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Torque problem -- designating center of mass as pivot point

  1. Dec 9, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform horizontal beam of length 8.00 m and weight 200 N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0 degrees with the horizontal. If a 600-N man stands 2.00 m from the wall, find the tension in the cable and the force exerted by the wall on the beam at the pivot.



    2. Relevant equations
    ∑Fx = 0
    ΣFy = 0
    Σtorque = 0
    R = force exerted by wall at the pivot
    T = tension from cable


    3. The attempt at a solution
    ∑Fx = R cos θ - T cos 53° = 0
    ΣFy = R sin θ + T sin 53° - (weight of man) - (weight of beam) = 0
    → R sin θ + T sin 53° - 600 N - 200 N = 0
    Σtorque = (weight of man) * (d1) + R(d3) - T(d2) = 0
    → (600) * (2 m) + R (4 sin θ) - T (4 sin 53°) = 0
    up = positive
    down = negative
    → = positive
    ← = negative
    clockwise = positive
    counterclockwise = negative
    d1 = ⊥ distance from the line of action of force (weight of man) to the pivot point
    d2 = ⊥ distance from the line of action of force (tension - T) to the pivot point
    d3 = ⊥ distance from the line of action of force (force - R) to the pivot point
    Total length of beam = 8 m so 8m/2 = 4 m
    since man is 2 m from the wall, and half of 8 m is 4 m, therefore, the ⊥ distance from weight of man to center of mass (pivot) should be 2 m.
    Through trig., d3 = 4 sin θ and d2 = 4 sin 53°

    I am trying to solve this problem by designating the pivot point as the center of mass of the horizontal beam.
    I have solved this problem by choosing the pivot at the pin connection. However, I am having trouble arriving at the same answer if I choose the pivot at the center of mass of the beam. I rearranged the equations to find the unknowns; however, I do not get the correct answer. I know that the answer should be the same regardless of where the pivot point is located. Any help would be greatly appreciated. Thanks.
     
  2. jcsd
  3. Dec 9, 2014 #2

    SteamKing

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    If you choose the c.o.m. of the beam as the reference point, you must also include the moment due to the reaction at the pin in your moment calculations. By using the reference point as the pin, you simplify the moment calculation because the moment on the beam due to the reaction at the pin is zero. This is why, almost invariably, one of the reaction points on a beam is chosen as the moment reference, as opposed to the c.o.m. of the beam or some other point.
     
  4. Dec 9, 2014 #3
    Hi, I may be wrong, but I thought R (4 sin θ) was the moment due to the reaction force (R) at the pin connection.
    After rearranging the equations to solve for the unknowns,
    I get T (tension) = 687.5 N
    ⇒Σ Torque = 1200 + R (4 sin θ) - 3.2 T = 0
    and θ = 31.2°
    ⇒ΣFy = R sin θ + 0.8 T - 800 = 0
    ⇒R sin θ = 800 - 0.8 T
    ⇒plugging in the T (tension) = 687.5 N into R sin θ = 800 - 0.8 T yields R sin θ = 250 N
    ⇒ΣFx = R cos θ - T cos 53° = 0
    ⇒ 0.6 T = R cos θ
    ⇒plugging in the T (tension) = 687.5 N into R cos θ = 0.6 T yields R cos θ = 412.5 N
    ⇒ (R sin θ) / (R cos θ) = 250 N/ 412.5 N
    ⇒arc tan = 0.606
    ⇒θ = 31.2°

    I keep arriving to the wrong solution. Am I getting the ⊥ distance (d3) wrong? I thought it was 4 m since the center of mass of beam is 1/2 of the total length of 8 m.
     
  5. Dec 9, 2014 #4

    SteamKing

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    Your equation for the moment about the c.o.m. is incorrect:

    The moment due to the weight of the man produces a CCW moment about the c.o.m. of the beam

    d2 is not correct. The moment arm of the tension from the c.o.m. of the beam is 4 m. T sin 53° is the vertical component of the tension.
     
  6. Dec 9, 2014 #5

    haruspex

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    No, I think that bit was ok. bijou is saying the torque due to tension T is T*d2 = T*(4 sin 53°) m.
     
  7. Dec 9, 2014 #6

    SteamKing

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    I was trying to correct bijou's impression that the moment arm was equal to 4 sin 53°. Numerically, the moment is T*(4 sin 53°), but it is the vertical component of the tension, namely T sin 53°, which is the applied force producing the moment, and 4 m is the arm. I think it's best to learn these distinctions when you are starting out, than to make a mistake later on and have to learn from that.
     
  8. Dec 9, 2014 #7

    haruspex

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    I don't agree. It is equally valid to say that the moment arm is 4m and the force is T sin 53°, or that the moment arm is 4 sin 53° and the force is T.
    Are you requiring the moment arm to have a physical representation as a lever in that direction? I don't think that's required. After all, the physical lever could be curved.
     
  9. Dec 9, 2014 #8

    SteamKing

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    There is a beam for which this calculation is required, as described in the OP. We are not dealing with a hypothetical situation. I am not requiring anything; I am merely stating what the moment arm is according to the geometry of the beam. If you want to get into general cases or M = r x F, fine. But this appears to be someone's first attempt at solving a static equilibrium problem (as evidenced by using the c.o.m. of the beam instead of the pin as the moment reference), so IMHO, the simplest interpretation is to be preferred. Later on, after some experience is acquired, you can generalize, use r x F, whatever.
     
  10. Dec 9, 2014 #9

    haruspex

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    The first four hits I found on the net. They all support bijou's usage.
    From http://www.ptdirect.com/training-design/training-fundamentals/movement-mechanics-and-motor-learning/moment-arms-force-vectors-and-a-squat-analysis [Broken]:
    "A moment arm is simply the length between a http://www.ptdirect.com/training-design/training-fundamentals/movement-mechanics-and-motor-learning/movement-mechanics-2013-the-essentials-uncovered# [Broken] and the http://www.ptdirect.com/training-design/training-fundamentals/movement-mechanics-and-motor-learning/movement-mechanics-2013-the-essentials-uncovered# [Broken] acting on that joint."
    From http://www.aaronswansonpt.com/basic-biomechanics-moment-arm-torque/:
    "The moment arm (lever arm) of a force system is the perpendicular distance from an axis to the line of action of a force. "
    From http://muscle.ucsd.edu/musintro/ma.shtml:
    "the moment arm is defined as the perpendicular distance from line of force application to the axis of rotation."
    From http://web.mit.edu/4.441/1_lectures/1_lecture5/1_lecture5.html:
    "The moment arm or lever arm is the perpendicular distance between the line of action of the force and the center of moments."
    I didn't look any further.
     
    Last edited by a moderator: May 7, 2017
  11. Dec 11, 2014 #10
    Thank you so much for all your help! I was able to arrive to the correct solution with the axis of rotation as the center of mass. I know that designating the pin connection as the pivot is easier to solve and I have solved it before using that method; however, I still wanted to prove that I was able to arrive to the same solution with the pivot on any reference axis. Again, thank you so much ~
     
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