Question

A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is $5$ times the force required to just prevent it from sliding down. If we define $N =9\, \mu$, then $N$ is________.

Answer 1

Force while pushing up the block
$F _{1} = mg \sin 45^{\circ}+\mu R$
$= mg \sin 45^{\circ}+\mu\left( mg \cos 45^{\circ}\right)$
$=\frac{ mg }{\sqrt{2}}+\frac{\mu mg }{\sqrt{2}}$ ...(i)
Force to prevent block from sliding
$F _{2}= mg \sin 45^{\circ}-\mu R$
$=m g \sin 45^{\circ}-\mu\left(m g \cos 45^{\circ}\right)$
$=\frac{ mg }{\sqrt{2}}-\frac{\mu mg }{\sqrt{2}}$ ...(ii)
Given: $F_{1}=5 \,F_{2}$ ...(iii)
From (i), (ii) and (iii), we have,
$1+\mu=5-5\, \mu$
$\therefore 6\, \mu=4$
$\therefore \mu=\frac{2}{3}$
Given: $N =9\, \mu$
$\therefore N =6$