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Standard XII

Physics

Angle of Repose

A body of mas...

Question

A body of mass m is kept on an inclined plane of inclination $30^{0}$ to the horizontal. If the angle of repose is $45^{0}$ , static friction on the body is

\frac{m g}{\sqrt{2}}

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\frac{m g}{2}

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\sqrt{2} m g

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Solution

The correct option is B $\frac{m g}{2}$
$μ = tan 45^{0}$ .
As $μ = 1$
So, max static friction $= m g sin θ$
Friction force $= \frac{m g}{2}$

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A body of mass m is kept on an inclined plane of inclination 300 to the horizontal. If the angle of repose is 450, static friction on the body is

The correct option is B mg2μ=tan450.As μ=1So, max static friction =mgsinθ Friction force =mg2

A body of mass m is kept on an inclined plane of inclination $30^{0}$ to the horizontal. If the angle of repose is $45^{0}$ , static friction on the body is

The correct option is B $\frac{m g}{2}$
$μ = tan 45^{0}$ .
As $μ = 1$
So, max static friction $= m g sin θ$
Friction force $= \frac{m g}{2}$