Question

13. Figure shows a system of four pulleys with two
masses m = 3 kg and me = 4 kg. At an instant,
force acting on block A, if block B is going up
at an acceleration of 3 m/s2 and pulley Q is
going down at an acceleration of 1 m/s2 is
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(A) 7 N acting upward
(B) 7 N acting downward
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(C) 10.5 N acting upward
(D) 10.5 N acting downward.​

Answer 1

Answer:

here is your answer

Explanation:

7N  acting upward

7N acting downward

10.5N acting upward

10.5N acting downward

Answer :

D

Solution :

Writing constraint for the string

2xA+2xB−xQ+k=l

Differnting twice w.r.t time 2aA+2aB−aQ=0

Taking downward direction to be positive

2aA+2(−3)−1=0,2aA=7,aA=3.5m/s2

Positive sign indicates that block is acceleration at 3.5m/s2 in downward direaction.

So, force acting F=3×3.5=10.5N (downward) .

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Answer 2

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