Question

In the figure, a block of mass $M$ is placed on a table with rough surface having coefficient of friction ${\mu }_1$. Another block of mass $m$ is placed on vertical rough surface of this block with coefficient of ${\mu }_2$ (${\mu }_2$ may be greater than $1$). If system is released from rest, find the maximum mass of block $A$, so that the block of mass m does not slip.

• A. $\frac{(M+m)({\mu }_1{\mu }_2+1)}{({\mu }_2+1)}$
• B. $\frac{(M+2m)({\mu }_1{\mu }_2+1)}{({\mu }_2-1)}$
• C. None of these
• D. $\frac{(M+m)({\mu }_1{\mu }_2+1)}{({\mu }_2-1)}$

Answer 1

The correct option is D $\frac{(M+m)({\mu }_1{\mu }_2+1)}{({\mu }_2-1)}$

Acceleration of the block:

$a=\frac{m_{A}g-{\mu }_1(M+m)g}{(m_A+m+M)}\dots \left(i\right)$

If the block $m$ is not sliding acceleration of the system $F.B.D$ of $mw.r.t.M.$

If $m$ is not sliding

$f=mg\dots \left(ii\right)$

$N=ma\dots \left(iii\right)$

$f\le {\mu }_{2}N$

$mg\le {\mu }_{2}ma$

$g\le {\mu }_2\left[\frac{m_{A}g-{\mu }_1(M+m)g}{m_A+m+M}\right]$

$(m_A+m+M)\le {\mu }_2(m_A-{\mu }_1(M+m)$

$({\mu }_2-1)m_A\ge (M+m)+{\mu }_1{\mu }_2(M+m)$

$m_A\ge \frac{(M+m)[1+{\mu }_1{\mu }_2]}{({\mu }_2-1)}$

Final answer : Option(a)