question_answer

Block B lying on a table weighs W. The coefficient of static friction between the block and the table is \[\mu .\]Assume that the cord between B and the knot is horizontal. The maximum weight of the block A for which the system will be stationary is

A)  \[\frac{W\tan \theta }{\mu }\]

B)  \[\mu W\,\tan \theta \]

C)  \[\mu W\sqrt{1+{{\tan }^{2}}\theta }\]

D)  \[\mu W\sin \theta \]

Correct Answer: B

Solution :

 Let weight of A is W. From the free body diagram, For equilibrium of the system, \[T\,\cos \theta =\mu N=\mu W\] ?(i) \[T\sin \theta =W\] ?(ii) where, T = tension in the thread lying between knot and the support. On divindg Eq. (ii) by Eq. (i). we get \[\frac{T\sin \theta }{T\cos \theta }=\frac{W}{\mu W}\] \[\Rightarrow \] \[\tan \theta =\frac{W}{\mu W}\Rightarrow \mu W\tan \theta \]