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 $3$ times the force required to just prevent it from sliding down. If we define $N=10\mu$, then $N$ is _____.

Answer 1

$F_1=mg\sin \theta +\mu mg\cos \theta$

$F_2=mg\sin \theta -\mu mg\cos \theta$

Given that,

$F_1=3F_2$

$mg\sin \theta +\mu mg\cos \theta =3(mg\sin \theta -\mu mg\cos \theta )$

$2\sin \theta =4\mu \cos \theta$

$\mu =\frac{\tan \theta }{2}=\frac{\tan {45}^{\circ }}{2}=0.5$

$\therefore N=10\mu =5$.