Question

Four blocks are arranged on a smooth horizontal surface as shown. The masses of the blocks are given (see the figure). The coefficient of static friction between the top and the bottom blocks is ${\mu }_s$. What is the maximum value of the horizontal force $F$, applied to one of the bottom blocks ass shown, that makes all the four blocks move with the same acceleration?

• A. $F_{max}=2{\mu }_{s}mg\left(\frac{2m+M}{m+M}\right)$
• B. $F_{max}={\mu }_{s}mg\left(\frac{m+M}{2m+M}\right)$
• C. $F_{max}=2{\mu }_{s}mg\left(\frac{m+M}{2m+M}\right)$
• D. $F_{max}={\mu }_{s}mg\left(\frac{2m+M}{m+M}\right)$

Answer 1

The correct option is C $F_{max}=2{\mu }_{s}mg\left(\frac{m+M}{2m+M}\right)$

Let all the four blocks move with a common acceleration $a$, then $a=\frac{F}{2(m+M)}$.
Friction between $C$ and $D$ will be more than friction between $A$ and $B$.
Let this friction be $f.$ then

Writing the force equation for $A$, $B$ and $C$ together
$f=(m+m+M)a=\frac{(2m+M)F}{2(m+M)}$
For no slippinh : $f\le f_{max}$
$\frac{(2m+M)F}{2(m+M)}\le {\mu }_{s}mg\Rightarrow F\le \frac{2{\mu }_{s}mg(m+M)}{2m+M}$