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Mechanics Pulley Platform Problem

  1. Jan 23, 2015 #1
    1. The problem statement, all variables and given/known data

    Two construction workers each of mass m raise themselves on a hanging platform using pulleys as shown above. If the platform has a mass of 1.2⋅m, the initial distance between the pulleys and the platform is d, and the workers each pull with a force f on the ropes, what is the acceleration a of the workers? Assume the pulleys and ropes are massless.



    2. Relevant equations

    Tension = m(a + g) (possibly).
    F = ma



    3. The attempt at a solution

    To be wholly honest I have next to no idea how to proceed. My only real thought was to use F = ma. Thinking first that I'd need my total mass, I added (m + m + 1.2m) = 3.2m total (one mass for each worker plus the platform). Next I tried to consider the force, thinking we have two workers each exerting a tension force f, thus 2f being the total force upwards, then subtracting out the downward forces (namely gravity) to achieve (2f - 3.2mg). Plugging into F = ma I got (2f - 3.2mg) = a(3.2m) => a = (2f - 3.2mg)/(3.2m) which is marked incorrect. At this point I have hit a wall. My thanks for any help.


    (Attached is the image accompanying the problem).
     

    Attached Files:

    Last edited: Jan 23, 2015
  2. jcsd
  3. Jan 23, 2015 #2
    Why 3.4mg ?
     
  4. Jan 23, 2015 #3
    Sorry, that was a typo which persisted through my idiocy. I corrected it.
     
  5. Jan 23, 2015 #4
    I suggest you to think about the free body diagram.
    At first, think about the free body diagram of one of the men.
    He is exerting force f each on the rope, and so according to Newton's third law, the rope is also exerting force f (upward) on him. Let's assume, he is exerting a force R on the platform, and the platform is exerting the same force R on him in the upward direction. The other force on him is gravitational force which is equal to mg. So the net force on him is, ##F_{net} = R + f - mg ##; Again, ##F_{net} = ma##. So ## R + f - mg = ma##
    In this same way, you may find out, equations for the other man and the platform, and solving these equations, you may find out accelaration.
     
  6. Jan 23, 2015 #5
    Wouldn't the force R simply be equal the normal force the man is exerting on the platform and in turn then be equal to ##mg##, giving ##f = ma## out again since R would cancel with ##mg##.
     
  7. Jan 23, 2015 #6
    No. Suppose, you hang a ball with a rope. The rope is straight. Then you just touch the ball on a weighing machine keeping the rope straight. Will there be any weight found on the machine?
    upload_2015-1-24_10-42-47.png
     
  8. Jan 23, 2015 #7
    Another example. Suppose, a heavy stone hangs from a crane. And you touch your hand under the stone. Will you feel any force on your hand?
    upload_2015-1-24_10-53-5.png
     
  9. Jan 23, 2015 #8
    Alright, I think I understand what you mean.


    So would the equation for the other man just be ##R+f−mg=ma## since he is identical to the first man and the equation for platform would be ##2f -1.2mg = ma## Two f for the tensions forces acting on the block, minus the downward component of force?

    I'll work on it more tomorrow as well, if I can't reply any more tonight.
     
  10. Jan 23, 2015 #9
    You are missing the normal forces which are exerted on the platform by the men, and look, the mass of the platform is not m. When you are calculating with the forces on the platform, the net force equals to the mass of 'the platform' times the accelaration of 'the platform'.
     
  11. Jan 24, 2015 #10
    So it would then be: ##2f - 1.2mg - 2R = 1.2mg## (because since before you defined the force R to be the force the platform exerts on the men upwards, the normal force for the platform by the men should be twice the negative of R).
     
  12. Jan 24, 2015 #11
    "## = 1.2mg##" ? Probably, a typo.
     
  13. Jan 24, 2015 #12
    Should have been ##=1.2ma## I think.
     
  14. Jan 24, 2015 #13
    My eternal thanks good sir or madame. Your advice was instrumental in my finding the solution.
     
  15. Jan 25, 2015 #14
    Exactly!
    You're welcome. I also thank you for sharing this interesting problem.
     
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