Question

The friction coefficient between the two blocks shown in figure is µ but the floor is smooth $\left(a\right)$ what maximum horizontal force F can be applied without disturbing the equilibrium of the system$?\left(b\right)$ suppose the horizontal force applied is double of that found in part$\left(a\right),$ find the acceleration of the two masses.

• A. $\frac{25}{12}mg$
• B. $\frac{5}{3}mg$
• C. $\frac{2mg\mu }{\left(m+m\right)}$
• D. $\frac{4}{3}mg$

Answer 1

The correct option is C $\frac{2mg\mu }{\left(m+m\right)}$

$\left(a\right)$ $.$F =Fric. + T$

Now$,$ For $T,$

$T=fric.$

$\therefore F=fric.+fric.$

$\Rightarrow F=2\times friction.$

$\Rightarrow F=2\times \mu mg$

$\therefore F=2\mu mg$

$\left(b\right)$ If the Horizontal force applied will becomes twice$,$ then$,$

$F=4\mu mg$

$\Rightarrow 4\mu mg-T-\mu mg=ma$

$\Rightarrow 3\mu mg-T=ma$

For $T,$

$T-\mu mg=Ma$

$\Rightarrow T=m\mu g+ma$

$\therefore 3\mu mg-m\mu g+Ma=ma$

$\Rightarrow 3\mu mg-m\mu g-Ma=ma$

$\Rightarrow a(M+m)=g\mu (3m-m)$

$\therefore a=2mg\mu /(M+m)$

Hence,

option $\left(C\right)$ is correct answer.