Question

Help! I know this question is not too hard but I don't know how to approach it!
A 5.1kg cat and a 2.5kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. How far to the left of the pivot must a 4.0kg cat stand to keep the seesaw balanced?

Answer 1

First, we sum the masses of the entire system:

5.1 kg + 2.5 kg + 4.0 kg = 11.6 kg

Next, we weight each of the masses with the position coordinate:

(5.1 kg) * (0 m) + (2.5 kg) * (4 m) + (4.0 kg) * (x m) = (10 + 4x) kg m

Dividing the moments by the entire mass of the system gives the CM. We want this coordinate to be in the middle of the see-saw at x = 2 m.

Xcm = (moments) / (total mass)
2 = (10+4x) / (11.6)
23.2 = 10 + 4x
13.2 = 4x
x = 3.3 m

Now, this x value we found is measured from where we set x = 0. We set x=0 to correspond to the 5.1 kg cat. That means we need to place the 4.0 kg cat 3.3 m from the 5.1 kg cat.

That would be at x = 3.3 m. However, the question asks how far to the side of the pivot point. Since the pivot point is at 2 m, we can find the difference.

3.3 m - 2 m = 1.3 m

So we should place the 4.0 kg cat 1.3 m to the left of the pivot point, given that the tuna fish is on the left end of the see-saw and the other cat is on the right end of the see-saw.

Answer 2

Thank you so much @Pompeii00