A block of mass $10\, kg$ 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? (take $g = 10\, ms^{-2}$) ........ $N$
JEE MAIN 2019, Medium
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For equilibrium of the block net force should be zero.
Hence we can write. $m g \sin \theta+3=p+$ friction $m g \sin \theta+3=p+\mu m g \cos \theta$.
After solving, we get, $p=32 N$.
Hence we can write. $m g \sin \theta+3=p+$ friction $m g \sin \theta+3=p+\mu m g \cos \theta$.
After solving, we get, $p=32 N$.
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