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 Author Topic: Another Pulley Problem  (Read 23164 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Bitupon
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 « Embed this message on: December 14, 2005, 03:56:15 pm »

[size=18:825bcc8bd7][b:825bcc8bd7]Two blocks A and B of mass 1kg and 2kg respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taut. At moment t=0 ,a force F=20t N starts acting on the pulley along vertically upward direction as shown in the figure. Calculate
(a) Velocity of A and B when loses contact with the floor.
(b) Height raised by the pulley upto that instant and
(c) Work done by the force upto that instant
(Consider g=10m per sec sq.)[/b:825bcc8bd7] [/size:825bcc8bd7]

[size=24:825bcc8bd7][color=red:825bcc8bd7]Solution:
(a) Let T be the tension in the string .

2T=20t
Or, T=10t N

Let the block A loses its contact with the floor at time t=t¡¦.
This happens when the tension in the string becomes equal to the weight of A.Thus,
T=mg
Or,10t¡¦=10
Or, t¡¦=1sec
Similarly for block B,we have
10 t¡¦¡¦=20
Or, t¡¦¡¦=2sec
i.e. the block B loses contact after 2secs.
For block A , at time t such that t¡¦<t, let a be the acceleration in upward direction. Then
10t-10=a=(dv/dt)
Or, dv=10(t-1)dt
Integrating we get,
v=5(t^2)-10t+5
At t=2sec , v=5m/s

[/size:825bcc8bd7]
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Fu-Kwun Hwang
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 « Embed this message Reply #1 on: December 28, 2005, 12:46:13 pm » posted from:Taipei,T'ai-pei,Taiwan

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When the force F is large enough both blocks will be lifted, but if the force is not large enough , the heavy one will be supported by the base.
You can study the following simulation.

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