Question

A board of mass '$M$' is placed on a rough inclined plane and a man of mass '$m$' walks down the board. If the coefficient of friction between the board and inclined plane $\mu$, the acceleration of the man, such that plank does not slip, is given by:

• A. $a\le \left(\frac{M-m}{m}\right)(\cos \theta +\mu \sin \theta )g$
• B. $a\ge \left(\frac{M+m}{m}\right)(\sin \theta +\mu \cos \theta )g$
• C. $a\le \left(\frac{M+m}{m}\right)(\sin \theta +\mu \cos \theta )g$
• D. $a=\left(\frac{m}{M+m}\right)(\sin \theta +\mu \cos \theta )g$

Answer 1

Answer: (C)

$a\le \left(\frac{M+m}{m}\right)(\sin \theta +\mu \cos \theta )g$

Let $F$ be the interaction force in between the board and the man , $f$ is the friction force in between the board and inclined plane.
Here the value of $F$ is $Mg\sin \theta -\mu (M+m)g\cos \theta \le F\le Mg\sin \theta +\mu (M+m)g\cos \theta ...\left(1\right)$
and

$f=\mu N=\mu (M+m)g\cos \theta$

also, $ma=F+mg\sin \theta \Rightarrow F=ma-mg\sin \theta$

now from (1), $Mg\sin \theta -\mu (M+m)g\cos \theta \le ma-mg\sin \theta \le Mg\sin \theta +\mu (M+m)g\cos \theta$

or $(M+m)g\sin \theta -\mu (M+m)g\cos \theta \le ma\le (M+m)g\sin \theta +\mu (M+m)g\cos \theta$

or$\left(\frac{M+m}{m}\right)(g\sin \theta -\mu g\cos \theta )\le a\le \left(\frac{M+m}{m}\right)(g\sin \theta +\mu g\cos \theta )$