Torque with see-saw

1. Apr 19, 2015

goonking

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I set up my problem as :

(12kg x .5m) + ( 1/3 x 4kg) = (1.05m x Mass) + (2/3 x 4kg)

then , i just solve for mass, is that correct?

actually, should I change the (1/3 x 4kg) to ( 1m x 4kg) and the (2/3 x 4kg) to ( 2m x 4kg)?

2. Apr 19, 2015

mukundpa

Second terms on each sides is only force not torque. The best way is to take center of mass of the plank under consideration.

3. Apr 19, 2015

goonking

which equation do you mean? the first or second one?

4. Apr 19, 2015

sun18

You are trying to balance torques about the pivot point of the seesaw. What you have written is mass*length, which does not have units of Newton-meters.
You are close to the right idea, but the second term on each side of the equality is not correct. As mukundpa says, find the center of mass of the board and treat it as a point mass; what torque does that point mass produce, and which side of the pendulum is it on?

5. Apr 19, 2015

goonking

the center of mass is at the middle of the board, which is 1.5 meters.
the pivot point is to the right of the center of mass.

6. Apr 19, 2015

sun18

That's right, so now think about the problem: you have 2 masses on the left of the pivot point, and one on the right. Now try to balance the torques and solve for M.

7. Apr 19, 2015

goonking

what do you mean 2 ? are you counter the board and the orange cat?

if you did, shouldn't there be 2 masses on the right of the pivot point too?

8. Apr 19, 2015

sun18

Imagine that the board is completely massless, except for one concentrated mass at the center of mass of the board. Then, the only masses in the system are the orange cat on the left, the mass of the board on the left, and the blue cat on the right.

9. Apr 19, 2015

AlephNumbers

I believe there would also be a force exerted by gravity on the mass of the portion of the board that is to the right of the pivot. Actually, I need to learn to read.

Last edited: Apr 19, 2015
10. Apr 19, 2015

haruspex

There are two ways you can deal with the board. You can consider each side separately, exerting opposite torques, or the simpler way sun18 suggests: consider the whole board's mass to be concentrated at its mass centre. That gives you only 3 masses to worry about.

11. Apr 19, 2015

AlephNumbers

Oh okay. I did not realize that that was what sun18 was suggesting. That is indeed simpler.

12. Apr 19, 2015

goonking

so now I need to find how far away the center of mass of board is from the pivot point?

13. Apr 19, 2015

haruspex

Yes.

14. Apr 19, 2015

goonking

center of mass is 0.5m away from pivot

(orange cat mass x 1.05m) + (4kg x 0.5m) = 12kg x 0.5m

mass orange cat = 3.809 kg.

correct?

15. Apr 19, 2015

goonking

i'm interested on how you would do this problem the 'less simple' way. can you show me the math?

16. Apr 19, 2015

haruspex

Looks right (but you can't justify that many significant digits).
Much the same... just treat the plank as two planks, one each side. It's more complicated because you have to calculate the mass and mass centre of each piece.

17. Apr 19, 2015

goonking

so you mean like this?

(12kg x .5m) + ( 1/3 x 4kg) = (1.05m x Mass) + (2/3 x 4kg)

18. Apr 19, 2015

haruspex

You've missed out the distances to the mass centres of the two pieces of plank.

19. Apr 19, 2015

goonking

ok i see

so it is :

(12kg x .5m) + ( 1/3 x 4kg x .5m) = (1.05m x Mass) + (2/3 x 4kg x 1m)

the center of mass of the plank of the right side is .5 meters away from pivot

and the center of pass of the plank of the left side is 1 meter away from pivot.

correct?

20. Apr 19, 2015

haruspex

looks right. Do you get the same answer?