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A pulley system with two objects stacked on one another

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Block B, with mass 5.00kg, rests on block A, with mass 8.00kg , which in turn is on a horizontal tabletop. There is no friction between block A and the tabletop, but the coefficient of static friction between block A and block B is 0.750. A light string attached to block A passes over a frictionless, massless pulley, and block C is suspended from the other end of the string.



    2. Relevant equations
    F=ma
    FFriction = μN



    3. The attempt at a solution
    What makes sense for me to do is to find the tension in the string which I think is equal to the force due to gravity on block C. What throws me off is that the force of friction between B and A is in the same direction of the Tension force of the string.

    I tried to set up a net force equation of B and got

    F = FgravityonC - Ffriction = 0

    But then the answer I get (3.75kg) is wrong. I'm not convinced that FgravityonC is supposed to be in the equation. And I haven't addressed block A at all.

    Please give me a prod in the right direction!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 1, 2011 #2

    PeterO

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    That which I coloured red in interesting.

    The friction force between B and A is in both directions - It depends whether you are talking about the force of A on B or the force of B on A. One of them will be in the same direction as T, the other in the opposite direction to T.
    One thing for you to work out is which is which.

    Also, if the Tension was EQUAL to the weight of C, then C won't move [due to the balanced forces] so if you think anything is going to move you need to re-think the size of Tension.

    Draw a good diagram including all forces on each body - especially forces in the direction of any possible motion.
     
  4. Oct 1, 2011 #3
    Hmmmmmmmmm... my new approach is to find out the maximum acceleration the box on top can handle before it breaks free. I tried solving with T = FgravityonC but you're right, the system won't move if that is the case. But how does this work when the table is frictionless? What is actually stopping the system from moving? Would the system accelerate if mass of c = 0.00001kg?
     
  5. Oct 1, 2011 #4

    PeterO

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    Yes, but you would possibly have to come back tomorrow to notice that it had moved.
    That is theoretically speaking.
    If the mass was that small, the weight of the string and the tiny friction in the pulley would be of greater significance.
     
  6. Oct 1, 2011 #5
    Interesting. Anyhow, is there any more hints you could give me to solve this? I feel confident I can solve this if I had an expression for T.
     
  7. Oct 1, 2011 #6

    SammyS

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    Draw a Free-Body-Diagram for each of the three objects, as PeterO suggested (although he used other words).

    The all three block have the same acceleration, although the acceleration of block C is in a different direction.

    You should have two equations each blocks A & B, one for x-components, one for y-components. You only need a y-component equation for block C.
     
  8. Oct 1, 2011 #7

    PeterO

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    The tension is "applied" to A. The friction between it and B has to transfer sufficient of that tension to accelerate the pair, without slippage.

    AN example with different figures, as this post doesn't have the original figures in it

    Suppose A is 2 kg sitting on top of B of 10 kg, with coefficient of friction 0.5 I will use g = 10 fro simplicity - you probably better use 9.8 or 9.81

    Normal reaction force [B on A] is 20 N [mg of A]. Using coefficient of friction of 0.5 that means maximum friction transfer can be is 10N

    That 10 N force will cause B to accelerate at 1 m/s.

    That means the tension can only accelerate the pair of masses at 1 m/s without slippage.

    SO T = 12 x 1 = 12N [acceleration 1 m/s; combined mass 12 kg

    Now try with the real figures, and you will have the real Tension.
     
  9. Oct 2, 2011 #8
    Why wasn't the mass of A used to determine the max acceleration possible? What I mean is why 10/10 and not 10/(10+2)
     
  10. Oct 2, 2011 #9

    PeterO

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    I was only looking at the maximum acceleration possible for Block B. Block B is accelerated by a friction force from block A. Remember, the string is not tied to Block B. The tension force is transferred vis Block A in the form of the Friction force.
     
  11. Oct 2, 2011 #10

    ehild

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    It is difficult to speak about Physics problems without a picture, so I show one. I drew all forces which act in the possible direction of motion, which is on the right for the boxes A and B and downward for box C. They move with acceleration of the same magnitude.

    There are the weight Wc and the tension T of the rope acting on box C.

    a*Mc=Wc-T.

    Only one force acts on box B: the static friction between A and B, Fs. This force counteracts any relative motion of the surfaces in contact, so it points forward as A moves forward.

    a*Mb=Fs

    ehild

    The tension pulls box A forward but the static friction from B acts backwards, according to Newton's Third Law.

    a*Ma=T-Fs


    The tension and the friction are internal forces, they act in pairs, and have opposite direction for the interacting objects.

    If you add all three equations, the internal forces cancel, and only the external force(s) remain. It is Wc now:

    a(Ma+Mb+Mc)=Wc, that is a=g*Mc/(Ma+Mb+Mc)

    Having obtained the acceleration, you need to check if the static friction can provide enough force to accelerate box B.

    Fs≤μ*NAB=μ*Wb.

    To accelerate B, F=a*Mb force is needed. So check if a*Mb≤μ*Wb, or eliminating Mb,

    a≤μ*g
     

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  12. Oct 2, 2011 #11

    PeterO

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    Having seen the diagram drawn by echild I realise two things:

    #1 I mis-interpretted which block was where [A & B]

    #2 There is no question mark in your original problem statement, clearly showing that you gave a lot of information but told no-one what you might be trying to calculate.????
     
  13. Oct 2, 2011 #12

    ehild

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    Hi, Peter

    That is the reason why we always suggest to draw a diagram first. If the OP does not do it, we try and he/she may agree or not. We are from of all parts of the world and our English can be quite different. It is easy to misinterpret each other, but pictures speak on the same language. You even misread my name, I am very far from a child :wink: My name is ehild.

    This problem is very frequent and the usual question is : How much can be the weight/mass of the hanging object (or the coefficient of the static friction) so as the top object does not slip. Determine the acceleration of the system.

    The mass of C is not given here so that can be the question. Get the maximum acceleration from a=μg, and use the other equation to determine Mc.


    ehild
     
  14. Oct 2, 2011 #13

    PeterO

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    Thanks ehild, I had guessed that the question was along those lines, but wanted to indicate to OP that under the section "The problem statement ..etc" it would be useful to actually state the problem.
    Peter
     
  15. Oct 2, 2011 #14

    SammyS

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    ehild, Thanks for the diagram and thanks for "we always suggest to draw a diagram first". I hope I'm part of that "we"

    I always look forward to reading your posts!

    Sam
     
  16. Oct 2, 2011 #15
    Wow that was very silly of me. Yes, my question was what is the maximum weight object C can be. I've learned a lot by this problem, namely that object B actually 'pulls' object A in the opposite direction on T. Thanks for all the help.
     
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