Laws Of Motion Question 51
Question: A block of mass m lying on a rough 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 [Haryana CEE 1996]
Options:
A) $ \frac{(P+Q\sin \theta )}{(mg+Q\cos \theta )} $
B)$ \frac{(P\cos \theta +Q)}{(mg-Q\sin \theta )} $
C) $ \frac{(P+Q\cos \theta )}{(mg+Q\sin \theta )} $
D) $ \frac{(P\sin \theta -Q)}{(mg-Q\cos \theta )} $
Show Answer
Answer:
Correct Answer: A
Solution:
By drawing the free body diagram of the block for critical condition $ F=\mu R $ therefore$ P+Q\sin \theta $
$ =\mu (mg+Q\cos \theta ) $
$ \therefore $ $ \mu =\frac{P+Q\sin \theta }{mg+Q\cos \theta } $
