Torque w/ See-Saw: Solve for Mass

[goonking]

Homework Statement

Homework Equations

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)?

 

[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.

 

[goonking]

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

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

 

[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?

 

[goonking]

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?

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.

 

[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.

 

[goonking]

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.

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?

 

[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.

 

[AlephNumbers]

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.

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.

 

[haruspex]

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.

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.

 

[AlephNumbers]

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.

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

 

[goonking]

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

 

[haruspex]

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

Yes.

 

[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?

 

[goonking]

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.

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

 

[haruspex]

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?

Looks right (but you can't justify that many significant digits).

how you would do this problem the 'less simple' way

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.

 

[goonking]

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.

so you mean like this?

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

 

[haruspex]

so you mean like this?

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

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

 

[goonking]

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

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?

 

[haruspex]

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?

looks right. Do you get the same answer?

 

[goonking]

looks right. Do you get the same answer?

yep! thanks!

 

[sun18]

Just to point out, the units of torque are N*m, and you're writing kg*m.
In the end, it won't matter because the factor of g will cancel out, but if you had to show your work, you would need to include (mass)*(gravity)*(length)