Question

Find the magnitudes and directions of frictional forces at contact points.

Answer 1

The set of forces for the cylinder and the acceleration are $F_1$ is frictional force between the body and cylinder. $F_2$ is frictional force of the cylinder and ground $a_1$ acceleration of the plank and $a_2$ acceleration and $\alpha$ is the angular acceleration of cylinder.

For the Plank

$F-F_1=m_2{a}_1⟶\left(i\right)$

For the cylinder

$F_1+F_2=m_1{a}_2⟶\left(ii\right)$

For the rotational motion of cylinder

$(F_1-F_2)r=I\beta ⟶\left(iii\right)$

$I=$ moment of inertia of cylinder about $O=\frac{1}{2}{M}_1{r}^2$ here frictional forces on cylinder due to plank acts toward right wherea for plank it acts toward.

Again we have

$a_1=a_2+\beta r⟶\left(iv\right)$

and $\beta =\frac{a_2}{r}⟶\left(v\right)$(As the cylinder rotates without slip)

so i have,

$A_1=2a_2$

Putting these values in equation (iii) we get,

$(F_1-F_2)r=I\frac{a_2}{r}$

$\Rightarrow (F_1-F_2)=\frac{1}{2}{m}_1{r}^2\times \frac{a_2}{r^2}⟶\left(vi\right)$

From $\left(ii\right)$ and $vi$ we get

$F_1=\frac{3}{4}{a}_2.m_1$

$F_2=\frac{1}{4}{a}_2.m_1$

Putting the value of $F_1$ and $a_1$ in equation $\left(i\right)$ we get

$F-\frac{3}{4}{a}_2.m_1=m_2\times 2a_2$

$\Rightarrow F=a_2(2m_1+\frac{3}{4}{m}_1)$

$\Rightarrow a_2=\frac{4F}{8m_2+3m_1}$

so, $F_1=\frac{3F{m}_1}{8m_2+3m_1}$

$F_2=\frac{F{m}_1}{8m_2+3m_1}$