Question

A bar of mass m is placed on a triangular block of mass M as shown in Fig. The friction coefficient between the two surfaces is $\mu$ and ground is smooth. Find the minimum and maximum horizontal force F required to be applied on a block so that the bar will not slip on the inclined surface of block.

Answer 1

Here if both the masses are moving together, acceleration of the system will be a= F/(M+m). If we observe the mass m relative to M. It experiences a pseudo force ma towards left. Along the inclined it experiences two forces, $mgsin\theta$ is more than $macos\theta$, it has a tendency of slipping downwards, so friction on it will act in upward direction. Here if block m is euilibrium on inclined surface, we must have

$f=\mu N,$

Here $f=(mgsin\theta -macos\theta )$

and $N=(mgsin\theta +mgcos\theta ).$

$f=mgsin\theta -macos\theta \le \mu (mgcos\theta +masin\theta )$

or $a\ge \frac{sin\theta -\mu cos\theta }{cos\theta +\mu sin\theta }g$

$(M+m)a\ge \frac{sin\theta -\mu cos\theta }{cos\theta +\mu sin\theta }(M+m)g$

$F\ge \frac{sin\theta -\mu cos\theta }{cos\theta +\mu sin\theta }(M+m)g$ (i)

If force is more than the value obtained in Eq. (i), $macos\theta$ will increase on m and the static friction on it will decrease. At $a=gtan\theta$$(whenF=(M+m\left)gtan\theta \right)$, we know that the force $mgsin\theta$ will balnced $macos\theta$ at this acceleration no friction will act on it. If applied force will increase beyond this valued, $macos\theta$ will exceed $mgsin\theta$ and friction starts acting in downward direction. Here if block m is equilibrium, we must have $f=(macos\theta -mgsin\theta )$, $N=(mgcos\theta +masin\theta )$, and $f\le \mu N.$

$macos\theta -mgsin\theta \le \mu (mgcos\theta +masin\theta )$

$\Rightarrow a\le \frac{sin\theta +\mu cos\theta }{cos\theta -\mu sin\theta }g$

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

$\Rightarrow F\le \frac{sin\theta +\mu cos\theta }{cos\theta -\mu sin\theta }(M+m)g$

Hence, $F_{min}\le \frac{sin\theta -\mu cos\theta }{cos\theta +\mu sin\theta }(M+m)g$

and $F_{min}\le \frac{sin\theta +\mu costheta}{cos\theta -\mu sin\theta }(M+m)g$