Question

A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q in Grilled an at an angle $\theta$ to the vertical. The minimum value of the coefficient of friction between the block and the surface for which the block will remain in equilibrium is:

• A. $\frac{P+Qsin\theta }{mg+Qcos\theta }$
• B. $\frac{Pcos\theta +Q}{mg-Qsintheta}$
• C. $\frac{P+Qcos\theta }{mg+Qsin\theta }$
• D. $\frac{Psin\theta -Q}{mg-Qcos\theta }$

Answer 1

The correct option is A $\frac{P+Qsin\theta }{mg+Qcos\theta }$

In order for the block to remain in equilibrium, the net force acting on it should be 0.
The forces acting on the block are:
1. mg (downwards)
2. $Q\cos \theta$ (downwards)
3. P (to the right)
4. $Q\sin \theta$ (to the right)
5. friction (to the left)
Let the coefficient of friction be $\mu$ then the maximum friction that can act = $\mu (mg+Q\cos \theta )$
For equilibrium,
$P+Q\sin \theta =\mu (mg+Q\cos \theta )$
=> $\mu =\frac{P+Q\sin \theta }{mg+Q\cos \theta }$