Page 149 - Schaum's Outline of Theory and Problems of Applied Physics

134 EQUILIBRIUM [CHAP. 11

CENTEROFGRAVITY
The center of gravity of a body is that point at which the body’s entire weight can be regarded as being
concentrated. A body can be suspended in any orientation from its center of gravity without tending to rotate. In
analyzing the equilibrium of a body, its weight can be considered as a downward force acting from its center of
gravity.

SOLVED PROBLEM 11.6
(a) Under what circumstances is it necessary to consider torques in analyzing an equilibrium situation?
(b) About what point should torques be calculated when this is necessary?

(a) Torques must be considered when the various forces that act on the body are nonconcurrent, that is, when their
lines of action do not intersect at a common point.
(b) Torques may be calculated about any point whatever for the purpose of determining the equilibrium of the
body. Hence it makes sense to use a point that minimizes the labor involved, which usually is the point through
which pass the maximum number of lines of action of the various forces; this is because a force whose line of
action passes through a point exerts no torque about that point.

SOLVED PROBLEM 11.7
A beam 3 m long has a weight of 200 N at one end and another weight of 80 N at the other end. The
weight of the beam itself is negligible. Find the balance point of the beam.
When the beam is supported at its balance point, the torques of the two weights cancel, and the beam has no
tendency to rotate. The supporting force F exerts no torque since it acts through the balance point. If the balance
point is the distance x from the 200-N weight, as in Fig. 11-8, it is the distance 3 m − x from the 80-N weight. Since
the beam is horizontal, the moment arms of the weights are, respectively, x and 3 m − x, and the torques the weights
exert about the balance point O are
τ 1 = w 1 L 1 = 200x N τ 2 =−w 2 L 2 =−80(3m − x) N

Fig. 11-8

The torque τ 1 is positive because it tends to cause a counterclockwise rotation; the torque τ 2 is negative because it
tends to cause a clockwise rotation. The condition for rotational equilibrium yields

τ = τ 1 + τ 2 = 0
200x N − 80(3m − x) N = 0
200x N = 240 N·m − 80x N
280x N = 240 N·m
x = 0.86 m = 86 cm