question_answer

The pushing of the block will be possible along the surface if

A) \[\tan \theta \ge \mu \]             

B) \[\cot \theta \ge \mu \]

C) \[\tan \frac{\theta }{2}\ge \mu \]                    

D)   \[\cot \frac{\theta }{2}\ge \mu \]

Correct Answer: C

Solution :

As is clear from the figure, vertical component of applied force increases the weight by mg\[\cos \theta .\]Therefore, normal reaction increases to \[R=mg+mg\cos \theta =mg(1+\cos \theta )\] Force of friction, \[f,=\mu R=\mu mg(1+cos\theta )\] \[=\mu mg.2{{\cos }^{2}}\frac{\theta }{2}.\] The block can be pushed only when horizontal component of applied force \[\ge f\] \[mg\sin \theta \ge \mu \,mg\,2{{\cos }^{2}}\frac{\theta }{2}.\] \[2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\ge \mu 2{{\cos }^{2}}\frac{\theta }{2}\,or\,\tan \theta /2\ge \mu \] Hence, the correction option is [c].