Balancing a Seesaw: Finding Mass with Torque

[Katelyn]

Homework Statement

The fulcrum of a uniform 20-kg seesaw that is 4.0 m long is located 2.5 m from one end. A 26-kgchild sits on the long end.

Determine the mass a person at the other end would have to be in order to balance the seesaw.

Homework Equations

Torque = force x distance from fulcrum
Force = ma

The Attempt at a Solution

(26 + 12.5) (g) (2.5) = (7.5 +m) (g) (1.5)
Acceleration due to gravity cancels out.
38.5 (2.5) = (7.5 + m) (1.5)
96.25 = 11.25 + 1.5m
85 = 1.5m
57 m

I tried to enter this answer (supposed to be rounded to two significant figures) but it came up as wrong. What is wrong with my equation?

 

[DoItForYourself]

Hi,

Edit Note: Sorry, I now realized that the fulcrum is not at the center of the seesaw. Just think where exactly (at which point) the weight of the one side and the weight of the other side applies? This point is not at the end of the seesaw.

 

[Chandra Prayaga]

The fulcrum is not at the center of the seesaw. So the weight of the seesaw, which acts at its center, will have a torque about the fulcrum.
Katelyn, you are assuming that the portion of the seesaw on each side of the fulcrum acts at the corresponding end of the seesaw. That is a wrong assumption. The correct way is to assume that the entire weight of the seesaw acts at its center of mass, which in this case is at the center of the seesaw. Then that weight will have its own torque about the fulcrum. You should draw a diagram of the seesaw, with each force acting at the appropriate position, and then you can figure out the individual torques.

 

[DoItForYourself]

I agree, this is the complete and right way to proceed with this kind of problems.

But as long as we have uniformly distributed mass, we can take each side (with its respective mass) separately and consider the torque of the mass of each side (which applies in the centre of each side). This is what I meant previously and this will also lead to the same result.

 

[Chandra Prayaga]

Absolutely. Whether you consider the entire weight of the seesaw to be acting at its center of mass, or think of he weight of each side of the fulcrum as acting at the center of mass of that side, it is entirely equivalent. This is where I would always advise the student to draw a diagram. It is a diagram, rather than looking for some formula or a set of equations, that clears up all the doubts.

 

[Katelyn]

Ok, so should I consider the center of mass for the seesaw at 2 m, and then calculate that as a separate torque? Maybe like this:
T1+T2=T3
26 kg (2.5 m) + (20 kg) (.5 m) = (m) (1.5 m)

 

[haruspex]

Ok, so should I consider the center of mass for the seesaw at 2 m, and then calculate that as a separate torque? Maybe like this:
T1+T2=T3
26 kg (2.5 m) + (20 kg) (.5 m) = (m) (1.5 m)

Yes.

 

[Katelyn]

Thank you!