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Torque has been a difficult chapter

  1. Nov 16, 2009 #1
    I'm currently working on two homework problems that are giving me a bit of trouble. Torque has been a difficult chapter for me. I'm going to worry about just this one for now:

    Question 1:
    Calculate FA and FB for the beam shown in Fig. 9-49. The downward forces represent the weights of machinery on the beam. Assume the beam is uniform and has a mass of 210 kg.

    Fig. 9-49 shows:
    http://grab.by/Byt

    I know that the net external force must be zero: SUM of Forces = 0
    I also know that the net external torque must be zero: SUM of Torque = 0

    So I've set up my equation for Torque, where T is Torque:

    T = Force x Distance from the Pivot

    So if I make my pivot the point at FA, shouldn't I have:

    0 = - T(4300N x 2.0M) - T(3100N x 6.0M) - T(2200N x 9.0M) + Tboard + TFB

    Solving for the Torque of the Board:
    Tboard = 210Kg * 9.8 m/s2 * 5M = 10290

    Plugging in all the Torques, 10FB being 10 (distance from FA) x Force (FB)
    0 = -8600 - 18600 - 19800 + 10290 + 10FB

    When I solve for FB however, and plug that force back into the equation, I'm not getting zero. I'm lost as to where I'm going wrong.
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Nov 16, 2009 #2

    PhanthomJay

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    Re: Torque

    You missed the minus sign on T_board. The torque is in the same direction as the machinery loads torque. The plus or minus sign will bite you every time.
     
  4. Nov 16, 2009 #3
    Re: Torque

    Okay, I see what I did there. So then:

    57290 = 10FB
    so 5729 = FB


    edit: I figured it out:
    Then I just go to the other side and make the FB pivot at zero, so that:

    0 = (2200 x 1M) + (3100 x 4M) + (4300 x 8M) + T_Board - 10FA
    0 = 2200 + 12400 + 34400 + 10290 - 10FA
    FA = 5929

    Thanks for the help!
     
    Last edited: Nov 16, 2009
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