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Exam question problem: moments

  1. Mar 27, 2012 #1
    1. The problem statement, all variables and given/known data
    I am having problem calculating this question. I even looked at the answer sheet and worked back, which gave me some luck, but still do not fully grasp the question.(i) Calculate the weight W of the workman.


    2. Relevant equations
    moment clockwise = moment anticlockwise


    3. The attempt at a solution
    My attempt pre look at question: I did 0.20-0.03= 0.17, 0.20+0.03= 0.23 so far so good.
    I the did 0.20+0.50=0.70 so middle Q which then gave me 0.67 and 0.73. I then took the W*0.17=160*0.73.

    But when I looked at the answer I got the first part right but the second part wrong. Do I have to include R and S into the equation? I would appreciate the help if possible.

    Also here is what the answer says: W × 0.17/0.20/0.23 = 160 × 0.72/0.75/0.78
    W × 0.17 = 160 × 0.78 or 600 N
    730/734 N

    I really cant see where the 0.72 ect come from unless you included distance from R S but the weight is in the middle.
     

    Attached Files:

  2. jcsd
  3. Mar 27, 2012 #2

    tms

    User Avatar

    The first thing to do, for this and all problems, is to solve it symbolically; forget there are numbers until the very last step. You'll never learn the physics if you plug in numbers from the beginning.

    Second, your "relevant equation" is not only ambiguous (what moment are you talking about), it is wrong. The system is in equilibrium, so the net force is zero and the net torque is also zero. That is your starting point. Write down those equations, then solve for W.
     
  4. Mar 27, 2012 #3
    could you please expand, I really cant see where you are coming from. I have looked in my study books and the point me in the directions of levers and moments.
     
  5. Mar 27, 2012 #4

    tms

    User Avatar

    The torque of the man plus his segment of the board and the torque of the rest of the board must be equal. Recall that [itex]\mathbf\tau = \mathbf{r} \times \mathbf{F}[/itex]. Since the angles of both torques are 90 degrees, this reduces to [itex]\tau = rF[/itex].
     
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