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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION
हिन्दीA non-uniform bar of weight \( W \) and length \( L = 2.0 \text{ m} \) is suspended horizontally at rest by two light strings attached to its ends. The strings make angles \( 36.9^\circ \) and \( 53.1^\circ \) with the vertical respectively (given \( \tan 36.9^\circ = 0.75 \), \( \tan 53.1^\circ = 1.33 \)). The distance \( d \) of the centre of gravity of the bar from its left end is approximately:
Correct answer · Option 3
\( 0.72 \text{ m} \)
Explanation
For horizontal equilibrium, \( T_1 \sin 36.9^\circ = T_2 \sin 53.1^\circ \implies T_1(0.6) = T_2(0.8) \implies T_1 = \frac{4}{3}T_2 \). Torque about CG: \( T_1 \cos 36.9^\circ \cdot d = T_2 \cos 53.1^\circ \cdot (2-d) \implies (\frac{4}{3}T_2)(0.8) d = T_2 (0.6)(2-d) \implies \frac{3.2}{3} d = 0.6(2-d) \implies 3.2 d = 1.8(2-d) = 3.6 - 1.8d \implies 5.0 d = 3.6 \implies d = 0.72 \text{ m} \).
