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Equilibrium Problem

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    A 62.5kg painter stands 1.35m from one end of a 45.0 kg scaffold that is 4.50m long. She has 11.0kg supply of paint 1.00m from the end she is nearest to. Find the reaction forces at the supports if the scaffold is supported at each end.


    2. Relevant equations

    F=ma
    N1+N2-w(mg)=0

    3. The attempt at a solution
    Well I know that sum of forces equal to zero and the newton's second law formula F=ma. So I thought of calculating all the weights to get force (N) and then I got stuck. Please help me on this.
     
    Last edited: Mar 22, 2010
  2. jcsd
  3. Mar 22, 2010 #2
    Hint: Is the system rotating?
     
  4. Mar 22, 2010 #3
    Yes, so the axis of rotation I choose in the left end of scaffold so the right end point will go counter clockwise (+) direction? Also I placed the painter on end of left side of scaffold as well.

    http://img404.imageshack.us/img404/7128/equilibrium.png [Broken] . Here's the diagram I drew.
     
    Last edited by a moderator: May 4, 2017
  5. Mar 22, 2010 #4
    Fpainter= (62.5kg*9.81m/s)= 613.125N
    Fsupply= (11.0kg*9.81m/s)= 107.91N
    Fstands= 45.0kg*9.81m/s) = 441.45N
    Total force Fpainter+Fsupply+Fstands= 1162.485N = N1+N2
    Then 613.125N*1.35m+107.91*1.00m/4.50m= N2= 207.9175N
    N1+N2= 1132.485N+207.9175N= 1370N
    so am I got it right? I don't get it. Please help
     
  6. Mar 22, 2010 #5

    rl.bhat

    User Avatar
    Homework Helper

    Then 613.125N*1.35m+107.91*1.00m/4.50m= N2= 207.9175N
    In this equation you have left one term.
    F(stand) acts at the center of the scaffold. So the torque due to this force is
    45.0kg*9.81m/s) *2.25m.
    Now find N2.
     
  7. Mar 22, 2010 #6
    I got N2 as 429N. Do I add N1(1132.485N)? or subtract to get the other force?

    EDIT: Nevermind, I got it. I had the Total force calculation got wrong. Thank you!
     
    Last edited: Mar 22, 2010
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