Find the acceleration of the blocks A and B in the three situations shown in figure (5−E17).
Answers
Given :
5a + T − 5g = 0
From free-body diagram (1),
T = 5g − 5a …..(i)
1/2T-4g-8a=0
⇒ T − 8g − 16a = 0
(1)
From free-body diagram (2),
T = 8g + 16a ……(ii)
From equations (i) and (ii), we get:
5g − 5a = 8g + 16a
⇒21a=-3g-a=-9/7
So, the acceleration of the 5 kg mass is 97 m/s2 upward
and that of the 4 kg mass is
2a=2g/7 downward.
(b) From free body diagram-3,
4a-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
⇒a=5g/13 downward
Acceleration of mass 2 kg is
2a=10/13 (g)and 5 kg is 5g/13.
(c) T + 1a − 1g = 0
From free body diagram-5
T = 1g − 1a …..(i)
Again, from free body diagram-6,
T/2-2g-4a=0
⇒ T − 4g − 8a = 0 …..(ii)
From equation (i)
1g − 1a − 4g − 8a = 0
⇒a=g/3 downward
∴Acceleration of mass 1 kg is g/3 upward,
Acceleration of mass 2 kg is 2g/3 downward
5a + T − 5g = 0
From free-body diagram (1),
T = 5g − 5a …..(i)
1/2T-4g-8a=0
⇒ T − 8g − 16a = 0
(1)
From free-body diagram (2),
T = 8g + 16a ……(ii)
From equations (i) and (ii), we get:
5g − 5a = 8g + 16a
⇒21a=-3g-a=-9/7
So, the acceleration of the 5 kg mass is 97 m/s2 upward
and that of the 4 kg mass is
2a=2g/7 downward.
(b) From free body diagram-3,
4a-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
⇒a=5g/13 downward
Acceleration of mass 2 kg is
2a=10/13 (g)and 5 kg is 5g/13.
(c) T + 1a − 1g = 0
From free body diagram-5
T = 1g − 1a …..(i)
Again, from free body diagram-6,
T/2-2g-4a=0
⇒ T − 4g − 8a = 0 …..(ii)
From equation (i)
1g − 1a − 4g − 8a = 0
⇒a=g/3 downward
∴Acceleration of mass 1 kg is g/3 upward,
Acceleration of mass 2 kg is 2g/3 downward