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Physics · LAWS OF MOTION
हिन्दीA mass of \( 6\text{ kg} \) is suspended by a rope of length \( 2\text{ m} \) from a ceiling. A horizontal force of \( 50\text{ N} \) is applied at the midpoint P of the rope. If the system is in equilibrium, what is the angle \( \theta \) that the upper part of the rope makes with the vertical? (Take \( g = 10\text{ m s}^{-2} \))
Correct answer · Option 2
\( \tan^{-1}(\frac{5}{6}) \)
Explanation
At point P: \( T_1 \cos\theta = W = 6 \times 10 = 60\text{ N} \), and \( T_1 \sin\theta = 50\text{ N} \). Dividing gives \( \tan\theta = 50/60 = 5/6 \implies \theta = \tan^{-1}(5/6) \).
