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. $\frac{(P\sin \theta -Q)}{(Mg-\cos \theta )}$
• B. $\frac{(P-Q\sin \theta )}{(Mg+Q\cos \theta )}$
• C. $\frac{(P\cos \theta +Q)}{(Mg-Q\cos \theta )}$
• D. $\frac{(P+Q\sin \theta )}{(Mg+Q\cos \theta )}$

Answer 1

The correct option is B $\frac{(P-Q\sin \theta )}{(Mg+Q\cos \theta )}$


From FBD,
Normal reaction, $N=(Mg+Q\cos \theta )$
therefore, frictional force
$f=\mu N=\mu (Mg+Q\cos \theta )$
For block to be in equilibrium, we have
$P=Q\sin \theta +f$
$(P-Q\sin \theta )=\mu (Mg+Q\cos \theta )$
$\mu =\frac{(P-Q\sin \theta )}{(Mg+Q\cos \theta )}$
So, option B is correct.