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Rotational motion of bicep

  1. Oct 11, 2007 #1
    would someone be able to tell me if this is right or what i've done wrong? thankyou
    1. The problem statement, all variables and given/known data

    a) What torque does the bicep muscle have to apply to hold the forearm of mass 1.10 kg and length 30 cm horizontally? Assume the arm is a rod of uniform mass density.
    b) What torque does the bicep need to apply if the forearm is to hold a 20 kg weight in one hand at an angle of 20[tex]\circ[/tex] below horizontal?

    2. Relevant equations

    [tex]\tau[/tex] = Fx = mgx
    [tex]\tau[/tex] = Fsin[tex]\theta[/tex]
    [tex]\Sigma[/tex][tex]\tau[/tex] = 0

    3. The attempt at a solution

    a) m = 1.1kg x = 15cm = 0.15m
    [tex]\tau[/tex] = Fx = mgx = 1.1 x 9.81 x 0.15
    = 1.61865
    = 2Nm

    b) m[tex]_{o}[/tex] = 20kg [tex]\theta[/tex] = 20[tex]\circ[/tex]
    [tex]\Sigma[/tex][tex]\tau[/tex] = 0
    [tex]\tau[/tex][tex]_{1}[/tex] + [tex]\tau[/tex][tex]_{2}[/tex] = 0
    [tex]\tau[/tex][tex]_{a}[/tex] + [tex]\tau[/tex][tex]_{o}[/tex] = 0 a=arm o=object
    [tex]\tau[/tex][tex]_{o}[/tex] = -[tex]\tau[/tex][tex]_{a}[/tex]
    Fsin[tex]\theta[/tex]r = -mgx
    F x sin20 x 0.3 = -(1.1 x –9.81 x 0.15)
    0.1026F = 1.61865

    F = [tex]\frac{1.61865}{0.1026}[/tex]
    = 15.775
    = 16Nm
     
  2. jcsd
  3. Oct 11, 2007 #2

    Astronuc

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    Staff: Mentor

    Part a is correct. One uses the CM of the arm to determine the moment about the elbow joint.

    The logic on part b is not correct. The moment of the object and the moment of the arm work together against the bicep. The torque provided by the bicep must equal the sum of torques of the arm and object.
     
  4. Oct 11, 2007 #3
    sorry i forgot about the 3rd force
    [tex]\Sigma[/tex][tex]\tau[/tex] = 0
    [tex]\tau[/tex]1 + [tex]\tau[/tex]2 + [tex]\tau[/tex]3 = 0
    [tex]\tau[/tex]1 + [tex]\tau[/tex]2 = -[tex]\tau[/tex]3

    but i'm a little unsure about the force of [tex]\tau[/tex]2 (which i have as the object)

    is it: -[tex]\tau[/tex]3 = [tex]\tau[/tex]1 + [tex]\tau[/tex]2
    = m1gx1 + m2gx2
    = 1.1 x 9.81 x 0.15 + 20 x 9.81 0.3
    = 60.47865N
    [tex]\tau[/tex]3 = -60.47865N
    = -6 x 10[tex]^{2}[/tex]N
     
  5. Oct 11, 2007 #4

    Astronuc

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    Staff: Mentor

    Be careful with units, torque is N-m, as opposed to force which uses N.

    Don't forget the angle 20°, which influences the force normal to the moment arm.
     
  6. Oct 11, 2007 #5
    ok so
    [tex]\tau[/tex]1 = Fsin[tex]\theta[/tex]
    = mgsin[tex]\theta[/tex]
    = 1.1 x -9.81 x sin 20 x 0.15
    = -0.5536Nm

    [tex]\tau[/tex]2 = Fsin[tex]\theta[/tex]
    = mgsin[tex]\theta[/tex]
    = 20 x -9.81 x sin 20 x 0.3
    = -20.131Nm

    -[tex]\tau[/tex]3 = [tex]\tau[/tex]1 + [tex]\tau[/tex]2
    = -0.5536 + -20.131
    = -20.685Nm

    therefore [tex]\tau[/tex]3 = 20.685Nm
    or 20.685Nm in an anticlockwise direction
     
  7. Oct 11, 2007 #6

    Astronuc

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    Staff: Mentor

    I believe one wants the cosine of the angle (between the weight and the normal to the moment arm) in this case or the sine of the angle between the force and the moment arm (which would be 90° + 20°).
     
  8. Oct 11, 2007 #7
    so if i change sin 20 to cos 20 (or sin 110) i get:

    [tex]\tau[/tex]1 = Fsin
    = mgsin
    = 1.1 x -9.81 x cos 20 x 0.15
    = -1.521Nm

    [tex]\tau[/tex]2 = Fsin
    = mgsin
    = 20 x -9.81 x cos 20 x 0.3
    = -55.310Nm

    -3 = 1 + 2
    = -1.521 + -55.310
    = -56.831Nm

    therefore 3 = 56.831Nm
    or 56.831Nm in an anticlockwise direction

    is this the correct answer?
    (thankyou for all your help)
     
  9. Oct 11, 2007 #8

    Astronuc

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    Staff: Mentor

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