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Torque and Forces

  1. Nov 19, 2006 #1
    A light triangular plate OAB is in a horizontal plane. The 3 forces, F1 = 7 N, F2 = 3 N, F3 = 8 B, act on the plate, which is pivoted about a vertical axes through point O. Consider the counterclockwise sense as positive. The sum of the torques about the vertical axis through point O, acting on the plate due to forces F1, F2, and F3, is closest to:

    a. 0.30 N*m-----------b. 1.2 N*m---------c. –0.30 N*m------d. –1.2 N*m-------e. zero

    In the attachment, please note that the angle between F1 and the dotted line is 30 degrees. The angle between the dotted line and F3 is 45 degrees.
    Angle 1 = 30 degrees
    Angle 3 = 45 degrees

    Hypotenuse = 1.0 m

    If you cannot see the attachment, please tell me.

    Since the plate is rotated around point O,

    The torque where F3 acts is 0 m*N since r = 0 m??

    Therefore, only forces 1 and 2 are left.

    Is Force 2 counterclockwise, which makes it positive? Torque 2 = F*r*sin theta = 3N*0.8 m*sin 90 = +2.4 m*N??

    Is Force 1 clockwise, which makes it negative? Torque 1 = F*r*sin theta = 7N*sin 90*0.6 m = -2.1 m*N?

    Sum of torques = 2.4 m*N – 2.1 m*N = 0.30 m*N??


    Attached Files:

  2. jcsd
  3. Nov 19, 2006 #2


    User Avatar
    Homework Helper

    Unfortunately, I cannot see the attachment, but it is possibly due to my computer [:grumpy:]. But yes, if the force vector intersects with the axis, the torque produced by that force must equal zero.
  4. Nov 19, 2006 #3
    I tried saving it as a Word document. Hopefully, it is accessible now.

    Attached Files:

  5. Nov 20, 2006 #4
    Are my signs for the torque correct?

    Thanks again.
  6. Nov 21, 2006 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, your signs are correct.
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