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 point on AB where the center of mass is located. Using the formula:

$Center\:of\:mass=\frac{\int xdm}{\int dm}$

where $x$ is the distance from one end of the rod, $dm$ is the mass element, and the integral is taken over the entire length of the rod, we can simplify the expression since the rod has uniform mass:

$Center\:of\:mass=\frac{\int xdm}{\int dm}=\frac{\int xdx}{\int dx}=\frac{\frac{1}{2}(1.2)^2}{1.2}=0.6\:m$

So the center of mass is located at a distance of 0.6 m from end A.

Next, we need to find the tension in string AC. Since the rod is in equilibrium, the tension in string AC must be equal to the weight of the rod. So:

$T_{AC}=16\:N$

Now we can use the condition that the rod is in rotational equilibrium about point B. The torque about point B due to the weight of the rod and the tension in string AC must be balanced by the torque about point B due to the tension in string BD and the weight of the block P. Taking counterclockwise torques as positive:

$(16\:N)(0.6\:m)=(96\:N)(0.9\:m)-T_{BD}(1.2-0.3\:m)$

Solving for $T_{BD}$, we get:

$T_{BD}=32\:N$

So the answer is C.

A Uniform Rod AB Is 1