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Objects in equilibrium problem

  1. Apr 10, 2014 #1
    The problem is a uniform plank of length 6 m and mass 30 kg rests horizontally on a scaffold, with 1.5 m of the plank hanging over one end of the scaffold. How far can a 70 kg person walk on the overhanging part of the plank before it tips?

    Relevant equations
    Sum of all the forces = 0
    Sum of the torque must= 0


    Attempt-
    (1) I make the pivot point at the point of overhang so 1.5 of of the board is over the pivot point and -4.5 is to the left of the pivot point.
    (2) 1.5 is .25 of the length so it is .25 of the mass so the mass over the pivot point is 7.5. 4.5 is .75 of the length so it is .75 of the mass so the mass to the left of the pivot point is 22.5 kg.
    (3) since the sum of all the torque is 0 then 22(-4.5)+7.5(1.5)+70(x)= 0
    (4) solving for x I got 1.29 the correct answer is 0.643.
     
    Last edited: Apr 10, 2014
  2. jcsd
  3. Apr 10, 2014 #2
    First you calculate the torque you need to lift the bar
    Torque=F*d then T=300 * 1.5 = 450 N*m
    now you do sum of the torque must be = 0
    -450 + 700 *x = 0
    solve for x = 0,643
     
  4. Apr 10, 2014 #3
    What do you mean by "lift" the bar?
     
  5. Apr 10, 2014 #4
    Tip the plank down. my bad xD
     
  6. Apr 10, 2014 #5
    Ok I understand. Where did you get the 300 in calculating the torque needed to lift the bar?
     
  7. Apr 10, 2014 #6
    its 30kg * 10 m/s²
     
  8. Apr 10, 2014 #7
  9. Apr 10, 2014 #8
    Ok thanks a lot!
     
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