# How long is the board?

1. Apr 13, 2014

### confused111

1. The problem statement, all variables and given/known data
A 7kg cat walks to the end of a 15 kg board resting on two saw horses as shown below.
When the cat is at the 1m position as shown, the board “just” lifts off of
the left hand support. How long is the board?

2. Relevant equations

T = D x F

3. The attempt at a solution
I know that the torque on both sides has to be equal, but I have no idea how to approach this without knowing the length!

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2. Apr 13, 2014

### CWatters

Perhaps I'm misreading the question but it does say that "A 7kg cat walks to the _end_ of a 15 kg board".

So you can calculate how far the centre of mass of the board is from the right hand horse (call it L).

3. Mar 5, 2016

### CyberneticTitan

Hello,

Sorry for grave-digging this thread, but I have the exact same problem as you. Perhaps you were also a NCDES Physics 12 student with the same workbook aha. I may have figured out the solution, for all the NCDES Physics 12 students googling this question aha. No where in our lessions were we shown how to do this with just T=Fd

Because the board is about to lift, there is now weight on the left hand side and that saw horse can be removed so there is just one pivot point.

http://hyperphysics.phy-astr.gsu.edu/hbase/cmms.html (m1gL1=m2gL2)

Plug in the values and solve for L2, then find the total distance.

4. Mar 5, 2016

### haruspex

So what answer do you get?

5. Mar 6, 2016

### CyberneticTitan

Using m1gL1=m0gL2

(7kg)(9.8m/s2)(1m)=(15kg)(9.8m/s2)(L2)

Solve for L2 results in 0.4666....meters, which is the distance from the pivot to the center of gravity.

Add 1 meter to find the distance from the cat to the center of gravity, which is 1.4666...meters.

Double to find the total length of the board, which is 2.9333..meters before sigfigs.

This is right according to: http://bclearningnetwork.com/LOR/media/ph/learning_guides/PH12_2013/PH12_U2_notes.pdf (Page 19, answers in the back.)