Question

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

Answer 1

Along horizontal direction, $N_1=ma$

Along vertical direction, $mg-({\mu }_1{N}_1+T)=m\left(2a\right)$ (i)

After substitution value of $N_1$ in Eq. (i), we have

$Mg-({\mu }_{1}ma+T)=m\left(2a\right)$ (ii)

For the motion of Block M:

Along vertical direction $\sum F_v=0$,

or $N_2=T+{\mu }_1{N}_1+Mg$ (iii)

Along horizontal direction,

$2T-(N_1+{\mu }_2{N}_2)=Ma$ (iv)

From equation (i) ans (iii), we get

$2T-[N_1+{\mu }_2(T+{\mu }_1{N}_1+Mg\left)\right]=Ma$ (v)

or $2T-[ma+{\mu }_{2}T+{\mu }_1{\mu }_2(ma)+{\mu }_{2}Mg]=Ma$

Now solving equations (ii) and (iv), we get

$a=\frac{2m-\mu (M-m)g}{M+m[5+2({\mu }_1-{\mu }_2\left)\right]}$