Question

Find the mass M in the situation shown in the figure such that m remains at rest on the front surface of $M_1$.The coefficient of friction between the front surface of $M_1$ and that of m is $\mu$.

Answer 1

$Mg-T=Ma$ .......................(1)

$T-(m+M_1)g\sin \theta =(m+M)a$ .............(2)

From eq. (1) and (2);

$(M-(m+M_1\left)\sin \theta \right)g=(M+M_1+m)a$

$a=\frac{(M-(m+M_1\left)\sin \theta \right)g}{(M+M_1+m)}$

Now, consider block $m$ and frictional force $f$ is

$f=\mu N=\mu ma$

$m$ rest on $M_1$ then

$f=mg\cos \theta$

$\Rightarrow \mu a=g\cos \theta$

$\Rightarrow \mu \frac{(M-(M_1+m\left)\sin \theta \right)}{(M+M_1+m)}g=g\cos \theta$

$\Rightarrow \mu M-\mu (M_1+m)\sin \theta =M\cos \theta +(M_1+m)\cos \theta$

$\Rightarrow (\mu -\cos \theta )M=(M_1+m)(\cos \theta +\mu \sin \theta )$

$\Rightarrow M=\frac{(M_1+m)(\cos \theta +\mu \sin \theta )}{\mu -\cos \theta }$