Consider a block kept on an inclined plane (inclined at $45 °$ ) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(µ)$ is equal to :
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Question

Consider a block kept on an inclined plane (inclined at $45 °$ ) as shown in the figure. If the force required to just push it up the incline is $2$ times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane $(µ)$ is equal to :

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TestHub · Accepted Answer

The forces acting on block is shown in above figure.

Here, normal force, $N = m g \cos 45 °$ and frictional force, $f = μ N$ .

Balancing the force along incline, we have

$F_{1} = m g \sin 45 ° + f = m g \sin 45 ° + μ N$

$\Rightarrow F_{1} = \frac{m g}{\sqrt{2}} + μ m g \cos 45 °$

$\Rightarrow F_{1} = \frac{m g}{\sqrt{2}} (1 + μ)$

Now, the forces acting on block to just push it up on inclined plane is shown below.

Balancing the force along incline, we have

$F_{2} = m g \sin 45 ° - f = m g \sin 45 ° - μ N$

$= \frac{m g}{\sqrt{2}} (1 - μ)$

Given, $F_{1} = 2 F_{2}$

$\Rightarrow \frac{m g}{\sqrt{2}} (1 + μ) = 2 \frac{m g}{\sqrt{2}} (1 - μ)$

$\Rightarrow 1 + μ = 2 - 2 μ$

$\Rightarrow μ = \frac{1}{3} = 0.33$

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