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Question

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

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Solution

## (a) 5a + T - 5g = 0 From free body diagram -1, T = 5g -5a ...(i) Again (12)T−4g−8a=0 ⇒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 ⇒21a=−3g−a=−97 So, acceleration of 5 kg mass is 97 (upward) and that of 4 g mass is 2a = 2g7 (downward) (b) From the body diagram -3, 4a - T2=0 ⇒ 8a - T = 0 ⇒ T = 8a Again, T + 5a -5g = 0 From free body diagram -4, 8a + 5a -5g = s0 ⇒ 13a -5g =0 ⇒ a =5g13 (downward) Acceleration of mass 2 kg is 2a = 1013 (g) and 5 kg is 5g13 (c) T + 1a -1g = 0 From free body diagram-5, T = 1g -1a ...(i) Again. from free body diagram -6, T2−2g−4a=0 ⇒ T - 4g -8a = 0 ...(ii) From equation (i), 1g - 1a-4g -8a = 0 ⇒ a = g3 (downward). Acceleration of mass 1 kg is g3 (upward), Acceleration of mass 2 kg is 2g3 (downward),

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