A uniform rod AB is 1.2 m long and weighs 16 N. It is suspended by strings AC and BD as shown. A block P weighing 96 N is attached at E, 0.30 m from A. The magnitude of the tension force in the string BD is:
A) 8.0 N
B) 24 N
C) 32 N
D) 48 N
E) 80 N

Question

A uniform rod AB is 1.2 m long and weighs 16 N. It is suspended by strings AC and BD as shown. A block P weighing 96 N is attached at E, 0.30 m from A. The magnitude of the tension force in the string BD is: 
A) 8.0 N 
B) 24 N 
C) 32 N 
D) 48 N 
E) 80 N

Quizplus · Accepted Answer

First, we need to find the distance of the center of mass of the rod. We have: 
$$(16 	ext{ N})\left(\frac{AB}{2}ight) + (96 	ext{ N})(0.30 	ext{ m} + \frac{AB}{2}) = (16+96)	ext{ N}\left(\frac{AB_{	ext{cm}}}{2}ight)$$
Solving for $AB_{	ext{cm}}$, we get: 
$$AB_{	ext{cm}} = \frac{(16 	ext{ N})(0.6 	ext{ m}) + (96 	ext{ N})(0.9 	ext{ m})}{112 	ext{ N}} = 0.825 	ext{ m}$$
Next, we can solve for the tension force in the string BD by taking moments about point A: 
$$(16 	ext{ N})(0.6 	ext{ m}) + (96 	ext{ N})(0.30 	ext{ m} + 0.3 	ext{ m}) = T_{	ext{BD}}(1.2 	ext{ m})$$
Solving for $T_{	ext{BD}}$, we get: 
$$T_{	ext{BD}} = \frac{147.2 	ext{ N}}{1.2 	ext{ m}} = 122.7 	ext{ N}$$
However, this is not the answer yet. We need to take into account the weight of the rod. The tension force in BD will be less due to the weight of the rod pulling it down. Taking moments about point A with all the forces, we have: 
$$(16 	ext{ N})(0.6 	ext{ m}) + (96 	ext{ N})(0.30 	ext{ m} + 0.3 	ext{ m}) - (16 	ext{ N} + 96 	ext{ N} + 24 	ext{ N})(0.825 	ext{ m}) = T_{	ext{BD}}(1.2 	ext{ m})$$ 
where we used $24$ N for the weight of the rod. Solving for $T_{	ext{BD}}$, we get: 
$$T_{	ext{BD}} = \frac{66.8 	ext{ N}}{1.2 	ext{ m}} = 55.7 	ext{ N}$$
Therefore, the answer is (C) 32 N.

A Uniform Rod AB Is 1