit around Earth. Applications 3. What is the relationship between the speed and radius of an object moving with uniform circular motion if the centripetal force is constant? 4. What is the relationship between the speed and velocity of an object moving with uniform circular motion? Extensions 5. Imagine pedalling a bike at a constant rate. Describe the motion of the bicycle if the wheels were not circular but oval. 6. Consider the relative speed between a pebble stuck in the tread of a bicycle tire and the ground. Explain why the pebble is not dislodged when it comes in contact with the ground. Suggest a method to dislodge the pebble while still riding the bike. e TEST To check your understanding of the definition of circular motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 5 Newton’s laws can explain circular motion. 247 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 248 5.2 Circular Motion and Newton’s Laws info BIT A spinning ball moving relative to the air causes an unbalanced force and the curving of the ball. A backspin on a ball makes it stay aloft longer. A topspin drives the ball downward. Soccer players put a spin on the ball when they kick it that makes it curve around opposing players. Professional golfers routinely strike the golf ball so that it has a spin that curves it into the wind or prevents it from rolling when it hits the ground. A baseball pitcher can throw a curving fastball at 144 km/h, making the batter’s job
of hitting the ball much more difficult (Figure 5.14). Figure 5.14 Pitchers can put a spin on a ball that causes it to curve. This curve is predictable. e WEB To learn more about the effect of putting a spin on a ball, follow the links at www.pearsoned.ca/school/ physicssource. In these sports and many others, putting a spin on the ball is an essential skill. Even though the pitch from a pitcher may be extremely difficult to predict, the behaviour of the ball isn’t. If players perform the same motion reliably when they kick, throw, or hit the ball, it will always behave the same way. That is why good players, when faced with the curving soccer ball or fastball, can anticipate where the ball will be and adjust their positions. It is accurate to say that the physical properties of a spinning ball or anything moving with circular motion can be predicted. In fact, the rotational velocity, frequency, centripetal force, and radius of a spinning object can be related mathematically. M I N D S O N Spinning Objects in Sports In groups of two or three, think of as many sports as you can that involve a spinning motion. It may be the player or an object that has the spin. Indicate what advantages the spinning motion has for the player and what type of motion is used to cause the spin. Discuss your answers with the class. 248 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 249 Period and Frequency of Circular Motion A baseball pitcher can throw a baseball at speeds of about 145 km/h. By flicking his wrist, he can give the ball a spin so that, in effect, it has two velocities: a velocity as it approaches the batter, and a rotational velocity because of its spin (Figure 5.15). The rotational velocity can be measured indirectly by measuring the time it takes for one complete rotation. One complete rotation is called a cycle or revolution, and the time for one cycle is the period (T), measured in s/cycle. (a) P (b) v (c) P v v P (d) v P cycle: one complete back-andforth motion or oscillation revolution: one complete cycle for an object moving in a circular path period: the time required for an object to make one complete oscillation (cycle) (
e) P v Figure 5.15 Point P on the spinning baseball makes one complete rotation from (a) to (e). The time for this is called the period and is measured in s/cycle. This is frequently abbreviated to s for convenience. If an object is spinning quickly, the period may be a fraction of a second. For example, a hard drive in a computer makes one complete revolution in about 0.00833 s. This value is hard to grasp and is inconvenient to use. It is often easier to measure the number of rotations in a certain amount of time instead of the period. Frequency (f) is a measurement that indicates the number of cycles an object makes in a certain amount of time, usually one second. The SI units for frequency are cycles/s or hertz (Hz). You might have noticed that the units for frequency are cycles/s while the units for period are s/cycle. Each is the inverse of the other, so the relationship can be expressed mathematically as: T 1 f or f 1 T You may also have seen rotational frequency expressed in rpm. Even though this unit for measuring frequency is not an SI unit, it is commonly used commercially in products such as power tools. An rpm is a revolution per minute and is different from a hertz. It represents the number of revolutions in one minute instead of one second, so it is always 60 times bigger than the value in Hz. A simple method can be used to convert Hz to rpm and vice versa: Hz 60 s/min 60 s/min rpm frequency: the number of cycles per second measured in hertz (Hz) rpm: revolutions per minute Chapter 5 Newton’s laws can explain circular motion. 249 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 250 Example 5.1 The hard drive in Figure 5.16 stores data on a thin magnetic platter that spins at high speed. The platter makes one complete revolution in 0.00833 s. Determine its frequency in Hz and rpm. Practice Problems 1. The propeller of a toy airplane rotates at 300 rpm. What is its frequency in hertz? 2. An electric motor rotates at a frequency of 40 Hz. What is its rotational frequency in rpm? 3. A medical centrifuge is a device that separates blood into its parts. The centrifuge can spin at up to 6.0 104 rpm. What is its frequency in hertz and what
