question_answer

In the arrangement shown in the figure. The mass of wedge A and that of the block B are 3m and in respectively. Friction exists between A and B only. The mass of the block C is m.

The force F = 19.5 m x g is applied on the block C as shown in the figure. The minimum coefficient of friction (\[\mu \]) between A and B so that B remains stationary with respect to wedge A will be

 

A) \[\frac{2}{3}\]                          

B) \[\frac{1}{10}\]

C) \[\frac{2}{5}\]                          

D) \[\frac{1}{4}\]

Correct Answer: D

Solution :

Block C\[19.5mg+mg\sin {{30}^{o}}-T=ma\]

\[20mg-T=ma\]                       ... (1)

Block (B + A) T = 4 m a           ...(2)

From (1) and (2),

\[20mg=5mg\Rightarrow \]           f = mg

N = ma = 4 mg

for\[{{\mu }_{\min }},f={{f}_{\max }}=\mu N=mg\]\[\mu =\frac{1}{4}\]