Question #31590

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Question #31590

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Answer 1 · 2016-10-24T19:08:10

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Forces acting on the body are shown in the diagram above

Sum of the forces along the inclined plane:
$F_{net} = m g sin θ - f$ $F_"net" = mg sin theta -f$
where $m g$ $mg$ is the weight of the body and $f$ $f$ is force of friction.

There is no acceleration along the inclined plane, $F_{net} = 0$ $F_"net" = 0$ , therefore
$m g sin θ = f$ $mg sin theta =f$ ......(1)

Similarly in a direction perpendicular to the inclined plane:
$F_{net} = N - m g cos θ$ $F_"net" = N -mg cos theta$
where $N$ $N$ is normal reaction.

As per given condition above there is no acceleration in this direction either, $F_{net} = 0$ $F_"net" = 0$ , and therefore
$N = m g cos θ$ $N =mg cos theta$ ......(2)

If $μ_{k}$ $mu_k$ is coefficient of kinetic friction, then
$f = µ_kN$

Using (1) and (2) we get
$mg sin theta=mu_kxxmg cos theta$

Solving for $mu_k$
$mu_k=tan theta$
$mu_k=tan 30^@=1/sqrt3$

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Question #31590

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