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Work with pulleys

  1. Nov 5, 2006 #1
    In Figure 7-48 (see attatched), a cord runs around two massless, frictionless pulleys. A canister with mass m = 11 kg hangs from one pulley, and you exert a force F on the free end of the cord:

    (a) What must be the magnitude of F if you are to lift the canister at a constant speed?
    (b) To lift the canister by 1.9 cm, how far must you pull the free end of the cord?
    (c) During that lift, what is the work done on the canister by your force (via the cord)?
    (d) What is the work done on the canister by the gravitational force on the canister?

    I don't like pulleys. Can I just pretend that the applied force acts directly on the canister and points straight up? I'm fairly certain that's OK for parts (a) and (c), but I'm not so sure about part (b).

    Attached Files:

  2. jcsd
  3. Nov 5, 2006 #2
    W-oah. I have the problem from WebAssign; I finally flipped through my textbook (Halliday, 7e) in the vague hope that the book would provide inspiration. It adds a helpful little line at the end of the problem: "(Hint: When a cord loops around a pulley as shown, it pulls on the pulley with a net force that is twice the tension in the cord.)"
    Was I supposed to figure that out by myself?
    And does that mean that I can indeed imagine the applied force as acting directly on the canister, but it points up and has twice the magnitude of the pulling-on-rope force?

    BTW, my attatchment still says that it's pending approval, but I can download it no problem.
    Last edited: Nov 5, 2006
  4. Nov 5, 2006 #3
    Hokey-dokey. I got it all right, and I even know how I got it all right. But I'm not quite sure why it all worked.

    Take part (b): just multiply the distance by two. But why? How would I have figured that out other than looking in the back of the book and working backwards to figure out how they manipulated their numbers to get their answer?

    For part (c), W=Fxcosphi=Fx*1. Why is x the distance you pull rather than the distance the can moves? Same for part (d)
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