Question

A block is on an inclined plane making an angle ${45}^{\circ }$ with the horizontal and having coefficient of friction $\mu .$ The force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down. If we define $N=10\mu ,$ then value of $N$ is

Answer 1

Let $F_1$ be the force required to just push it up the incline. Then direction of friction is downwards along the inclined plane.


For equilbrium , balancing the forces
$R=\frac{mg}{\sqrt{2}}$
$F_1=\frac{mg}{\sqrt{2}}+f_L$
$F_1=\frac{mg}{\sqrt{2}}+\mu \frac{mg}{\sqrt{2}}$$(\because f_L=\mu R)$
$F_1=\frac{mg}{\sqrt{2}}(1+\mu )$

Let $F_2$ be the force required to just prevent it from sliding down. Then direction of friction is upwards along the inclined plane.

For equilbrium , balancing the forces
$R=\frac{mg}{\sqrt{2}}$
$F_2+f_L=\frac{mg}{\sqrt{2}}$
$F_2=\frac{mg}{\sqrt{2}}-\mu \frac{mg}{\sqrt{2}}$
$F_2=\frac{mg}{\sqrt{2}}(1-\mu )$
Given $F_1=3F_2$

$\Rightarrow \frac{mg}{\sqrt{2}}(1+\mu )=3\times \frac{mg}{\sqrt{2}}(1-\mu )$
$\Rightarrow 1+\mu =3-3\mu$
$\Rightarrow 4\mu =2$
$\Rightarrow \mu =0.5$

$\Rightarrow N=10\mu =10\left(0.5\right)=5$