Question

If the tension in the horizontal wire shown in the figure is 30 N, then the weight \( W \) and tension in the wire \( OA \) are respectively:

• A. \( 30\sqrt{3} \) N, 30 N
• B. \( 30\sqrt{3} \) N, 60 N
• C. \( 60\sqrt{3} \) N, 30 N
• D. \( 60\sqrt{3} \) N, 60 N

Hint:

For equilibrium problems involving inclined tensions, always resolve forces into horizontal and vertical components before solving.

Answer 1

The Correct Option is B

Step 1: Analyzing Forces in the System From the given figure, the tension in the horizontal wire is \( T = 30 \) N. Resolving forces at point \( O \), we consider the equilibrium condition. The wire \( OA \) makes an angle of \( 60^\circ \) with the horizontal. Step 2: Resolving Forces The vertical component of tension in \( OA \) balances the weight \( W \): \[ T_{\text{OA}} \sin 60^\circ = W \] Similarly, the horizontal component balances the tension in the horizontal wire: \[ T_{\text{OA}} \cos 60^\circ = 30 \] Using \( \cos 60^\circ = \frac{1}{2} \), \[ T_{\text{OA}} \times \frac{1}{2} = 30 \] Solving for \( T_{\text{OA}} \): \[ T_{\text{OA}} = 60 \text{ N} \] Now solving for \( W \): \[ W = 60 \times \sin 60^\circ \] Using \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), \[ W = 60 \times \frac{\sqrt{3}}{2} = 30\sqrt{3} \text{ N} \] Conclusion Thus, the weight \( W \) is \( 30\sqrt{3} \) N and the tension in wire \( OA \) is \( 60 \) N. The correct answer is: \[ 30\sqrt{3} \text{ N}, 60 \text{ N} \]