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Homework Help: Pulling yourself up with a pulley

  1. Feb 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A window washer pulls herself up using a bucket that she stands in and a pully system that she pulls down on a cord to go up.
    How hard must she pull in order to go up slowly at a constant speed?
    assuming pulley frictionless and massless
    2. Relevant equations

    3. The attempt at a solution
    I set up a FBD but apparently it was wrong and i don't fully understand why.

    I set it up as the bucket+human = T - mg = ma
    and the other side of the pull the cord which is Fh - T = ma
    Fh = Force by hand
    In the book they just say its 2T - mg = ma which I get my Fh is = to T but I don't see how the pulley makes it 2T. It does make some sense as it is easier to pull down then pull up but.. i can't wrap my head around it lol any help?
  2. jcsd
  3. Feb 23, 2016 #2


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    There is tension in both ropes and there is a force from the other rope on the bucket (and a force from the bucket on the cleaner).
  4. Feb 23, 2016 #3
    It is asking us to find the amount of force applied Fa in one end in order to pull up the human in the bucket 'slowly' at constant speed. It basically is asking for the amount of force required for there to be no net force.

    Let's first indicate positive directions as upward, for the end where the human is in the bucket, and downward for the end where the rope is being pulled.
    The tension in one end is the same as the tension in the other. The acceleration is also the same for both ends.These are property of this type of system (known as atwood's machine).

    Account for the forces acting on each ends and list them out.
    Fnet = T - mg = ma = 0
    Fnet = FA - T = ma = 0

    Now use your knowledge in what these quantities mean and you can re-write this in the way your book wants you to.
    Last edited: Feb 23, 2016
  5. Feb 23, 2016 #4
    I get that, but my book wants it to look like 2T = mg how do I get that by getting rid of Fh
  6. Feb 23, 2016 #5
    I thought they were asking you to find the applied force? In any case, the tension is not equal to half of the force of gravity of one side.
    If you derive tension from the equations above, the tension is equal to mg. Not only that, but force applied is as well. They want you to equalize the force of gravity of an object with an equal force on the opposite end.

    The tension is equal and opposite in both ends to the force downward. If it were not the case, and instead 2T = mg, then both ends would accelerate downward which is impossible.

    Edit: here this should help http://www.drcruzan.com/AtwoodsMachine.html
  7. Feb 23, 2016 #6
    I am asking for her applied force. According to my book you are wrong. It shows an FBD of mg the girl in the bucket pointing down and two forces of T pointing up.
    It says you only show the forces acting on the bucket holding the girl. So because she pulls down her Force is = T so on both sides T points up. Lol doesnt really make sense but thanks anyways.

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  8. Feb 24, 2016 #7


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    I suggest you make a free body diagram of a system which includes the girl and the bucket and the rope up to some height above the girl.

    What forces are acting on this system?
    What is then the tension in the rope?

    Once you have that answered you can start asking yourself what the pulling force is.
  9. Feb 24, 2016 #8
    I made an FBD and I know their are only 4 forces cuz string is massless and pulley is frictionless. But I just don't get how the force she pulls down only has to be half her weight.
  10. Feb 24, 2016 #9
    I guess when pulling on a system like that you only consider your tension force on the one side because that is the reaction force to your pulling and there is actually no mass there. hmm makes sense i think
  11. Feb 24, 2016 #10
    Sorry, I did not consider treating the pulley system as a whole. The human is directly affected through newton's third law by both ropes. This system can be thought of as one mass held together by two strings. In that case, T = 1/2 mg. Because the forces on the system is
    Fnet = 2T - mg = ma = 0
    On individual ends the net force can be written
    Fnet = T - Fa = ma = 0
    Fnet = T - mg =ma = 0.
    There is one rope with tension mg/2 connected to the bucket. What then must be the force applied to to the other rope?
    Sorry about formatting I'm on my phone
    Last edited: Feb 24, 2016
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