Question

A block of mass m, lying on a horizontal plane, is acted upon by a horizontal force P and another force Q, inclined at an angle $\theta$ to the vertical. The block will remain in equilibrium if the coefficient of friction between it and the surface is (assume P > Q)

• A. $(Psin\theta -Q)/(mg-cos\theta )$
• B. $(P-Qsin\theta )/(mg+Qcos\theta )$
• C. $(Pcos\theta +Q)/(mg-Qcos\theta )$
• D. $(P+Qsin\theta )/(mg+Qcos\theta )$

Answer 1

The correct option is B $(P-Qsin\theta )/(mg+Qcos\theta )$

In vertical direction

$N=Qcos\theta +mg$

As $P>Q$

So block will try to slip in forward direction so friction force will act in backward direction

$f=\mu N$

$f=\mu (Qcos\theta +mg)$

As block remains in equilibrium so net force on block is 0 in horizontal direction

So

$P=f+Qsin\theta$

$P=\mu (Qcos\theta +mg)+Qsin\theta$

$\mu =\frac{P-Qsin\theta }{mg+Qcos\theta }$