Question

A block is moving on an inclined plane making an angle ${37}^{\circ }$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is $9$ times the force required to just prevent it from sliding down. If the normal force on the block is $N=10\mu$, then $N$ is

Answer 1

Let the force required to just prevent the block from sliding down be $F$.
Then, for equilibrium, perpendicular to the plane $N=mg\cos {37}^{\circ }$
Along the plane, $mg\sin {37}^{\circ }=F+\mu N$
$mg\sin {37}^{\circ }-\mu mg\cos {37}^{\circ }=F\dots \left(i\right)$
As given the force required to push the block is $9F$,
Then, using NLM along the plane,
$mg\sin {37}^{\circ }+\mu mg\cos {37}^{\circ }=9F\dots \left(ii\right)$
By dividing equation $\left(ii\right)$ by $\left(i\right)$
$\frac{\sin {37}^{\circ }+\mu \cos {37}^{\circ }}{\sin {37}^{\circ }-\mu \cos {37}^{\circ }}=9$
$\mu =0.6$
As given, normal reaction
$N=10\mu$
$N=10\left(0.6\right)$
$=6N$