Question

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

Answer 1

The block 'm' is in contact with block 'M'. So small block has also acceleration towards right.

Therefore it will experience two inertia forces.

From FBD-1,

$R_1-ma=0R_1=ma---------\left(1\right)$

$2ma+T-mg+{\mu }_1{R}_1=0=>T=mg-(2-{\mu }_1)m{a}_1--------\left(2\right)$

From FBD-2,

$T+{\mu }_1{R}_1+mg-R_2=0=>R_2=T+{\mu }_{1}ma+Mg$

$R_2=(mg-2ma-{\mu }_{1}ma)+{\mu }_{1}ma+Mg\therefore R_2=Mg+mg-2ma--------\left(3\right)$

Again from FBD-2,

$T+T-R-Ma-{\mu }_2{R}_2=0=>2T=(M+m)+{\mu }_2(Mg+mg-2ma)=0$

$2T=(M+m)+{\mu }_2(Mg+mg-2ma)-----\left(4\right)$

From equation (2) and (4),

$2T=2mg-2(2+{\mu }_1)mg=(M+m)a+{\mu }_2(Mg+mg-2ma)$

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