Example 7-11 An Ant on a Twig
A narrow, uniform twig is 15 cm long and has a mass of 2.2 g. The twig floats in a
swimming pool so that its left-hand end touches the edge of the pool (Figure 7-23a).
A 0.35-g ant, initially resting at the right-hand end of the twig, crawls along the twig
toward the left-hand end. Assuming that the twig experiences no resistive forces due
to the water, how far is the left-hand end of the twig from the edge of the pool when
the ant reaches that end (Figure 7-23b)?
(a) Initial
mA = 0.35 g
Ant
Twig
mT = 2.2 g
15 cm
(b) Final
Figure 7-23 ​How far does the twig move? When the ant walks to the left along the length of the
floating twig, how far to the right does the twig move?
Set Up
By assumption there are no horizontal resistive forces, and the water exerts an upward
buoyant force that balances the downward
gravitational force. So there is zero net external force on the system of ant and twig. (The
ant and twig exert forces on each other as
the ant walks. But these forces are internal to
the system and so aren’t included in the sum
s
a Fexternal on system.) So Equation 7-34 tells us
that the center of mass of the system, which
is initially at rest, does not accelerate and so
remains at rest at the same position given by
Equation 7-28. We’ll use this to determine
the final position of the ant.
d=?
How the net external force on a
system affects the center of mass:
B (buoyant force)
sCM
Dv
s
a Fexternal on system = Mtot Dt
sCM (7-34)
= Mtot a
side view
mA g
In this situation a Fsexternal on system = 0,
so the center of mass does not accelerate
and so remains at rest at the same position.
mT g
Position of the center of mass of a system:
m3
m1
m2
mN
x1 +
x2 +
x + c +
x
Mtot
Mtot
Mtot 3
Mtot N
1 N
(7-28)
=
mi xi
Mtot ia
=1
xCM =
Total mass of the system of two objects:
Mtot = m1 + m2
(7-27)
Length of twig is L = 15 cm = 0.15 m
Mass of ant: mA = 0.35 g
Mass of twig: mT = 2.2 g
Solve
First calculate the position of the center of
mass when the ant and twig are in their
initial positions. Since the center of mass
doesn’t move, this will also be its position
when the ant begins walking.
x=0
Choose the x axis to lie along the
length of the twig, with x = 0 at the
edge of the pool.
The ant is small enough to treat as a
point located at the end of the twig:
xA = L
L = 15 cm
xcm = 8.5 cm
x
xT =
x
L
2
xA = L
The twig is uniform, so its center of
mass is at its geometrical center:
xT =
L
2
Total mass of the system:
Mtot = mA + mT
Position of the center of mass:
mA L + mT L>2
mA
mT
x +
x =
xCM =
Mtot A
Mtot T
mA + mT
10.35 g2 115 cm2 + 12.2 g2 115 cm2 >2
=
= 8.5 cm
0.35 g + 2.2 g
The center of mass is between the center of the twig and the position of
the ant, and closer to the center of the more massive twig.
As the ant walks to the left, it pushes backward on the twig and so the twig moves
to the right. However, the center of mass
remains in the same position. Solve for the
final distance d from the edge of the pool to
the ant.
When the ant has completed its
walk to the left-hand end of the
twig, the new positions of the ant
and the center of the twig are
xA = d
xcm = 8.5 cm
x x
xA = d
xT = d +
x
L
2
L
2
The new position of the center of mass is
xT = d +
xCM =
mA d + mT 1d + L>22
mA
mT
xA +
xT =
Mtot
Mtot
mA + mT
This must be equal to the original position of the center of mass as
found above, so
mA d + mT 1d + L>22
mA + mT
=
mA L + mT L>2
mA + mT
Multiply through by mA + mT:
mA d + mT 1d + L>22 = mA L + mT L>2
The term mT L>2 is on both sides of the equation, and so cancels:
mAd + mTd = mAL
Solve for the distance d:
d =
10.35 g2 115 cm2
mA L
=
= 2.1 cm
mA + mT
0.35 g + 2.2 g
The ant and the left-hand end of the twig end up 2.1 cm from the edge
of the pool.
Reflect
The ant was initially a distance L = 15 cm from the edge of the pool and ends up a distance d = 2.1 cm from the edge.
Hence the ant moves a distance L 2 d = 12.9 cm. The twig, by contrast, moves away from the edge of the pool by only
d = 2.1 cm. The twig moves a much shorter distance than the ant because the twig is far more massive.