In the pulley system shown the movable pulleys A,B and C have mass m e

1.	In the pulley system shown the movable pulleys A,B and C have mass m each, D and E are fixed pulleys. The strings are vertical, light and inextensible. Then Pulleys A and B have acceleration each in downward direction and pulley C have acceleration in upward direction Then the value of N ispls follow me also
Answer» ANSWER: the tension throughout the string is the same and equals T=2mg3 . pulleys A and B have ACCELERATION g3 each in downward DIREACTION and PULLEY C has acceleration g3 in wpward direaction . pulleys A,B and C all have accleration g3 in downward direction . pulley A has acceleration g3 in downward direction and pulleys B and C have acceleration g3 each in upward direction . Answer : A::B::D Solution : Let T be the tension in string. −2TaC+TaB−TaA=0 aa+2aC=aB…..(i),mg−2T=maC….(II) T−mg=maB…..(ii),mg−T=maA….(iv)

In the pulley system shown the movable pulleys A,B and C have mass m each, D and E are fixed pulleys. The strings are vertical, light and inextensible. Then Pulleys A and B have acceleration each in downward direction and pulley C have acceleration in upward direction Then the value of N ispls follow me also

Answer»

the tension throughout the string is the same and equals T=2mg3 .

pulleys A and B have ACCELERATION g3 each in downward DIREACTION and PULLEY C has acceleration g3 in wpward direaction .

pulleys A,B and C all have accleration g3 in downward direction .

pulley A has acceleration g3 in downward direction and pulleys B and C have acceleration g3 each in upward direction .

Answer :

A::B::D

Solution :

Let T be the tension in string.

−2TaC+TaB−TaA=0

aa+2aC=aB…..(i),mg−2T=maC….(II)

T−mg=maB…..(ii),mg−T=maA….(iv)