Question

An object of mass $8 kg$ is hanging from one end of a uniform rod CD of mass $2 kg$ and length $1 m$ pivoted at its end $C$ on a vertical wall as shown in figure It is supported by a cable $AB$ such that the system is in equilibrium The tension in the cable is : (Take $g =10 m / s ^2$ )

• A. $240 N$
• B. $30 N$
• C. $300 N$
• D. $90 N$

Answer 1

The Correct Option is C

The correct answer is (C) : 300 N

Taking torque about point $C$
$\frac{T}{2}\times 60=20\times 50+80\times 100$
$\Rightarrow 3T=100+800$
$\Rightarrow =300N$

Answer 2

1. For equilibrium, sum of torques about point C = 0:
- Weight of the rod acts at its center of gravity, i.e., 0.5 m from C.
- Tension \(T\) in the cable acts at point B at an angle of \(30^\circ\).
2. Calculate torques:
- Torque due to the rod: \(10 \times 0.5 = 5 \, \text{Nm}\).
- Torque due to the hanging object: \(80 \times 1 = 80 \, \text{Nm}\).
- Torque due to tension: \(T \times \sin 30^\circ \times 1 = T \times 0.5 \, \text{Nm}\).
3. Equating torques: \[ 80 + 5 = T \times 0.5. \]
\[ T = \frac{85}{0.5} = 300 \, \text{N}. \]
Thus, the tension in the cable is 300 N. Equilibrium conditions require that the net torque and net force on the system are zero. Tension is calculated by balancing the clockwise and counterclockwise torques about the pivot.