Question

Block $1$ sits on top of block $ 2$ . Both of them have a mass of $1 \,kg$. The coefficients of friction between blocks $1$ and $2$ are $\mu_s=0.75$ and $\mu_k=0.60$. The table is frictionless. A force $\frac{P}{2}$ is applied on block $1$ to the left, and force $P$ on block $2$ to the right. Obtain the minimum value of $P=P_{\min }$ such that both blocks move relative to each other and fill $\frac{P_{\min }}{2}$ in OMR.

Answer 1

The acceleration of both bodies is the same.

For block 1
$P-f=maP-f=a......\left(\right.1\left.\right)$
For block 2,
$f-\frac{P}{2}=maf-\frac{P}{2}=a......\left(\right.2\left.\right)$
Subtract equation (1) and equation (2),
$P-f-f+\frac{P}{2}=0\frac{3 P}{2}=2fP=\frac{4 f}{3}$
$P=\frac{4 \mu _{s} m g}{3}P=\frac{4 \times 0 . 75 \times 1 \times 10}{3}=10$
So the value of
$\frac{P_{ \min}}{2}=\frac{10}{2}=5$