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Tractor and pulleys - expression of speed

  1. Dec 17, 2006 #1
    1. The problem statement, all variables and given/known data

    A tractor named A is used to lift a bale B with a system of pulleys.
    The tractor has an unknown speed called VA : find the expression of the bale's vertical speed VB in function of x (distance from the tractor to the pulley).
    The diameter of both pulleys can be neglected.

    2. Relevant equations

    I call T the tension in the rope, N the reaction of the ground and W=mg the weight, mA the tractor's mass and mB the bale's mass, aA the tractor's acceleration and aB the bale's acceleration.

    In A, I have : horizontally -T*(x/l) = mA * aA
    vertically :N - mA*g + N*(h/l) = 0
    In B : T - mB*g = mB*aB

    3. The attempt at a solution

    I've tried to put the problem into some equations, but I don't see how I can switch from accelerations to speed...moreover, I don't have any numeric data, so any piece of information is welcome!

    Thanks in advance!

    Attached Files:

  2. jcsd
  3. Dec 17, 2006 #2

    Doc Al

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    Staff: Mentor

    This is more of a trig problem than a physics problem. No need to consider forces. Hint: The rope doesn't stretch. Hint 2: How does VB depend on the rate that "l" increases?
  4. Dec 17, 2006 #3


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    Staff Emeritus
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    Gold Member

    Hint: there's no dynamics involved in this problem: it is entirely kinematical. This means that we don't care about forces and so on: it is just "geometry": for a given position of the tractor, and a given length of the rope, the bale will be at a certain height.

    Second hint: call the total rope length "L" and consider the 3 pieces of rope (going from the pulley to the bale, going from the bale up to the pulley again, and finally going from the pulley to the tractor) that make up the entire rope (and hence, whose sum must be equal to L).

    EDIT: I see that Doc Al was typing the same thing while I was doing so...
  5. Dec 17, 2006 #4


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    If I remember correctly, having this type of setup gives a force advantage to the tractor. Pulling rope (l) a distance of 10m will raise bail B by 5m. The force on (l) is equal to the force on each of ropes in the loop that make up (h-y), so an applied force of 10N on (l) will give a force of 20N on the bale.

    First step is thinking what the equation would be with no angles:
    Vb = (1/2)Va; where Vb and Va are parallel (the tractor would be going straight up at x=0)
    Now realistically, maximum velocity of Vb is when Va is parallel to rope (l), so we'll use cosine (because cosine starts at maximum). Since we're using cosine we also need to use an angle that starts at 0 (where cosine is max). Sticking with this idea of max speed, Va will be parallel to rope (l), which will also be parallel to ground (x). Having a parallel between measurable distances is what we want for our angle of 0, so our cosine is for the angle between rope (l) and ground (x).
    Vb = (1/2)(Va)cos(theta), alternatively written as:
    Vb = (1/2)(Va)(x/l)
    Last edited: Dec 17, 2006
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