Laws Of Motion Question 399

Question: A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force$ P $ and another force Q is 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

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:

The free body diagram of the block for critical condition is shown below.

$ F=\mu R\Rightarrow P+Q\sin \theta =\mu (mg+Q\cos \theta ) $

$ \therefore $ $ \mu =\frac{P+Q\sin \theta }{mg+Q\cos \theta } $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक