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?

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?

anonymous · Answer

This is torque problem. So it's Force time distance. 

Since this is a seesaw, the pivot point is going to be located at the center of the seesaw board, 2 meters from the end of the seesaw board.

The problem doesn't designate on which end of the board the 5.1kg cat and the fish are located. However, since we are using a 4kg cat positioned on the left of the board for balance, it probably makes sense to put the fish on the left end of the board.

You should be able to figure it out now.