In the system shown in fig., `m_(A)=4m`, `m_(B)=3m`, and `m_(c)=8m`. Friction is absent everywhere. String is inextensible and light. If the system is released from rest, then

Acceleration of Block B is

Options

A. `1.5 mg`
B. `5.8 mg`
C. `4.7 mg`
D. `3.2 mg`

Answer

A

Explanation

Let acceleration of block C be `a_(1)` (rightwards) and acceleration of block B be `a_(2)` (lefwards).
Then acceleration of A will be `(a_(1)+a_(2))` downwards and `a_(1)` rightwards.
Free body diagram of A is shown in fig.

Using `Sigma F_(x)=ma_(x)` and `Sigma F_(y)=ma_(y)`, we get
`N=4m(a_(1))` ..(i)
and `4mg-T=4m(a_(1)+a_(2))` ..(ii)
Free bocy diagram of B (showing horizontal force only) is shown in the figure.

Using `Sigma F_(x)=ma_(x)`, we get
`T=3ma_(2)` ...(iii)
Free body diagram of C (showing horizontal force only) is shown in fig.
Using `Sigma F_(x)=ma_(x)`, we get
`T-N=8ma_(1)`...(iv)
We have four unknowns `T, N, a_(1)`, and `a_(2)`. solving these four equations, we get
`a_(1)=g/8` and `a_(2) = g/2`, `a_(1)+a_(2)=5/8 g`
Thus, acceleration of A is `g//8` in horizontal direction and `5g//8` in vertical direction.
Acceleration of B is `g//2` in horizontal direction (leftwards) and acceleration of C is `g//8 in horizontal direction (rightwards).