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Equilibrium Problem

  • Thread starter kee23
  • Start date
  • #1
17
0

Homework Statement


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.


Homework Equations



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

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:

Answers and Replies

  • #2
421
1
Hint: Is the system rotating?
 
  • #3
17
0
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.
 
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  • #4
17
0
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
 
  • #5
rl.bhat
Homework Helper
4,433
5
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.
 
  • #6
17
0
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!
 
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