Question

At what maximum mass $M$ (in kg) of the block, the bars $m_1$ and $m_2$ will not slide over each other? If the coefficient of friction between the bars is equal to $\mu =\frac{1}{8}$, and the coefficient of friction on the plane is equal to zero. Strings are massless and inextensible, pulleys are ideal. (take $g=10m/s^2$)

Answer 1

If the bars are not sliding over each other, it means both the bars and the block have same acceleration $a$.
The given system can be reframed as shown below


The FBDs of the block and the bars are as shown

As the rough surface exists between bars $m_1$ and $m_2$, the friction force between them is given by
$f=\mu N=\mu \times 2g=\frac{1}{8}\times 2g=2.5N$
So, from the FBDs of the bars we have
$T+f=2a\Rightarrow T+2.5=2a$
$T-f=1a\Rightarrow T-2.5=1a$
From these two equations we have,
$a=5m/s^2$
and, $T=(2.5+5)=7.5N$
Now, from the FBD of the block we get
$Mg-2T=Ma$
$\Rightarrow M\times 10-2\times 7.5=M\times 5$
$\Rightarrow 5M=15$
$\Rightarrow M=3kg$