Find the acceleration of the block A and B in the three situations shown in figure.

(a) 5a + T - 5g = 0

From free body diagram -1, T = 5g -5a ...(i)

Again $\left(\frac{1}{2}\right)T-4g-8a=0$

$\Rightarrow T-8g-16a=0$

From free body diagram -2,

T = 8g + 16 a ....(ii)

From equations (i) and (ii) , we get

5g -5a = 8g + 16 a

$\Rightarrow 21a=-3g-a=-\frac{9}{7}$

So, acceleration of 5 kg mass is $\frac{9}{7}$ (upward)

and that of 4 g mass is 2a = $\frac{2g}{7}$ (downward)

(b) From the body diagram -3,

4a - $\frac{T}{2}=0$

$\Rightarrow$ 8a - T = 0

$\Rightarrow$ T = 8a

Again, T + 5a -5g = 0

From free body diagram -4,

8a + 5a -5g = s0

$\Rightarrow$ 13a -5g =0

$\Rightarrow$ a =$\frac{5g}{13}$ (downward)

Acceleration of mass 2 kg is 2a = $\frac{10}{13}$ (g) and 5 kg is $\frac{5g}{13}$

(c) T + 1a -1g = 0

From free body diagram-5,

T = 1g -1a ...(i)

Again. from free body diagram -6,

$\frac{T}{2}-2g-4a=0$

$\Rightarrow$ T - 4g -8a = 0

...(ii)

From equation (i),

1g - 1a-4g -8a = 0

$\Rightarrow$ a = $\frac{g}{3}$ (downward).

Acceleration of mass 1 kg is $\frac{g}{3}$ (upward),

Acceleration of mass 2 kg is $\frac{2g}{3}$ (downward),