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displacement As along the diagonal of a parallelepiped whose sides are Az, A#,
and Az. In terms of the velocity, the displacement Az is the #-component of the
velocity times A#, and similarly for A¿ and Az:
Az = uy At, AU = uy At, Az =0; At. (9.4)
9-3 Components of velocity, acceleration, and force
In Eq. (9.4) te heœue resolued the uelocitụ imito components by telling how fast
the object is moving in the #ø-direction, the -direction, and the z-direction. The
velocity is completely specifed, both as to magnitude and direction, IÝ we give
the numerical values of its three rectangular components:
U„ = dw/dt, 0y = dụ/dt, Uy = dz/dl. (9.5)
On the other hand, the speed of the object 1s
ds/đdt = |u| = viuà + 02 + tỷ. (9.6)
Next, suppose that, because of the action of a force, the velocity changes
to some other direction and a diferent magnitude, as shown in Fig. 9-2. We
--- Trang 181 ---
(T—#!
„ 1 Ị mm I
z1 l Tý Hi I
(-1---1 ~TrJ
I ———_—kL -
L7 Lự
Vh_____#
Fig. 9-2. A change in velocity in which both the magnitude and
direction change.
can analyze this apparently complex situation rather simply 1Ÿ we evaluate the
changes in the z-, -, and z-components of velocity. The change in the component
of the velocity in the z-direction in a time Af is Au„ = a„ At, where a„ is what
we call the #-component of the acceleration. 5imilarly, we see that Auy = ay At
and Aø; = ø; Ai. In these terms, we see that NÑewton?s Second Law, in saying
that the force is in the same direction as the acceleration, is really three laws, In
the sense that the component of the force in the z-, -, or z-direction is equal to
the mass times the rate of change of the corresponding component of velocity:
F„, = m(du„/dt) = m(dŠ#/dt?) = ma,
F„ = m(duy/dt) = m(d®u/dt?) = may, (9.7)
F, = m(du; /dt) = m(dŠz (dt?) = ma,.
Just as the velocity and acceleration have been resolved into components by
projecting a line sepment representing the quantity, and its direction onto three
coordinate axes, so, in the same way, a force in a given direction is represented
by certain components in the z-, -, and z-directions:
Tạ —= F'cos(œ, F),
Tụ = Fcos(u, `), (9.8)
Ty = Fcos(z,F),
--- Trang 182 ---
where #' is the magnitude of the force and (z, #) represents the angle between
the z-axis and the direction of Ƒ', etc.
Newton?s Second Law is given in complete form in Bq. (9.7). IÝ we know
the forces on an object and resolve them into z-, -, and z-components, then
we can find the motion of the object from these equations. Let us consider a
simple example. Suppose there are no forces in the - and z-directions, the only
force being in the z-direction, say vertically. Equation (9.7) tells us that there
would be changes in the velocity in the vertical direction, but no changes in
the horizontal direction. “This was demonstrated with a special apparatus In
Chapter 7 (see Eig. 7-3). A falling body moves horizontally without any change
in horizontal motion, while it moves vertically the same way as it would move
1f the horizontal motion were zero. In other words, motions in the z-, -, and
z-directions are independent If the ƒorces are not connected.
9-4 What is the force?
In order to use Newton”s laws, we have to have some formula for the force;
these laws say pay aœftenlion to the ƒorces. TỶ an object 1s accelerating, some
agency is at work; ñnd it. Our program for the future of dynamiecs must be to
imd the laus for the Ƒorce. Newton himself went on to give some examples. In the
case of gravity he gave a specifc formula for the force. In the case of other forces
he gave some part of the information in his Third Law, which we will study in
the next chapter, having to do with the equality of action and reaction.
Extending our previous example, what are the forces on objects near the
earth”s surface? Near the earth's surface, the force in the vertical direction due to
gravity is proportional to the mass of the object and is nearly independent of height
for heights small compared with the carths radius l: ' = GmM/R2 = mg,
where g = GM/R>? is called the acceleration oƒ graoit. Thus the law of gravity
tells us that weight is proportional to mass; the force is in the vertical direction
and is the mass times g. Again we find that the motion in the horizontal direction
1s at constant velocity. The interesting motion is in the vertical direction, and
Newton's Second Law tells us
mg = m(d°z/dt?). (9.9)
Cancelling the rm”s, we ñnd that the acceleration in the z-direction is constant
and equal to g. 'Phis is of course the well known law of free fall under gravity,
--- Trang 183 ---
EQUILIBRIUM
: x POSITION
Fig. 9-3. A mass on a spring.
which leads to the equations
U„ = 0o + g,
# = #o + 0of + šg2. (9.10)
As another example, let us suppose that we have been able to build a gad-
get (Eig. 9-3) which applies a force proportional to the distance and directed
oppositely—a spring. If we forget about gravity, which is of course balanced out
by the initial stretch of the spring, and talk only about ezcess forces, we see that
1f we pull the mass down, the spring pulls up, while if we push it up the spring
pulls down. This machine has been designed carefully so that the force is greater,
the more we pull it up, in exact proportion to the displacement from the balanced
condition, and the force upward is similarly proportional to how far we pull down.
Tf we watch the dynamies of this machine, we see a rather beautiful motion——up,
down, up, down, ... 'Phe question is, will Newton”s equations correctly describe
this motion? Let us see whether we can exactly calculate how it moves with this
periodic oscillation, by applying Newton”s law (9.7). In the present instance, the
equation 1s
— kœ& = rm(du„/dt). (9.11)
Here we have a situation where the velocity in the z-direction changes at a rate
proportional to z. Nothing will be gained by retaining numerous constants, so
we shall imagine either that the scale of time has changed or that there is an