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 center of mass of the rod and the block system. Since the rod is uniform and symmetrical, its center of mass will be at its midpoint, which is 0.6 m from A.

The center of mass of the block P will be at its midpoint, which is 0.15 m from A.

To find the tension in string BD, we can use the principle of moments. Taking the point of intersection of strings AC and BD as the pivot, the net clockwise moment about this point must be equal to the net anticlockwise moment.

Clockwise moments: 
- Weight of the rod = 16 N * 0.6 m = 9.6 Nm 
- Weight of the block = 96 N * 0.15 m = 14.4 Nm

Anticlockwise moments: 
- Tension in string AC = T_AC * 1.2 m (since it acts perpendicular to the rod and at its midpoint) 
- Tension in string BD = T_BD * 0.9 m (since it acts perpendicular to the rod and at a distance of 0.3 m from its end)

Setting the moments equal to each other: 
9.6 Nm + 14.4 Nm = T_AC * 1.2 m + T_BD * 0.9 m

Simplifying: 
24 Nm = 1.2T_AC + 0.9T_BD

Now, we need to consider the forces acting vertically. 
Taking upward forces as positive: 
- Tension in string AC - Weight of rod - Weight of block = 0

Substituting the weight of the rod (16 N) and the weight of the block (96 N): 
T_AC - 112 N = 0

Solving for T_AC: 
T_AC = 112 N

Substituting this value back into the moment equation: 
24 Nm = 1.2(112 N) + 0.9T_BD

Solving for T_BD: 
T_BD = 32 N

Therefore, the tension in string BD is 32 N.

The best choice is C.

A Uniform Rod AB Is 1