Question

In the system shown in the fig blocks A and B have mass $m_1$ = $2kg$ and $m_2$ =$\frac{26}{7}kg$ respectively. Pulley having moment of inertia $I=0.11kg{m}^{-2}$ can rotate without friction about a fixed axis. Inner and outer radii of pulley are $a=10cm$ and $b=15cm$ respectively. B is hanging with the thread wrapped around the pulley, while A lies on a rough inclined plane. Coefficient of friction being $\mu =\sqrt{\frac{3}{10}}$. Then, ($g=10m{s}^{-2}$)

• A. acceleration of block, $A$ is $2m{s}^{-2}$ (up the plane)
• B. acceleration of block, $B$ is $3m{s}^{-2}$ (downward)
• C. Tension in string $1$, $T_1=17N$
• D. Tension in string $2$, $T_2=26N$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A acceleration of block, $B$ is $3m{s}^{-2}$ (downward)
B Tension in string $2$, $T_2=26N$
C Tension in string $1$, $T_1=17N$
D acceleration of block, $A$ is $2m{s}^{-2}$ (up the plane)

The block $B$ has a downward acceleration $a_1$ in downward direction and block $A$

The frictional force acting on block $A$ is

$F_r=\mu N=\mu \cos {35}^o$ $\mu =$ coefficient of friction$=\sqrt{\frac{3}{10}}$

<The equation of motion of block $s$ are

$m_{2}g-T_2=m_2{a}_1⟶\left(i\right)$

$\Rightarrow T_1-m_{1}g\sin {35}^o-F_r=m_1{a}_1$

$\Rightarrow T_1-m_{1}g\sin {35}^o-\mu M_{1}g\cos {35}^o=m_1{a}_1⟶\left(ii\right)$

Let $\alpha$ be the angular acceleration of the motion

So, $\alpha .\beta =a_2⟶\left(iii\right)$

$\alpha .a=a_1⟶\left(ii\right)$

and,

$T_2.b-T_1.a=I\alpha$(For motion of pully)$⟶\left(v\right)$

Here,

$I=$ Moment of inertia of pully.

$a=$ In radius of pully $=10cm$

$b=$Outer radius of Pully$=15cm$

$m_1=02Kg=$ mass of block A

$m_2=03Kg=$ mass of block B

Putting these value in above equation we get

$10\alpha =a_1$

$15\alpha =a_2$

$\Rightarrow 5\alpha =a_2-a_1⟶\left(vi\right)$

$15T_2-10T_1=0.11\alpha ⟶\left(vii\right)$

Solving these equation we get

$a_1=2m{s}^{-1}$(upward direction along the plane)

$a_2=3m{s}^{-2}$(Downward direction)

$T_1=17N$

$T_2=26N$