Question

Choose the correct value(s) of $F$ which the blocks $M$ and $m$ remain stationary with respect to $M_0$

• A. $(M_0+M+m)\frac{g}{\mu }$
• B. $\frac{m(M_0+M+m)g}{M-\mu m}$
• C. $(M_0+M+m)\frac{mg}{M}$
• D. None of these

Answer 1

The correct option is C $(M_0+M+m)\frac{mg}{M}$

Here is range of force $F$ so thst $M$ and $m$ remains stationary.

for fig. 1;

$mg=f_1+M{a}_{min}$

$mg=\mu m{a}_{max}+M{a}_{min}$

$a_{min}=\frac{m}{(M+\mu m)}$

$f_{min}=(M_0+M+m)a_{min}$

$f_{min}=\frac{m(M_0+M+m)}{M+\mu m}$

for fig. 2;

$M{a}_{max}=mg+f_2$

$M{a}_{max}=mg+\mu \left(m{a}_{max}\right)$

$(M-\mu m)a_{max}=mg$

$a_{max}=\frac{mg}{(M-\mu m)}$

Force $F_{max}$ is given as

$F_{max}=\frac{m(M_0+M+m)}{(M-\mu m)}$

in fig. 3;

when frictional force is zero;

$mg=Ma$

$a=\frac{mg}{\left(M\right)}$

value of $f$ is

$F=(M_0+M+m)a$$=\frac{m(M_0+M+m)g}{M}$

So, option B and C are correct.