# Torque: Table tipping

#### henry3369

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.

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#### henry3369

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.

#### henry3369

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.

#### henry3369

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