Find the acceleration of the blocks A and B in the three situations shown in the following figure.
Solution
(a) 5a + T − 5g = 0
From free-body diagram (1),
T = 5g − 5a .....(i)
Again,
\[\left( \frac{1}{2} \right)T - 4g - 8a = 0\]
⇒ T − 8g − 16a = 0
From free-body diagram (2),
T = 8g + 16a ......(ii)
From equations (i) and (ii), we get:
5g − 5a = 8g + 16a
\[\Rightarrow 21a = - 3g - a = - \frac{9}{7}\]
So, the acceleration of the 5 kg mass is \[\frac{9}{7} m/ s^2 \left(\text{ upward }\right)\] and that of the 4 kg mass is
\[2a = \frac{2g}{7} \left(\text{ downward }\right)\]
\[4a - \frac{T}{2} = 0\]
⇒ 8a − T = 0
⇒ T = 8a
Again, T + 5a − 5g = 0
From free body diagram-4,
8a + 5a − 5g = s0
⇒ 13a − 5g = 0
\[\Rightarrow a = \frac{5g}{13} \left(\text{ downward }\right)\]
Acceleration of mass 2 kg is \[2a = \frac{10}{13} \left( g \right)\] 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,
\[\frac{T}{2} - 2g - 4a = 0\]
⇒ T − 4g − 8a = 0 .....(ii)
From equation (i)
1g − 1a − 4g − 8a = 0
\[\Rightarrow a = \frac{g}{3}\left(\text{ downward }\right)\]
Acceleration of mass 1 kg is \[\frac{g}{3} \left(\text{ upward }\right)\]
Acceleration of mass 2 kg is \[\frac{2g}{3} \left(\text{ downward }\right)\] 
