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Dynamics of Rotational Motion: Quarry Crane Problem

  1. Mar 31, 2013 #1
    1. The problem statement, all variables and given/known data
    A quarry crane is used to lift massive rocks from a quarry pit. Consider the simplified model of such a crane shown in the figure. (Figure 1) The ends of two poles are anchored to the ground at the same point (point O). From this point, one pole rises vertically and the second pole rises at an angle. The vertical pole has its free end connected to the ground via an unstretchable, massless rope labeled rope 1. A second rope, labeled rope 2, connects the free ends of the two poles. The angle between the tilted pole and rope 2 is θ. Both poles have length L and can be considered massless for the purposes of this problem. Hanging from the end of the second pole, via rope 3, is a granite block of mass m.

    Throughout this problem use g for the magnitude of the acceleration due to gravity.
    Part A) Find T3, the tension in rope 3.
    Express your answer in terms of some (or all) of the following quantities: ∏, m, g, L, θ, and ϕ.

    Part B) Find T2, the tension in rope 2.
    Express the tension in terms of some (or all) of the following quantities: ∏, m, g, L, θ, and ϕ.

    Part C) Find T1, the tension in rope 1.
    Express the tension in terms of some (or all) of the following quantities: m, g, L, θ, and ϕ.

    Figure 1:

    2. Relevant equations
    Ʃτ1,2,3 = 0

    3. The attempt at a solution

    Part A) Correct- T3 = mg

    Part B) Correct- T2 = mgsin(∏-2θ)/sin(θ)

    Part C) Incorrect -
    I know that the torque of rope 1 on the vertical pole in relation to the origin is τ1=T1*L*cos(ϕ)
    In my attempt I used the previously found values T2, and for l I used L*sin(2θ).
    Combining these I found
    τ2 = -Lmgsin(2θ)2/sin(θ) ;
    setting Ʃτ = 0, so τ1 = τ2, solving for T1 yields
    T1 = mg(sin(2θ))2/sin(θ)cos(ϕ);
    Originally I had used the value of (∏-2θ) rather than 2θ, but since the problem is the only one that does not call for the quantity of ∏, I replaced it with 2θ. I've been told it's incorrect, and it is not a problem with signs. I think I may be having a problem finding the moment arm with my geometry for the torque of rope 2, or may just be combining the equations incorrectly. I assumed that T2 would be usable.
    Any help would be very much appreciated.

    p.s. This problem has been posted and discussed on various places online, however none of them seem to go into the third part of the problem to find T1.
    Last edited: Mar 31, 2013
  2. jcsd
  3. Mar 31, 2013 #2


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    I would have thought the horizontal component of T1 would simply have to match the horizontal component of T2
  4. Nov 30, 2015 #3
    Check your trigonometry and mak sure to fuly simplify. You have the right equation, just remember, that when you divide sinθ^2/sinθ, you are left with just sinθ in the numerator. The correct,simplified, answer would be T1=[mgsin(2theta)] / cos(ϕ)

    LOL this question was asked in 2013 and you're getting an answer in 2015. Same questions still going around and messing everyone's minds!
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