Question

Find the acceleration of the block of mass M in the situation of figure (6−E10). The coefficient of friction between the two blocks is μ1 and that between the bigger block and the ground is μ2.

Answer 1

The acceleration of the block of mass M is  

$\begin{equation}\mathrm{a}=\frac{\left[2 \mathrm{m}-\mu_{2}(\mathrm{M}+\mathrm{m})\right] \mathrm{g}}{\left[\mathrm{M}+\mathrm{m}\left\{5+2\left(\mu_{1}-\mu_{2}\right)\right\}\right]}\end$

Explanation:

Let block M accelerate ' a ' to the right.  

So the ' m ' block has to go down with a ' 2a ' acceleration

Because the block m is in touch with the ' M ' block, it will also have a right acceleration. Therefore, as shown in the free body diagram, it will undergo two internal forces.

From the free body diagram: Consideration of forces on block m in the vertical direction.

$\begin{equation}\begin{array}{l}{m g-T-\mu_{1}(m a)=m(2 a)} \\{T=m g-\mu_{1}(m a)-m(2 a)}\end{array}$

Then look at the vertical forces on the bigger block M, see FBD.  

Downward forces are Mg weight and smaller block friction m =μ_1 (ma) and Tension T at the pulley.

So total downward force = Mg + µ1 . ma+T = Normal force by the ground R'  

So,frictional force = $\begin{equation}\mu_{2}\left(m g+\mu_{1}(m a)+T\right.$ .It's going to be contrary of ' a. '

Now look at the horizontal forces on the bigger block

$\begin{equation}\begin{array}{c}{M 2 T-R-\mu_{2}\left(\mathrm{Mg}+\mu_{1} \cdot \mathrm{ma}+\mathrm{T}\right)=m a} \\{2 \mathrm{mg}-\mu_{1} \cdot 2 \mathrm{ma}-4 \mathrm{ma}-\mathrm{ma}-\mu_{2}\left(\mathrm{Mg}+\mu_{1} \cdot \mathrm{ma}+\mathrm{mg}-\mu_{1} \cdot \mathrm{ma}-2 \mathrm{ma}\right)=m a}\end{array}$

(putting the value of T)

$\begin{equation}2 \mathrm{mg}-\mu_{2} \mathrm{Mg}-\mu_{2} \times \mathrm{mg}+\mu_{2} \times 2 \mathrm{ma}=\mathrm{Ma}+5 \mathrm{ma}+\mu_{1} \times 2 \mathrm{ma}$

                        $\begin{equation}\left[2 m-\mu_{2}(M+m)\right] g=M a+5 m a+\mu_{1} \cdot 2 m a-\mu_{2} \cdot 2 m a$

                        $\begin{equation}\left[2 m-\mu_{2}(M+m)\right] g=\left[M+m\left\{5+2\left(\mu_{1}-\mu_{2}\right)\right\}\right]$

                         $\begin{equation}\left[2 \mathrm{m}-\mu_{2}(\mathrm{M}+\mathrm{m})\right] \mathrm{g}=\left[\mathrm{M}+\mathrm{m}\left\{5+2\left(\mu_{1}-\mu_{2}\right)\right\}\right] a$

                                           $\begin{equation}\mathrm{a}=\frac{\left[2 \mathrm{m}-\mu_{2}(\mathrm{M}+\mathrm{m})\right] \mathrm{g}}{\left[\mathrm{M}+\mathrm{m}\left\{5+2\left(\mu_{1}-\mu_{2}\right)\right\}\right]}$

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

Please refer the above attachment