Page 150 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 11] EQUILIBRIUM 135

SOLVED PROBLEM 11.8
A 4-m wooden platform weighing 160 N is suspended from the roof of a house by ropes attached to its
ends. A painter weighing 640 N stands 1.2 m from the left-hand end of the platform. Find the tension in
each of the ropes.

With the left-hand end of the platform as the pivot point and the weight w 2 of the platform acting from its center
of gravity in the middle (Fig. 11-9), the condition for rotational equilibrium yields

τ =−w 1 L 1 − w 2 L 2 + T 2 L 3 = 0
−(640 N)(1.2m) − (160 N)(2m) + 4T 2 m = 0
4T 2 m = 768 N·m + 320 N·m = 1088 N·m
T 2 = 272 N
To find T 1 we proceed as follows:

T 1 + T 2 = w 1 + w 2
T 1 = 640 N + 160 N − 272 N = 528 N

Fig. 11-9

SOLVED PROBLEM 11.9
A cantilever is a beam that projects beyond its supports, such as a diving board. Figure 11-10 shows a
30-kg diving board 3.6 m long that has a 50-kg woman standing at its end. The board’s supports are 1.0 m
apart. Find the force that each support exerts.
First we calculate the downward force F 1 on the left-hand support. The easiest way to do this is to calculate
torques about the other support, since we do not yet know the value of F 2 . If the board is uniform, its center of gravity
is at its center, and so
2
x 1 = 1.0m w 1 = m 1 g = (30 kg)(9.8 m/s ) = 294 N
2
x 2 = 0.8m w 2 = m 2 g = (50 kg)(9.8 m/s ) = 490 N
x 3 = 2.8m

The three torques that act about the right-hand support are

τ 1 =+F 1 x 1 =+(F 1 )(1.0m) = 1.0F 1 m
τ 2 =−w 1 x 2 =−(294 N)(0.8m) =−235 N·m
τ 3 =−w 2 x 3 =−(490 N)(2.8m) =−1372 N·m