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Physics Torque Problem

  1. Nov 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose that the climber has mass m, has a center of gravity at a distance L1 from her feet, and holds the rope at a distance L2 from her feet. The climber rappels down a vertical cliff with her body raised θ1 above the horizontal. The rope makes an angle θ2 with the cliff face. Use the ‘usual’ coordinate system with +x to the right, +y up, and +z out-of-page. On the climber’s feet, call the horizontal and vertical components of the contact force H(vector) and V(vector) and begin by initially guessing that V (vector) points up. Give and expression for (a) the net force in the x-direction, (b) the net force in the y-direction, and (c) the net torque with respect to her feet. Solve for (d) the tension, (e) H, and (f) V.

    2. Relevant equations
    ∑F=0 (since there is no movement)
    ∑τ=0 (since there is no movement)

    3. The attempt at a solution

    This is my issue currently. Since all the values are arbitrary, then there is a scenario with the angles. For example, if the tension force makes an acute angle with respect to the angled climber then the same angle "formula" to determine the perpendicular force on the radius (we are treating the radius as the climber's body) would not be the same if the force makes an obtuse angle with respect to the angled climber. Picture included to explain what I mean. (the tension angle can vary because length of the body can vary since every value is arbitrary). Thank in advance for your time!

    To better explain what I am doing, since the tension force is at an angle, and the body is at an angle, I am finding the perpendicular component of the tension force with respect to the angled body. I hope this makes sense (since τ = R(vector) x F(vector)). I know I could also find the perpendicular radius component with respect to the force, but that would require the chosen force to be able to make a right triangle with the rotational point, which it can only do if the value is acute. Thanks!

    Attached Files:

  2. jcsd
  3. Nov 22, 2014 #2

    Simon Bridge

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    How can the climber "rappel down the cliff" without moving?

    ... then you either need a different angle formula, which can handle a wider range of angles, or you need to do the problem for both cases, or you could have a closer look at how climbers rappel: what sort of angle do they usually make?
    Mark in where the climber's feet and bottom would be in your diagram and see which makes sense.
  4. Nov 22, 2014 #3


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    Perhaps worth considering what happens to the vertical component at the feet if L2 is equal to, more, or less than L1.
  5. Nov 22, 2014 #4
    Simon Bridge, the problem calls to evaluate the forces when the climber is just sitting on the cliff. I actually used to rappel and if you rest, you would be rappelling without movement, as you would still be connected to your equipment and such, but just not moving. But none the less, we did a problem with this but given values already. So that way we knew the the angle was acute and they used Σf=0 to determine the frictional force at the feet (after determining tension using Στ=o).

    Also, again, a climber can make any angle with the cliff at any time. The usually is more close to a horizontal position, but like all things in physics, you can't just assume. So I am trying to create a formula that works for either scenario.

    CWatters, I guess I am slightly confused how that will help? Could you elaborate on that for me please? Thanks!
  6. Nov 22, 2014 #5


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    It might not help solve the problem but considering L1 vs L2 makes more sense to me than concentrating on the angle the rope makes to the body. The latter seems to be a rather arbitrary parameter to focus on.

    If L1=L2 then there is nothing to create a torque about the centre of mass. So the reaction force at the feet must pass through the centre of mass. This should make it easier for the climber. By contrast.. If L1<L2 there will be a tendency for the feet to fall. If L1>L2 there will be a tendency for the head to fall.

    I would take simon's advice and make a drawing. If you get the equations right they should really cope with all senarios.
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