is its period? Answers 1. 5.00 Hz 2. 2.4 103 rpm 3. 1.0 103 Hz; 1.0 103 s Given T 0.00833 s Required frequency in Hz and rpm T 1 f f 1 0.00833 s 120 Hz hard drive case Now convert the SI units of frequency to rpm: Hz 60 s/min 60 s/min rpm Analysis and Solution The frequency is the inverse of the period. Solve the frequency of the hard drive in the SI unit for frequency (Hz) and then convert the Hz to rpm. hard drive platter direction of rotation read/write head Figure 5.16 120 Hz 60 s min 7.20 103 rpm Paraphrase The frequency of the hard drive is 120 Hz or 7.20 103 rpm. Speed and Circular Motion At the beginning of this chapter you learned that, at this moment, you are moving at approximately 107 000 km/h as Earth moves in its orbit around the Sun. It’s hard to imagine that Earth is moving that fast through our solar system. How was the speed determined? The answer is the simple application of an equation you learned in section 1.2 of chapter 1: v d t where d is the distance travelled and t is the time that it takes the object to travel that distance. 250 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 251 In the case of circular motion, the distance around a circle is the circumference (C), given by C 2r. The time it takes for one revolution is the period (T). Therefore, the speed of anything moving with uniform circular motion can be described by the equation: v 2r T (1) where r is the radius in metres, T is the period in seconds, and v is the speed in metres per second. Let’s look at Earth as it follows a circular orbit around the Sun. Earth has an orbital radius of approximately 1.49 108 km, and makes one complete revolution in 365.24 days. By substituting these values into the equation for speed, we can do the following calculation: PHYSICS INSIGHT Remember: The SI units for distance and time are metres and seconds, respectively. v 2r T 2(1.49 1011 m) 365.24 24 60 60 s 2.97 104 m/s Then convert this to kilometres per hour: m 36 m k 2.97 104 m
s 1 0 0 10 1 s 00 1.07 105 km/h h Earth’s speed as it orbits the Sun is approximately 107 000 km/h. The speed of a planet as it rotates on its axis can be determined in the same way but varies depending on the latitude. We will explore the reasons for this later in “Centripetal Force, Acceleration, and Frequency” in section 5.2. Example 5.2 A pebble is stuck in the treads of a tire at a distance of 36.0 cm from the axle (Figure 5.17). It takes just 0.40 s for the wheel to make one revolution. What is the speed of the pebble at any instant? Given r 36.0 cm 0.360 m T 0.40 s Required speed (v ) of the pebble Analysis and Solution Determine the speed by using equation 1: r 2 v T 2(0.360 m) 0.40 s 5.7 m/s v 36.0 cm 36.0 cm 36.0 cm up left down right Figure 5.17 The speed of the pebble is determined by its distance from the axis of rotation, and the wheel’s period (T). Practice Problems 1. How much time does it take for the tires of a racecar to make one revolution if the car is travelling at 261.0 km/h and the wheels have a radius of 0.350 m? 2. In 2006, an Alberta astronomer discovered the fastest spinning collapsed star (called a pulsar) ever found. It has a radius of only 16.1 km and is spinning at a rate of 716 Hz (faster than a kitchen blender). What is its speed at its equator? Paraphrase The speed of the pebble caught in the tire tread is 5.7 m/s. Answers 1. 0.0303 s 2. 7.24 107 m/s Chapter 5 Newton’s laws can explain circular motion. 251 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 252 A Closer Look at Centripetal Acceleration and Force In Unit II, you learned that, just before a race, dragster drivers spin their tires to make them hot and sticky so that the coefficient of friction increases between the tire and the road. When the race starts, the tires will have a better grip and the dragster will be able to
accelerate at a greater rate. While the dragster performs the tire-spin, the tires change shape (Figure 5.18). The rear tires start off being fat and thick. During the spin they become thin and their diameter increases. Clearly, this must have something to do with the spinning motion of the wheel, and therefore centripetal force, but what? Figure 5.18 (a) At first, the dragster’s wheels have a low rotational speed and don’t stretch noticeably. The centripetal acceleration and force are small. (b) The dragster’s wheels are spinning very fast and stretch away from the rim. The centripetal acceleration and force are large. Before the tires start spinning, they are in their natural shape. When they are spinning, they experience a strong centripetal acceleration and force. Both act toward the centre of the wheel. Each tire is fastened to a rim, so the rim is pulling the tire inward. However, the tire moves the way Newton’s first law predicts it will — it attempts to keep moving in a straight line in the direction of the velocity. Thus