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Homework Help: PHysics 11 Pulley probem

  1. Nov 20, 2004 #1


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    PLZ HELP!! I am puzzled with this problem, and all i could find are that the forces equal, but the ans int eh back of the textbook clearly says the solution is 0.2 m/s2

    Question: This situation occurs in a pulley system. A 80.0 kg man inside a 40.0 kg dumb waiter (like an elevator) pulls down on the rope. At the end of the rope, it is attached to the dumb-waiter. At the other end, the person exerts a force. At that moment, the scale on which he is standing reads 200N. Determine the elevator's acceleration.

    THX for ur time!!
  2. jcsd
  3. Nov 20, 2004 #2
    Hmm.. thats wierd. I didn't get 0.2m/s^2

    Let me try to understand this better. The man is inside the dumbwaiter (which will pull up the elevator) has a mass of 80kg, the dumbwaiter is 40kg, so total mass of the dumbwaiter is 120kg. Correct?


    I got 8.1 m/s^2 as an answer. So I don't know if I'm wrong, or the book is.
    Last edited: Nov 20, 2004
  4. Nov 21, 2004 #3
    MAYBE this is the correct answer:





    Probably the 0.17 is rounded off.
    Last edited: Nov 21, 2004
  5. Nov 21, 2004 #4


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    sorry, Raza, nice try. the 1176 is the force, and not the mass, so u cannot use the force of 200N to divide that force, which results in some unknown unit number. THe answer just conincidently matches with the ans int he book, but i am sure the way u got it is wrong....again, thx for trying.

    physik, here's the link to the actual question and diagram from the textbk:


    again, thx for trying, maybe u can understand the problem better by this clear picture :)
    Last edited by a moderator: Apr 21, 2017
  6. Nov 21, 2004 #5
    Let us consider the dumb-waiter separately.
    The man standing on it has a mass 80 kg, but the force exerted by him on the elevator is 200N.
    He is applying some force on the rope, say F.
    You can solve this to know F.
    Now the weight of the elevator is 40 kg ang the force exerted by the man is 200N. Let the acc. of the elevator be a.
    Substituting F=80g-200
    we get 80g-200-200-40g=40a
    That is 0.2 m/s^2 downwards
  7. Nov 21, 2004 #6


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    thx alot, gauravkukreja!
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