Question

Block $1$ sits on top of block $2$ . Both of them have a mass of $1kg$. 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

Answer: 5

The acceleration of both bodies is the same.

For block 1
$P-f=maP-f=a......(1)$
For block 2,
$f-\frac{P}{2}=maf-\frac{P}{2}=a......(2)$
Subtract equation (1) and equation (2),
$P-f-f+\frac{P}{2}=0\frac{3P}{2}=2fP=\frac{4f}{3}$
$P=\frac{4{\mu }_{s}mg}{3}P=\frac{4\times 0.75\times 1\times 10}{3}=10$
So the value of
$\frac{P_{\min }}{2}=\frac{10}{2}=5$