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Acceleration due to *equal* weights on pulley system

  1. Nov 9, 2016 #1
    1. The problem statement, all variables and given/known data
    main-qimg-538caf7b76380d26e224c97725471535?convert_to_webp=true.gif

    Which of the following is correct?
    1. The acceleration of pulley B is g/11 downwards
    2. The acc. of pulley C is g/11 upwards
    3. Tension in string passing over pulley A is 12.46g N
    4. Tension in string passing over pulley A is 10g N

    2. Relevant equations
    F=m.a



    3. The attempt at a solution

    I wrote equations assuming pulley B to go down and pulley C to go up, with going down being positive.
    Here, T is the tensions in the strings. I think the tension in the string over pulley B and C are equal, that is 'T' here.
    a' is acceleration of pulley B.
    a'' is acceleration of pulley C
    a is acceleration of the weights with respect to pulley.
    My doubt is, in these equations below, should the RHS accelerations be WRT to ground frame or pulley frame? But here are my equations, though solving them doesnt give me answers matching to any of the options
    1. 8g - T = 8(a'-a)
    2. T - 6g=6(-a'-a)
    3. 10g - T = 10(a''+a)
    4. T - 4g = 4(a-a'')

    Any help would be appreciated.
     
  2. jcsd
  3. Nov 9, 2016 #2
    Are you saying that all four of the tension forces that act on the 4 masses (4, 6, 8, and 10 kg) are equal?

    Sometimes, I like to take things to extremes because it makes it easier for me to see if the assumption holds true. So is it true that if the two weights hanging off of pulley B add up to be the same as the two weights hanging off of pulley C, then the tensions in those strings will be the same - i.e. all 4 forces acting on the 4 masses are equal? What if on pulley B you had one weight at 7.001 kg and the other at 6.999 kg, and on pulley C you had one weight at 14.00000 kg and the other weight was a piece of dust? Would the two tensions be the same in that case?

    I believe the problem is asking for the absolute linear accelerations of pulleys B and C. I couldn't figure out you meant by RHS????

    Oh, by the way, welcome to Physics Forums.
     
  4. Nov 10, 2016 #3

    haruspex

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    The title says equal weights. Why?
    Your set of variables seems strange, or maybe unclear. Pulleys B and C must have equal and opposite linear accelerations at their centres, but different rotational accelerations. Likewise, there are three different tensions to consider.
    Each variable should have a unique name.
     
  5. Nov 10, 2016 #4
    Assuming the tension in the string over pulley B is T, then the tension in the string over the pulley A has to be 2T.
    Assuming the tension in the string over pulley C is T'. then the tension in the string over the pulley B has to be 2T'.
    Now 2T' has to be equal to 2T since it is the same string. So T=T'. Did I reason it out right?

    RHS means Right Hand Side of the equation. (Sorry!)

    Yes, the problem does ask that. So, the acceleration terms look right to me in these equations then.
    Thanks!
     
  6. Nov 10, 2016 #5
    Since the pulley's are massless, they wont have torques on them, right?
    And even though the pulleys have equal weights on both sides (10+4 and 6+8), they will move I think
     
  7. Nov 10, 2016 #6
    If T is the tension of string over the pulley B, then 2T is the force acting on pulley B as a result. However, if B is accelerating, then the tension in the string over pulley A cannot be equal to 2T. Otherwise, B would not be accelerating.
     
  8. Nov 10, 2016 #7
    So should I consider different tensions over the 3 pulleys?
    What will be the relation between them then?
     
  9. Nov 10, 2016 #8
    Yes, as haruspex mentioned below.
     
  10. Nov 10, 2016 #9
    Hi Shashank ,

    You only require tension in the strings connecting the masses . Will the tensions be same or different ?

    Do you believe acceleration of the masses (attached to pulley B ) with respect to pulley B is same as the acceleration of the masses (attached to pulley C) with respect to pulley C ?

    I would suggest you to work with these variables :

    a = magnitude of acceleration of pulleys B and C . Assume B goes down and C goes up .

    a1 = magnitude of acceleration of 8Kg and 6Kg masses with respect to pulley B .Assume 8 Kg goes down and 6 Kg goes up w.r.t pulley B .

    a2 = magnitude of acceleration of 10Kg and 4Kg masses with respect to pulley C .Assume 10 Kg goes down and 4 Kg goes up w.r.t pulley C .

    T = Tension in the strings connecting the four masses .

    Now think carefully and rewrite the above four equations in terms of the notation I have suggested . You will have four variables and four equations .That will give you the correct options .
     
    Last edited: Nov 10, 2016
  11. Nov 10, 2016 #10
    Is it really true that you don't need to consider the tension in the string of pulley A? Or did I misunderstand you?

    So the tension forces acting on the 4 weights (8 kg, 6 kg, 10 kg, and 4 kg) are all equal. Are you sure?
     
  12. Nov 10, 2016 #11

    haruspex

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    Ah, ok - your title meant equal total masses. Yes, there will be movement of the pulleys.
    As TomHart notes, you need to consider tensions in the strings over all three pulleys, and the three are likely all different.
     
  13. Nov 10, 2016 #12

    ehild

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    It is not stated in the problem, but the pulleys usually are assumed massless. Otherwise the masses should have been given.
    If the mass of pulley B is zero, the net force acting on it should be zero. So the tension in the string over A is 2T when the tension in the string over B is T.
     
  14. Nov 10, 2016 #13

    ehild

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    Yes, it is right.
     
  15. Nov 10, 2016 #14
    We only need tensions in the strings connecting the masses .

    Yes .

    All strings are massless and all pulleys are massless and frictionless .
     
  16. Nov 10, 2016 #15
    They are different but they bear a simple relationship between them since pulleys are massless .

    But as I said earlier , we only need tensions in the strings connecting the four masses to write Newton's II Law separately for the masses .These tensions are same .
     
  17. Nov 11, 2016 #16

    haruspex

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    I think we are at cross purposes here. Yes, you can figure out quite quickly what the relationship is between the tensions, but Shashank needs to develop that from standard equations, not just take it on trust.
     
  18. Nov 11, 2016 #17

    ehild

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    Shashank derived that relationship in his post #4.
     
  19. Nov 11, 2016 #18

    haruspex

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    Thanks, my oversight.
     
  20. Nov 11, 2016 #19
    Please see post#4 .
     
  21. Nov 11, 2016 #20
    Okay, I think these equations are right, because my answers match with the options.

    8g-T=8(a1+a)
    T-6g=6(a1-a)
    10g-T=10(a2-a)
    T-4g=4(a2+a)

    On solving, I got a1=100/77=1.29ms-2
    a2= 360/77=4.67ms-2
    T=4800/77=6.23g ms-2
    a=10/11 ms-2=g/11ms-2

    Tension over the pulley A is 2T=12.46g N

    Thanks a lot to everyone who helped me out!
     
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