Question

In the arrangement shown in the figure, the mass of wedge $A$ and that of the block $B$ are $3\text{m}$ and $\text{m}$ respectively. Friction exists between $A$ and $B$ only. The mass of the block $C$ is $m$. The force $F=19.5\text{m}g$ 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 $\frac{1}{x}$ then, find the value of $x$

Answer 1

Applying Newtons second law for block $C$:
$19.5mg+mg\sin {30}^{\circ }-T=ma$
$20mg-T=ma$ ... (1)
Applying Newtons second law for block $(B+A)$:
$T=4ma$ ... (2)
From (1) and (2)
$20mg=5ma\Rightarrow a=4\text{g}$

Applying Newtons second law for block B
$f=mg$
$N=ma=4mg$
$f=mg$
$f\le \mu N$
$mg\le \mu 4mg$
Hence
$\mu \ge \frac{1}{4}$