Question

Two blocks, $m_{1}=2 \,kg$ and $m_{2}=4 \,kg$, are connected with a light string that runs over a frictionless peg to a hanging block with a mass $M$ as shown in figure. The coefficient of sliding friction between block $m _{2}$ and the horizontal surface is $\mu_{ k }=0.2$. The coefficient of static friction between the two blocks is $\mu s =0.4$. What is the maximum mass $M$ for the hanging block if the block $m_{1}$ is not to slip on block $m_{2}$ while $m_{2}$ is sliding over the surface?

Answer 1

for not to slip $m_{1}$ pseudo force on $m_{1}$
should be equal to friction force $\left(f_{s}\right)$ on $m_{1}$.
$m_{1 a}=\mu_{s} m_{1 g}$
$a =\mu_{ s } g$
$a=0.4 \times 10$
$a=4 \,m / s ^{2}$
From fBD of $m _{2}$

Here $ f _{ k }=\mu_{ k }\left( m _{1}+ m _{2}\right) g$
$T - f _{ k }=\left( m _{1}+ m _{2}\right) a$
$=0.2 \times 6 \times 10$
$T -12=6 a =12 \,N$
$T -12 =6 \times 4 $
$T =36\, N$
For $M$
$Mg - T = Ma$
$10 M =36+4 M $
$6 M =36$
$\Rightarrow M =6\, kg $