Question 149 of 210

Consider a block kept on an inclined plane (inclined at $45^{\circ}$) 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 :
.

A$0.33$B$0.60$C$0.25$D$0.50$

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Answer: (A) $0.33$

The forces acting on block is shown in above figure.
Here, normal force, $N=mg\cos 45^{\circ}$ and frictional force, $f=\mu N$.
Balancing the force along incline, we have
${F}_{1}=mg\sin 45^{\circ}+f=mg\sin 45^{\circ}+\mu N$
$\Rightarrow {F}_{1}=\dfrac{mg}{\sqrt{2}}+\mu mg\cos 45^{\circ}$
$\Rightarrow {F}_{1}=\dfrac{mg}{\sqrt{2}}\left(1+\mu \right)$
Now, the forces acting on block to just push it up on inclined plane is shown below.

Balancing the force along incline, we have
${F}_{2}=mg\sin 45^{\circ}-f=mg\sin 45^{\circ}-\mu N$
$=\dfrac{mg}{\sqrt{2}}\left(1-\mu \right)$
Given, ${F}_{1}=2{F}_{2}$
$\Rightarrow \dfrac{mg}{\sqrt{2}}\left(1+\mu \right)=2\dfrac{mg}{\sqrt{2}}\left(1-\mu \right)$
$\Rightarrow 1+\mu =2-2\mu$
$\Rightarrow \mu =\dfrac{1}{3}=0.33$