Page 148 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 11] EQUILIBRIUM 133

We now let the x axis be in the direction of the boom with the y axis perpendicular to it, as in Fig. 11-5(b). Since
the boom is rigid, we need consider only the translational equilibrium of its upper end in the y direction. The angle
between T and the y axis is 90 − 75 = 15 , and so
◦
◦
◦
◦
T y = T cos 15 = 0.966T
The component of the weight w in the y direction is
◦
w y =−(8000 N)(cos 50 ) =−5142 N
At equilibrium
F y = T y + w y = 0 0.966T − 5142 N = 0 T = 5323 N

ROTATIONAL EQUILIBRIUM
When the lines of action of the forces that act on a body in translational equilibrium intersect at a common point,
they have no tendency to turn the body. Such forces are said to be concurrent. When the lines of action do not
intersect, the forces are nonconcurrent and exert a net torque that acts to turn the body even though the resultant
of the forces is zero (Fig. 11-6).

Fig. 11-6
A body is in rotational equilibrium when no net torque acts on it. Such a body remains in its initial rotational
state, either not spinning at all or spinning at a constant rate. The condition for the rotational equilibrium of a
body may therefore be written

τ = 0
where τ refers to the sum of the torques acting on the body about any point.
A torque that tends to cause a counterclockwise rotation when it is viewed from a given direction is considered
positive; a torque that tends to cause a clockwise rotation is considered negative (Fig. 11-7).

Fig. 11-7

To investigate the rotational equilibrium of a body, any convenient point may be used as the pivot point for
calculating torques; if the sum of the torques on a body in translational equilibrium is zero about some point, it
is zero about any other point.