Calculating Maximum Weight on a Tipping Table with a Cat: Torque Analysis

[henry3369]

Homework Statement

http://imgur.com/dMw79w1

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

Homework Equations

Net torque = 0

The Attempt at a Solution

First, I set the axis of rotation to be the bottom of the table (where it goes through F_x and F₀)
τ_{due to downward forces} = τ_{due to upward forces}
τ_vase + τ_table + τ_cat = τ_{due to Fy} (F_y (the solution to F_y is in the image)
W_vY + W_tL_y/2 + W_maxL_y = W_t/2 + W_vY/L_y

Rearrange for W_max:
W_max = W_t/2L_y + W_vY/L_y² - W_vY/L_y - W_t/2

The answer to the question is:
W_max = W_v(1 - X/L_x - Y/L_y)

I don't know how to obtain this answer from my answer above.

 

[henry3369]

It seems that I forgot to include the distance the cat is from the axis of rotation (L_y). After including it, I got W_max = 0.

 

[henry3369]

So here it the attempt once again:
W_vY + W_tL_y/2 + W_maxL_y = (W_t/2 + W_vY/L_y)(L_y)

Rearranging for W_max:
W_maxL_y = (W_t/2 + W_vY/L_y)(L_y) - W_vY - W_tL_y/2
W_max = W_t/2 + W_vY/L_y - W_vY/L_y - W_t/2
W_max = 0 N

 

[haruspex]

So here it the attempt once again:
W_vY + W_tL_y/2 + W_maxL_y = (W_t/2 + W_vY/L_y)(L_y)

Rearranging for W_max:
W_maxL_y = (W_t/2 + W_vY/L_y)(L_y) - W_vY - W_tL_y/2
W_max = W_t/2 + W_vY/L_y - W_vY/L_y - W_t/2
W_max = 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 W_maxL_y

 

[haruspex]

The cat exerts a downward force on the table though (the weight), which is shonw by W_maxL_y

Yes, but you used the expressions obtained for F_x etc. in the first part. There was no cat then. The forces will be different now.