Question

A stone of mass 2 kg is hung from the ceiling of the room using two strings. If the strings make an angle $ 60^\circ $ and $ 30^\circ $ respectively with the horizontal surface of the roof, then the tension on the longer string is:

• A. \( \sqrt{3}/2 \, \text{N} \)
• B. \( 10 \sqrt{3} \, \text{N} \)
• C. 10 N
• D. \( \sqrt{3} \, \text{N} \)

Hint:

When solving equilibrium problems with multiple forces at angles, use the vector components of the forces. Break them into horizontal and vertical components and solve the system of equations.

Answer 1

The Correct Option is C

We need to find the tension in the longer string. We know that the stone is hanging at equilibrium, so the forces in the vertical and horizontal directions must balance. 
Let the tensions in the two strings be \( T_1 \) (the shorter string) and \( T_2 \) (the longer string). 
The forces in the vertical direction are: \[ T_1 \sin(60^\circ) + T_2 \sin(30^\circ) = mg \] Substituting the known values: \[ T_1 \sin(60^\circ) + T_2 \sin(30^\circ) = 2 \times 10 \] \[ T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{1}{2} = 20 \] The forces in the horizontal direction must balance: \[ T_1 \cos(60^\circ) = T_2 \cos(30^\circ) \] \[ T_1 \cdot \frac{1}{2} = T_2 \cdot \frac{\sqrt{3}}{2} \] \[ T_1 = \sqrt{3} T_2 \] Substitute \( T_1 = \sqrt{3} T_2 \) into the vertical force equation: \[ \sqrt{3} T_2 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{1}{2} = 20 \] \[ \frac{3}{2} T_2 + \frac{1}{2} T_2 = 20 \] \[ 2 T_2 = 20 \] \[ T_2 = 10 \, \text{N} \] Thus, the tension in the longer string is \( T_2 = 10 \, \text{N} \).