Question

In the arrangement shown in the figure. The mass of wedge $A$ and that of the block $B$ are $3m$ and $m$ respectively. Friction exists between $A$ and $B$ only. The mass of the block $C$ is $m$. The force $F=19.5m\times g$ is applied on the block $C$ as shown in the figure. The minimum coefficient of friction $\left(\mu \right)$ between $A$ and $B$ so that $B$ remains stationary with respect to wedge $A$ will be:

• A. $\frac{4}{5}$
• B. $\frac{1}{10}$
• C. $\frac{2}{5}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D $\frac{1}{4}$

Block C. $19.5mg+mgsin{30}^o-T=ma$

$20mg-T=ma$ .....(1)
Block $(B+A)T=4ma$ ......(2)
from (1) and (2),
$20mg=5ma⟹a=4g$
$f=mg$
$N=ma=4mg$
for ${\mu }_{min},f=f_{max}=\mu N=mg$
$\mu =\frac{1}{4}$