# Homework Help: Find the mass of a cat as it walks across a plank

1. Jun 21, 2013

### agentlee

1. The problem statement, all variables and given/known data
A cat walks along a uniform plank that is 4.00m long and has a mass of 7.00kg. The plank is supported buy two sawhorses, one .440m from the left end of the board and the other 1.50m from its right end. When the cat reaches the right ened, the plank just begins to tip. What is the mass of the cat?

2. Relevant equations
ƩFx = 0
ƩFy = 0
Ʃτ = 0

τ=I∂=Fr

3. The attempt at a solution
This is a solved example from my book, but I don't understand how they do it.
They're saying to calculate the torque about the right sawhorse, so they do
Mg(.500m) - mg(1.50)m = 0

This is probably an obvious one but, I have no idea where the .500m came from. Can someone explain this to me?

2. Jun 21, 2013

### TSny

What point do you think they're choosing for the axis of rotation?

3. Jun 21, 2013

### agentlee

I forgot to mention they labeled F1=0, so that's the axis of rotation right?

4. Jun 21, 2013

### barryj

Where is the center of mass of the plank?

5. Jun 21, 2013

### TSny

No. The reason F1 = 0 is because the plank is about to tip and is therefore losing contact with the sawhorse on the left. But they are not choosing the sawhorse on the left as the axis of rotation.

As the plank begins to tip, about what point does it rotate?

6. Jun 21, 2013

### agentlee

Where the cat is positioned?

7. Jun 21, 2013

### TSny

No, that's not it either.

8. Jun 21, 2013

### agentlee

I see what you mean now, okay. So through a valiant use of process of elimination, the second sawhorse

9. Jun 21, 2013

### TSny

So, when you want to find the torque due to the weight of the board, at what point of the plank do you put the weight of the board?

10. Jun 21, 2013

### agentlee

Oh wow...The weight's at halfway and the plank is at 1.5m so 2-1.5...wow lol
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