Question

Find the acceleration of the block of mass $M$ in the situation of figure. The coefficient of the friction between the two blocks is ${\mu }_1$ and that between the bigger block and the ground is ${\mu }_2$.

Answer 1

$\mu 1.ma=m.2a$

$\to T=mg-\mu 1.ma-2ma$

Now consider the vertical forces on the larger block $M$

Downward forces are weight $Mg$ and friction by smaller block $m=\mu 1.ma$ and Tension $T$ at the pulley.

So total downward force $=Mg+\mu 1.ma+T=$ Normal force by the ground $R^{\prime }$

So friction force $=\mu 2(Mg+\mu 1.ma+T)$ its direction will be opposite to $a$

Now consider the horizontal forces on the larger block

$M2T-R-\mu 2(Mg+\mu 1.ma+T)=Ma$

$\to 2mg.\mu 1.2ma-4ma-ma-\mu 2(Mg+\mu 1.ma+mg-\mu 1.ma-2ma)=Ma$ (putting the value of $T$)

$\to 2mg+\mu 2Mg-\mu 2.mg+\mu 2\ast 2ma=Ma+5ma+\mu 1.2ma$

$\to [2m-\mu 2(M+m\left)\right]g=Ma+5ma+\mu 1.2ma-\mu 2.2ma$

$\to [2m-\mu 2(M+m\left)\right]g=[M+m\{5+2(\mu 1-\mu 2\left)\right\}]a$

$\to a=[2m-\mu 2(M+m\left)\right]g/[M+m\{5+2(\mu 1-\mu 2\left)\right\}]$