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Vinod_khatanhar asked this question on Class XI-Science » Physics » Laws Of Motion

i have been trying to solve this question but i am unable to do so. please explain me with diagrams(FBD). The question is from the book Concepts of physics by H C VERMA and chapter name is Laws of Motion. i cannot explain the question verbally as it has a diagram to be seen so that we understand the question.the question number is 28 .

m1=1kg,m2=2kg, m3=3kg. Masses m2 and m3 are attached to a single movable pully which is attached to a fixed pully having m1 on the other side. i calculated the equations for T(tension) but i think they are wrong.

(i)T = a1 + g (ii) T/2 = 2g + 2 a2 - 2 a1 (iii)T/2 = 3g - 3 a1 - 3 a2 . please correct me and tell me its solution

Asked by Vinod_khatanhar(student), on 28/7/12
Answers

see the following figure:

On second pulley:

Tension T acts upwards and and tension T' in lower string. So,

T = T'+T'

T' = T/2

For mass m1:

1. Tension T upwards,

2. weight m1g downwards,

3. Let the mass moves upwards with acceleration a thus,

..........(1)

For mass m2:

1. Tension T' upwards,

2. Weight m2g downwards,

3. mass moves upward with acceleration a' and downward with acceleration a. Now,

................(2)

For mass m3:

1. Tension T' upwards,

2. Weight m3g downwards,

3. downward acceleration a and a'. Now,

................(3)

By solving above three equations we get,

So the acceleration on mass 1 is (19g/29) upwards, on mass 2 (17g/29) downwards and on mass 3 (21g/29) downwards.

Posted by Ruhhi Ralhan(MeritNation Expert), on 30/7/12
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