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A block of mass $10 k g$ is kept on a rough inclined plane as shown in the figure. A force of $3 N$ is applied on the block. The coefficient of static friction between the plane and the block is $0.6$ . What should be the minimum value of force $P$ such that the block does not move downward?
$(T a k e g = 10 m s^{- 2})$

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$25 N$

$23 N$

$18 N$

$32 N$

answer is C.

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Detailed Solution

Limiting friction $f_{s} = μ m g \cos 45^{0}$

$= 0.6 \times 10 \times 10 \times \frac{1}{\sqrt{2}} = 30 \sqrt{2} N = 42.43 N$

When block starts to slide downward, the downward force on the block is

$F = 3 + m g \sin 45^{0} = 3 + 10 \times 10 \times \frac{1}{\sqrt{2}} = 3 + 50 \sqrt{2} = 73.71 N > f_{s}$

Block will not move if $P = F - f P = 73.71 - 42.43 = 31.38 N ≃ 32 N$

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