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Physics · LAWS OF MOTION
हिन्दीA block of mass \( m = 10\text{ kg} \) rests on a rough horizontal floor with static friction coefficient \( \mu_s = \frac{1}{\sqrt{3}} \). A pulling force \( F \) is applied at an angle \( \theta \) above the horizontal. Calculate the minimum force magnitude \( F_{\min} \) necessary to set the block in motion and the corresponding optimal pulling angle \( \theta \). (Take \( g = 10\text{ m/s}^2 \))
Correct answer · Option 1
\( F_{\min} = 50\text{ N} \) at \( \theta = 30^\circ \)
Explanation
Minimum force formula is \( F_{\min} = \frac{\mu_s m g}{\sqrt{1 + \mu_s^2}} \). Optimal angle is given by \( \tan\theta = \mu_s = \frac{1}{\sqrt{3}} \implies \theta = 30^\circ \). Substituting values gives \( F_{\min} = \frac{\frac{1}{\sqrt{3}} \times 100}{\sqrt{1 + \frac{1}{3}}} = \frac{100/\sqrt{3}}{2/\sqrt{3}} = 50\text{ N} \).
