A block of mass $m = 2 k g$ is resting on a rough inclined plane of inclination 37⁰ as shown in Fig. The coefficient of friction between the block and the plane is $μ = 0.5$ . What minimum force F (in newton) should be applied perpendicular to the plane on the block, so that the block does not slip on the plane? $(g = 10 m / s^{2})$

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Detailed Solution

(8)Since $m g \sin 37 ° > μ m g \cos 37 °$ the block has a tendency to slip downwards.
Let F be the minimum force applied on it, so that it does not slip, then,
$N = F + m g \cos 37 °$
$∴ m g \sin 37 ° = μ N = μ (F + m g \cos 37 °)$
$o r F = \frac{m g \sin 37 °}{μ} - m g \cos 37 ° = \frac{(2) (10) (3 / 5)}{0.5} - (2) (10) (\frac{4}{5}) = 8 N$

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