Page 146 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 11] EQUILIBRIUM 131

A free-body diagram of the forces acting on the box is shown in Fig. 11-3(a), and the forces are resolved into
components in Fig. 11-3(b). The x and y components of the forces are

T x =−T sin θ T y = T cos θ
F = 20 lb w =−50 lb

Fig. 11-3

Applying the conditions for equilibrium yields
20 lb
F x = T x + F = 0 −T sin θ + 20 lb = 0 sin θ =
T
50 lb
F y = T y + w = 0 T cos θ − 50 lb = 0 cos θ =
T
If we divide the expression for sin θ by that for cos θ, the T ’s cancel to give
sin θ 20 lb/T 20 lb
= tan θ = = = 0.40
cos θ 50 lb/T 50 lb
The angle whose tangent is nearest to 0.40 is 22 , and so θ = 22 .
◦
◦

SOLVED PROBLEM 11.4
To move a heavy crate across a floor, one end of a rope is tied to it and the other end is tied to a wall
10 m away. When a force of 400 N is applied to the midpoint of the rope, the rope stretches so that the
midpoint moves to the side by 60 cm. What is the force on the crate?
The first step is to find the angle θ between either part of the rope and a straight line between the crate and the
point of attachment of the rope to the wall. With the help of Fig. 11-4(a) we find that
0.6m
tan θ = = 0.12 θ = 6.8 ◦
5m
Figure 11-4(b) is a free-body diagram of the forces acting on the midpoint of the rope; T 1 and T 2 are the tensions
in the two parts of the rope, and F =−400 N is the applied force. These forces are resolved in Fig. 11-4(c). Since
T 1 = T 2 = T ,
◦
T 1y = T 2y = T sin θ = T sin 6.8 = 0.118T
At equilibrium
F y = T 1y + T 2y + F = 0
(2)(0.118T ) − 400 N = 0
T = 1695 N