Page 151 - Schaum's Outline of Theory and Problems of Applied Physics

136 EQUILIBRIUM [CHAP. 11

F 2

w 1
w 2

x 1 x 2
x
F 1 3
Fig. 11-10

The condition for rotational equilibrium then gives
τ = τ 1 + τ 2 + τ 3 = 0
1.0F 1 m − 235 N·m − 1372 N·m = 0
1607 N·m
F 1 = = 1607 N
1.0m
To find F 2 the upward force exerted by the right-hand support, we consider the translational equilibrium of the
system of board plus woman:

F =−F 1 + F 2 − w 1 − w 2
F 2 = F 1 + w 1 + w 2 = 1607 N + 294 N + 490 N = 2391 N

SOLVED PROBLEM 11.10
A horizontal boom 2.4 m long is attached to a wall at its inner end and is supported at its outer end by
a cable that makes an angle of 30 with the boom. The boom weighs 200 N, and a load of 1500 N is
◦
attached to its outer end. Find (a) the tension in the cable and (b) the compression force in the boom.
(a) Four forces act on the boom, as in Fig. 11-11: the load w 1 at its outer end, the boom’s own weight w 2 that acts
from its center, the tension T in the cable, and the force F that the wall exerts on the inner end of the boom. We
can disregard F by calculating torques about the inner end of the boom. It simplifies matters to use the vertical
component T y of the tension in the cable instead of T, since the moment arm of T y is the boom’s length L 1 . The
horizontal component of the tension T x exerts no torque about O because its line of action passes through O. The
torques exerted by the load w 1 , the boom’s weight w 2 , and the vertical component of tension T y are, respectively,
τ 1 = w 1 L 1 = (1500 N)(2.4m) = 3600 N·m
τ 2 = w 2 L 2 = (200 N)(1.2m) = 240 N·m
◦
τ 3 =−T y L 1 =−2.4T sin 30 m =−1.2T m
For the boom to be in equilibrium,

τ = τ 1 + τ 2 + τ 3 = 0
3600 N·m + 240 N·m − 1.2T m = 0
3840 N·m
T = = 3200 N
1.2m