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

Answer: 6

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
$F_2=mg\sin 45^{\circ }-\mu R$
$=mg\sin 45^{\circ }-\mu \left(mg\cos 45^{\circ }\right)$
$=\frac{mg}{\sqrt{2}}-\frac{\mu mg}{\sqrt{2}}$ ...(ii)
Given: $F_1=5F_2$ ...(iii)
From (i), (ii) and (iii), we have,
$1+\mu =5-5\mu$
$\therefore 6\mu =4$
$\therefore \mu =\frac{2}{3}$
Given: $N=9\mu$
$\therefore N=6$