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Calculating equilibrium force

  1. Nov 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A human holds a ball of mass 8kg at his hand, so that the forearm is perpendicular to the upperarm.
    The distance between the elbow and the center of mass of the forearm is 0.15 meters, the distance between the elbow and the muscle is 0.05 meters and the mass of the forearm is 2 kg. The ball is held 0.35 meters from the elbow (as depicted below).
    Xyk50Z.png
    Find the force of the mustle given equilibrium.
    2. Relevant equations
    $$\sum_{i=1}^n \tau_i=0$$
    $$F_{m} = ma$$
    3. The attempt at a solution
    $$r_{mustle}*F_{mustle} - r_{arm}*F_{arm} - r_{ball}*F_{ball}=0$$
    $$F_{mustle} = \frac {r_{arm}*F_{arm} + r_{ball}*F_{ball}}{r_{mustle}}$$
    Hence, $$F_{mustle} = \frac {r_{arm}*m_{arm}+r_{ball}*m_{ball}}{r_{mustle}} *g $$
    Setting the given data yields $$\mathbf F = 608.22\mathbf y N$$
    Here is the part I don't understand: If I am right, the forearm does not move and hence does not accelerate.
    But $$F-m_{arm}*g-m_{ball}*g = 608.22 - 2*9.81 - 8*9.81 = 510.12 N$$, hence the sum of forces on the formarm is not zero.
     
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  3. Nov 13, 2016 #2

    I like Serena

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    Hi pixietree! :oldsmile:

    That is correct.
    There is an additional upward force in the shoulder, that was left out, to ensure equilibrium of the vertical forces.
    However, it doesn't contribute to the torques, since the elbow is on the same line as that force.
     
  4. Nov 13, 2016 #3

    rock.freak667

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    Shouldn't there be an additional weight of the muscle downwards at a distance of rmuscle ?
     
  5. Nov 13, 2016 #4

    I like Serena

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    We're looking at the free body diagram of only the forearm.
    The force from the shoulder is transmitted to the forearm in the elbow, where it rotates.
    The muscle itself is part of the upper arm, which will have some upward force on it from the shoulder, and some torque as well, to keep it in equilibrium.
     
  6. Nov 13, 2016 #5

    rock.freak667

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    Ah yes that makes more sense.
     
  7. Nov 13, 2016 #6
    If I understand you correctly rock.freak667, I got the torques right hence there is some downward force of 510.12 newtons acted at the elbow. But why does the elbow feel this force downwards? I didn't understand your explanation.
     
  8. Nov 14, 2016 #7

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    If we look at just the forearm, we have a couple of downward forces on the right (ball and arm weight), and an upward force on the left (muscle).
    With just that the arm would tilt down.
    To compensate, we need an downward force at the elbow to counteract the muscle, that has to pull extra strong.
    Consequently all forces cancel out (which they must to be in equilibrium), and then the canceled out torque ensure the forearm doesn't tilt.

    So we have a downward force on the elbow.
    And according to "action = - reaction", that means we have an equal and opposite upward force from the upper arm on the elbow, which gets then transmitted through the upper arm to the shoulder.
     
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