Find the realation between the velocity of block $A$ and velocity of block $B$ in the shown figure.

8 v_{A} + 2 v_{B} = 4 v

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4 v_{A} + v_{B} = 2 v

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8 v_{A} + v_{B} = 6 v

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8 v_{A} + 3 v_{B} = 2 v

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Solution

The correct option is C $8 v_{A} + v_{B} = 6 v$
In the given condition, let the velocities of block $A$ and $B$ are denoted as $v_{A}$ and $v_{B}$ respectively.
Also, let the velocities of block $A$ , block $B$ , pulley $P_{1}$ and pulley $P_{2}$ be upward respectively and taking upward positive we have

For pulley $P_{2}$ we have
$v_{P 2} = \frac{v_{D} + v_{E}}{2} = \frac{- v - 3 v}{2} = - 2 v$
Thus, the velocity of pulley $P_{2}$ will be $2 v$ , downward.

Now, as we know that, the string's end attached to the fixed pulley will have zero velocity. So, considering the points $1$ to $6$ in the figure, we have
$v_{P 2} - v_{A} - v_{A} - v_{A} - v_{A} - v_{P 1} = 0$
$\Rightarrow 2 v - v_{P 1} = 4 v_{A} \Rightarrow v_{P 1} = 2 v - 4 v_{A}$ ......(i)

Again, for pulley $P_{1}$ we have
$v_{P 1} = \frac{v_{B} + v_{C}}{2} = \frac{v_{B} - 2 v}{2}$ .....(ii)
From equation (i) and (ii) we have
$2 v - 4 v_{A} = \frac{v_{B} - 2 v}{2}$
$\Rightarrow 8 v_{A} + v_{B} = 6 v$
Hence, the relation between the velocities of block $A$ and block $B$ is $8 v_{A} + v_{B} = 6 v$

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