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