Page 145 - Schaum's Outline of Theory and Problems of Applied Physics

130 EQUILIBRIUM [CHAP. 11

SOLVED PROBLEM 11.2
A 500-kg load is suspended from the end of a horizontal boom, as in Fig. 11-2(a). The angle between the
boom and the cable supporting its end is 45 . Assuming that the boom’s mass can be neglected compared
◦
with that of the load, find (a) the tension in the cable and (b) the inward force the boom exerts on the wall.

Fig. 11-2

(a) The three forces that act on the end of the boom are the weight w of the load, the tension T in the cable, and
the outward force F exerted by the boom. A free-body diagram of these forces is shown in Fig. 11-2(b). The x
and y components of these forces have the magnitudes
◦
T x =−T cos θ =−T cos 45 =−0.707T
T y = T sin θ = T sin 45 = 0.707T
◦
2
w =−mg =−(500 kg)(9.8 m/s ) =−4900 N
F = ?
The condition for translational equilibrium in the y (vertical) direction yields

F y = T y + w = 0
0.707T − 4900 N = 0
4900 N
T = = 6930 N
0.707
(b) To find the inward force the boom exerts on the wall, we start with the condition for equilibrium in the x
(horizontal) direction:
F x = T x + F = 0

−0.707T + F = 0
F = 0.707T = (0.707)(6930 N) = 4900 N

The inward force on the wall must have the same magnitude as the outward force on the load; hence, the inward
force is also equal to 4900 N.

SOLVED PROBLEM 11.3

A 50-lb box is suspended by a rope from the ceiling. If a horizontal force of 20 lb is applied to the box,
what angle will the rope make with the vertical?