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Finding Tensions and Accelerations with Moveable Pulley

  • Thread starter ryao3688
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(Moved to the homework section by moderator.)

Please see the image. My confusion is with a moveable pulley. Can I just do my balances on both 1kg and 2kg masses disregarding the acceleration of the pulley? Here are my eqns. Object numbers 1, 2 and 3 denote the 3kg, 1kg and 2 kg weights. 4 is the moveable pulley. T1 is the string on the upper left, T2 for the lower right string tension. Let's say up is neg, down is positive.

ma = sum of F's for all

m1a1 = m1g - T1
m2a2 = m2g - T2
m3a3 = m3g - T2
m4a4 = 2T2 - T1 + m4g

But I know that a1= -a4 and a2= -a3 since if the pulley moves one way with an object, the other one must travel in opposite direction with same acceleration.

You sub that in, you get four eqns and four unknowns and I can easily solve. But my confusion is really with that moveable pulley. I included its weight in the eqn. Do I need to include its acceleration into objects 2 and 3? Did I make any mistake? Thanks.
 

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  • #2
BvU
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Can I just do my balances on both 1kg and 2kg masses disregarding the acceleration of the pulley?
Yes and no. Write down the various tensions. That are the things that are the same on both sides of a pulley (why ?!). Tensions and gravity determine accelerations (there are no other forces acting).

You can practice on the easier case (3kg on left, 1 mass of 4kg on right).
 
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  • #3
haruspex
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a2= -a3
Hmm.... Your accelerations are all in the lab frame, right?

(I do not like this question. It includes a pulley of significant mass, which will accelerate rotationally, but the radius is not given. It would be better if it were a light pulley with a 1kg mass suspended from the pulley's axle.)
 
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  • #4
BvU
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Haru's Hmm... is a well-placed hint. (Much more useful than my yesno :-)

I agree with his criticism on the exercise composer, but it may be understandable if the moments of inertia and/or the angular accelerations need to be ignored -- for example because they haven't been treated yet at the current location in the textbook.
 
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Sorry for so many accelerations - I think I realized now that you could just set up your axes for your coordinates system differently on each side of the string, rather than just set up one axis and change the sign of the acceleration (right now I treat everything down as pos, up as neg).

I think I've got it now Haru. The a's in the second and third equation should have both itself and acceleration from the pulley since the whole frame is moving?
Keep the first and fourth eqn. Replace those second and third equations with:
m2(a2+a4)=m2g-T2
m3(a3+a4)=m3g-T2

Does this look better? And here again, a2= -a3, a weight should go up and the other should go down evenly since the length of string is 1:1, acceleration is directly proportional to length. Or I could've just flipped the directions around instead of dealing with so many accelerations.
 
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haruspex
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Sorry for so many accelerations - I think I realized now that you could just set up your axes for your coordinates system differently on each side of the string, rather than just set up one axis and change the sign of the acceleration (right now I treat everything down as pos, up as neg).

I think I've got it now Haru. The a's in the second and third equation should have both itself and acceleration from the pulley since the whole frame is moving?
Keep the first and fourth eqn. Replace those second and third equations with:
m2(a2+a4)=m2g-T2
m3(a3+a4)=m3g-T2

Does this look better? And here again, a2= -a3, a weight should go up and the other should go down evenly since the length of string is 1:1, acceleration is directly proportional to length. Or I could've just flipped the directions around instead of dealing with so many accelerations.
Yes, that's the way.
 

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