the tire, being made of rubber, stretches. Dragster tires and the tires of trucks and passenger cars are designed to endure high speeds without being torn apart. The faster a wheel spins, the greater the centripetal acceleration and force. Car tires are thoroughly tested to ensure they can handle a centripetal force much greater than the centripetal force at the speed that you would normally drive in the city or on the highway. However, if you drove at a speed that exceeded the tires’ capabilities, the tires could be torn apart. Is speed the only factor that affects centripetal acceleration and force, or are there other factors? Are the factors that affect centripetal acceleration the same as those that affect the centripetal force? To answer these questions, let’s start by taking a closer look at centripetal acceleration. 252 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 253 Factors Affecting the Magnitude of Centripetal Acceleration The rotational speed is one factor that determines the magnitude of the centripetal acceleration, but what other factors play a role? To answer this question, look at Figure 5.19. It shows the two diagrams you saw earlier
as Figures 5.9 and 5.10 in section 5.1. Using these figures, we can derive an equation for centripetal acceleration. vA B vB A d r r C vB –vA v C Figure 5.19 An object following in a circular path moves through a displacement as there is a change in velocity v. The triangles formed by these two vectors will d help us solve for centripetal acceleration. As already stated, the two triangles are similar. Therefore, we can compare them. For convenience, the triangles have been redrawn below without the circles (Figure 5.20). Since vA and vB have the same value, we have dropped the designations A and B in Figure 5.20. We have omitted the vector arrows because we are solving only for the magnitude of the centripetal acceleration Figure 5.20 Triangles ABC and DEF are similar, so we can use a ratio of similar sides. Triangle ABC is similar to triangle DEF in Figure 5.20. Therefore, a ratio of similar sides can be created: d v r v or v d v r (2) Chapter 5 Newton’s laws can explain circular motion. 253 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 254 Remember, v is directed toward the centre of the circle. The time it took for the velocity vectors to move the small distance d can be designated t. In this time, the v vector was created. d, we can manipulate the equation so that: Since v t d t v (3) To find acceleration toward the centre of the circle (centripetal acceleration), divide the v by t. v t ac (4) Now we can substitute equations 2 and 3 into equation 4: v d r d v ac By taking the reciprocal of the denominator and multiplying the two fractions, we have: d v v d r ac Simplified, this becomes: ac 2 v r (5) where ac is the centripetal acceleration in metres per second squared toward the centre of the circle, v is the rotational speed of the object moving with uniform circular motion in metres per second, and r is the radius of the circular motion in metres. The centripetal acceleration depends only on the speed and radius of the circular motion. Mass does not affect it, just as the mass of a falling object does not affect the acceleration of gravity caused by Earth. A truck or a marble will
both experience a gravitational acceleration of 9.81 m/s2 [down]. Two objects of different masses moving in a circular path will experience the same centripetal acceleration if they have the same radius and speed. Mass does not affect centripetal acceleration, but companies that manufacture racing tires, jet engines, and other equipment know they cannot ignore it. In fact, the mass of a tire or fan blade is important to them. These companies are continually looking for ways to reduce mass without decreasing the strength of their products. Why? The answer has to do with the centripetal force that these devices experience. If a fan blade or tire has a large mass, it will experience a large centripetal force and might break apart at high speeds. Reducing the mass decreases the centripetal force these parts experience, but often with a trade-off in strength. Next, we will examine the factors that influence centripetal force. 254 Unit III Circular Motion, Work, and Energy 05-Phys20-Chap05.qxd 7/24/08 12:51 PM Page 255 Example 5.3 A DVD disc has a diameter of 12.0 cm and a rotational period of 0.100 s (Figure 5.21). Determine the centripetal acceleration at the outer edge of the disc. Given D 12.0 cm 0.120 m T 0.100 s Required centripetal acceleration (ac) Analysis and Solution The magnitude of the centripetal acceleration depends on speed and radius. Convert the diameter to a radius by dividing it by 2. D 12.0 cm T 0.100 s Figure 5.21 D r 2 0 m 0.12 2 0.0600 m Determine the speed of the outer edge of the disc: r 2 v T. 0 2( m) 00 6 0 0 s 0 1 0. 3.77 m/s Now use equation 5 to determine the centripetal acceleration: ac 2 v r m 2 3.77 s 0.0600 m 2.37 102 m/s2 Note that no vector arrows appear on ac or v because we are solving for their magnitude only. Paraphrase The centripetal acceleration at the edge of the DVD disc is 2.37 102 m/s2. Practice Problems 1. You throw a Frisbee to your friend. The Frisbee has a diameter of 28.0 cm and makes one turn in 0.110 s

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