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 draw a free body diagram of the rod and block.

![image.png](attachment:image.png)

From the diagram, we can see that the forces acting on the rod are its weight (16 N) acting at its center of gravity (CG), and the tension forces in strings AC and BD, both acting at an angle of 45 degrees to the horizontal.

To find the tension in BD, we need to consider the equilibrium of the block P. The forces acting on the block are its weight (96 N) acting vertically downwards, and the tension force in string BD acting at an angle of 45 degrees to the horizontal. We can resolve these forces horizontally and vertically as follows:

Horizontally:
Tension in BD = Force acting to the right - Force acting to the left
Tension in BD = 0.3Psin45 - 0.6Wsin45  (since the horizontal components of string AC and the weight of the rod cancel out)
Tension in BD = 0.3(96)/√2 - 0.6(16)/√2 = 31.5 N

Vertically:
Net force = 0
Tension in BDsin45 = Force acting downwards
Tension in BD = W + Pcos45
Tension in BD = 16 + 0.3(96)/√2 = 48 N

Therefore, the magnitude of the tension force in string BD is 48 N, which is closest to choice C (32 N).

A Uniform Rod AB Is 1