Question

The situation shown in the figure, a block $A$ of mass $2\, kg$ is placed inside a box $B$ of mass $2\, kg$. The coefficient of static and kinetic friction between the blocks and also between box and ground are $\mu_{s}=0.6$ and $\mu_{ k }=0.4$. If a force $F$ is applied on the box $B$, the block $A$ begins to slide relative to the box. The value of $F$ (in newton) is____.

Answer 1

The maximum static friction on block $A$ is
$f_{1 \max }=\mu_{s} m_{1} g=0.6 \times 2 \times 10=12\, N$
The maximum static friction between ground and the box $B$
$f_{2 \max }=\mu_{s}\left(m_{1}+m_{2}\right) g$
$=0.6 \times 4 \times 10=24\, N$
Assume both blocks move together.
$F-\mu_{k}\left(m_{1}+m_{2}\right) g=\left(m_{1}+m_{2}\right) a$
Or $F-0.4 \times 4 \times 10=4 a$
$\because F-16=4 a$
For block $A, f_{1}=2 a=2\left(\frac{F-16}{4}\right)$
as $f_{1}=\frac{F-16}{2}$
For just sliding, $f_{1}=f_{1 \max }$
$\Rightarrow \frac{F-16}{2}=12$
$\therefore F=40\, N$