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Torque 2

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data
    A circular-shaped object of mass 9 kg has an
    inner radius of 7 cm and an outer radius of
    20 cm. Three forces (acting perpendicular
    to the axis of rotation) of magnitudes 12 N,
    27 N, and 16 N act on the object, as shown.
    The force of magnitude 27 N acts 29 below
    the horizontal. Find the magnitude of the net torque on
    the wheel about the axle through the center
    of the object.
    Answer in units of N · m

    2. Relevant equations
    t1 = r * F
    t2 = r * F
    t3 = r * F

    3. The attempt at a solution
    net torque = r1F1 + r1F2 + r3F3
    = (.2)(12) + (0.2)(16) + (0.07)(27)
    = 7.49 N * m
    But it's wrong... why...
     
  2. jcsd
  3. Oct 20, 2011 #2
    You mention 'as shown'. I don't see where it is shown.
     
  4. Oct 20, 2011 #3
    here is the picture that i draw. i hope it's good enough.

    ok so i think i got it but check it for me.

    If torque is going clockwise, it's negative. If it's counterclockwise, it's positive.

    So two torques are negative because if I pull the radius, the circle will turn clockwise.
    12(0.2) and 16(0.2) are negative torques which is -5.6.

    But the third one is positive because it is when I pull the radius, it will spin clockwise.
    27(0.07) which is 1.09

    So net torque is -5.6 + 1.09 is -3.71.

    But since it is asking for magnitude, I can put absolute value on net torque which is 3.71. Is that right?
     

    Attached Files:

    Last edited: Oct 20, 2011
  5. Oct 20, 2011 #4
    Yes, think of magnitude as an absolute value.
     
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