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A block of mass 15 kg kept on a rough inclined plane as shown in the figure. A force 5 N is applied on the block. The coefficient of static friction between the plane and the block is 0.5. What should be the minimum value of force P, such that it does not move downward? (Round off to nearest integer)
$[Take g = 10 {ms}^{- 2}]$

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answer is 15.

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Detailed Solution

$F + m g \sin θ = P + μ m g \cos θ$

$∴ 5 + (15 \times 10 \times \frac{1}{2}) = p + (0 .5 \times 15 \times 10 \times \frac{\sqrt{3}}{2})$
$∴ 80 = p + \frac{75 \sqrt{3}}{2}$

$∴ p = 80 - \frac{75 \sqrt{3}}{2}$

$∴ p = \frac{160 - 75 \sqrt{3}}{2}$

$∴ p = 15.04 N \approx 15 N$
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