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Torque: Table tipping

  • Thread starter henry3369
  • Start date
194
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1. Homework Statement
http://imgur.com/dMw79w1 [Broken]

The image contains the problem as well as the answers to the previous problems.

2. Homework Equations
Net torque = 0

3. The Attempt at a Solution
First, I set the axis of rotation to be the bottom of the table (where it goes through Fx and F0)
τdue to downward forces = τdue to upward forces
τvase + τtable + τcat = τdue to Fy (Fy (the solution to Fy is in the image)
WvY + WtLy/2 + WmaxLy = Wt/2 + WvY/Ly

Rearrange for Wmax:
Wmax = Wt/2Ly + WvY/Ly2 - WvY/Ly - Wt/2

The answer to the question is:
Wmax = Wv(1 - X/Lx - Y/Ly)

I don't know how to obtain this answer from my answer above.
 
Last edited by a moderator:
194
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It seems that I forgot to include the distance the cat is from the axis of rotation (Ly). After including it, I got Wmax = 0.
 
194
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So here it the attempt once again:
WvY + WtLy/2 + WmaxLy = (Wt/2 + WvY/Ly)(Ly)

Rearranging for Wmax:
WmaxLy = (Wt/2 + WvY/Ly)(Ly) - WvY - WtLy/2
Wmax = Wt/2 + WvY/Ly - WvY/Ly - Wt/2
Wmax = 0 N
 

haruspex

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So here it the attempt once again:
WvY + WtLy/2 + WmaxLy = (Wt/2 + WvY/Ly)(Ly)

Rearranging for Wmax:
WmaxLy = (Wt/2 + WvY/Ly)(Ly) - WvY - WtLy/2
Wmax = Wt/2 + WvY/Ly - WvY/Ly - Wt/2
Wmax = 0 N
The upward forces were calculated in the absence of a cat.
 
194
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The upward forces were calculated in the absence of a cat.
The cat exerts a downward force on the table though (the weight), which is shonw by WmaxLy
 

haruspex

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The cat exerts a downward force on the table though (the weight), which is shonw by WmaxLy
Yes, but you used the expressions obtained for Fx etc. in the first part. There was no cat then. The forces will be different now.
 

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