Question #4fdc1

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Answer 1 · 2017-10-07T04:11:27

drawn

The situation as described in the problem has been shown in the free body diagram shown above .

Here

The frictional force acting on the block when it slides downward is
$f = μ m g cos θ$ $f=mumgcostheta$

Considering the equilibrium condition at the position when the block tends to slide downward under gravity, we can write

$μ m g cos θ = m g sin θ$ $mumgcostheta=mgsintheta$

$\Rightarrow μ = tan θ = tan 30 = \frac{1}{\sqrt{3}}$ $=>mu=tantheta=tan30=1/sqrt3$