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14.1 Statics of Rigid Bodies 431

From this discussion, we conclude that for the purposes of static equilibrium,
any line through the body or any line passing at some distance from the body can be thought
of as a conceivable axis of rotation, and the torque about every such axis must be zero.This
means we have complete freedom in the choice of the axis of rotation, and we can make
whatever choice seems convenient. With some practice, one learns to recognize which
choice of axis will be most useful for the solution of a problem in statics.
The force of gravity plays an important role in many problems of statics.The force
of gravity on a body is distributed over all parts of the body, each part being subjected
to a force proportional to its mass. However, for the calculation of the torque exerted
by gravity on a rigid body, the entire gravitational force may be regarded as acting on the
center of mass.We relied on this rule in Fig. 14.1, where we assumed that the weight acts
at the center of mass of the bat.The proof of this rule is easy: Suppose that we release
some arbitrary rigid body and permit it to fall freely from an initial condition of rest.
Since all the particles in the body fall at the same rate, the body will not change its
orientation as it falls. If we consider an axis through the center of mass, the absence of
angular acceleration implies that gravity does not generate any torque about the center
of mass. Hence, if we want to simulate gravity by a single force acting at one point of
the rigid body, that point will have to be the center of mass.
Given that in a rigid body the force of gravity effectively acts on the center of
mass, we see that a rigid body supported by a single force acting at its center of mass
or acting on the vertical line through its center of mass is in equilibrium, since the
support force is then collinear with the effective force of gravity, and such collinear
forces of equal magnitudes and opposite directions exert no net torque. This pro-
vides us with a simple method for the experimental determination of the center of
mass of a body of complicated shape: Suspend the body from a string attached to a
point on its surface (Fig. 14.2); the body will then settle into an equilibrium position
such that the center of mass is on the vertical downward prolongation of the string
(this vertical prolongation is marked dashed in Fig. 14.2). Next, suspend the body
from a string attached at another point of its surface, and mark a new vertical
downward prolongation of the string. The center of mass is then at the intersection
of the new and the old prolongations of the string.

(a) (b)

To find center of mass,
suspend body by a
string from a point on
its surface.

Center of mass
will be along vertical
prolongation of string.

Any two such lines
must intersect at
center of mass.

FIGURE 14.2 (a) Bicycle suspended by
a string attached at a point on its “surface.”
(b) Bicycle suspended by a string attached
at a different point.