9781118230725.pdf

9-2 NEWTON’S SECOND LAW FOR A SYSTEM OF PARTICLES 223

Checkpoint 2

Two skaters on frictionless ice hold opposite ends of a pole of negligible mass. An axis

runs along it, with the origin at the center of mass of the two-skater system. One skater,

Fred, weighs twice as much as the other skater, Ethel. Where do the skaters meet if (a)

Fred pulls hand over hand along the pole so as to draw himself to Ethel, (b) Ethel pulls

hand over hand to draw herself to Fred, and (c) both skaters pull hand over hand?

Sample Problem 9.03 Motion of the com of three particles

If the particles in a system all move together, the com moves
with them—no trouble there. But what happens when they
move in different directions with different accelerations?
Here is an example.
The three particles in Fig. 9-7aare initially at rest. Each
experiences an externalforce due to bodies outside the
three-particle system. The directions are indicated, and the
magnitudes are F 1 6.0 N,F 2 12 N, and F 3 14 N. What
is the acceleration of the center of mass of the system, and in
what direction does it move?

KEY IDEAS

The position of the center of mass is marked by a dot in the
figure. We can treat the center of mass as if it were a real
particle, with a mass equal to the system’s total mass M16 kg.
We can also treat the three external forces as if they act at the
center of mass (Fig. 9-7b).

Calculations: We can now apply Newton’s second law
to the center of mass, writing
(9-20)

or

so (9-21)

Equation 9-20 tells us that the acceleration of the
center of mass is in the same direction as the net external force
on the system (Fig. 9-7b). Because the particles are ini-
tially at rest, the center of mass must also be at rest. As the
center of mass then begins to accelerate, it must move off in
the common direction of and
We can evaluate the right side of Eq. 9-21 directly on
a vector-capable calculator, or we can rewrite Eq. 9-21 in
component form, find the components of and then find
Along the xaxis, we have



6.0 N(12 N) cos 4514 N

16 kg

1.03 m/s^2.

acom,x

F 1 xF 2 xF 3 x

M

a:com.

a:com,

F

:

a net.

:

com

F

:

net

a:com

:acom

F 1

:

F 2

:

F 3

:

M

.

F 1

:

F 2

:

F 3

:

Ma:com

F

:

netMa

:

com

(F

:

netma

:)

Figure 9-7(a) Three particles, initially at rest in the positions shown,

are acted on by the external forces shown. The center of mass (com)

of the system is marked. (b) The forces are now transferred to the

center of mass of the system, which behaves like a particle with a

massMequal to the total mass of the system. The net external force

F and the acceleration :acomof the center of mass are shown.

:

net

x

y

3

2

1

0

–1

–2

–3

–3 –2 –1 1 2 3 4 5

x

y

3

2

1

0

–3 –2 –1 1 2 3 4 5

45 °

com 8.0 kg

4.0 kg

4.0 kg

com

θ

M = 16 kg

(b)

(a)

F 1 F 2

F 3

F 3

F 1

F (^2) Fnet
acom
The com of the system
will move as if all the
mass were there and
the net force acted there.
Additional examples, video, and practice available at WileyPLUS
Along the yaxis, we have
From these components, we find that has the magnitude
(Answer)
and the angle (from the positive direction of the xaxis)
tan^1 (Answer)
acom,y
acom,x

 27 .

1.16 m/s^2 1.2 m/s^2

acom 2 (acom,x)^2 (acom,y)^2

:acom



0 (12 N) sin 45 0

16 kg

0.530 m/s^2.

acom,y

F 1 yF 2 yF 3 y

M