Question:

Consider a seesaw (given below). The board of the seesaw is 1 foot wide and 8 feet long and weighs 36 pounds. But instead of the fulcrum being in the center of the board like a traditional teeter-totter, this fulcrum is actually 2 feet from one end and 6 feet from the other end. On the short side, there is a 1-foot by 1-foot 48-pound weight right at the end of the board. The problem is, what weight do you need on the other side of the board to balance the seesaw?

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Solution: The first thing to note is that the board has weight and must be considered in the weight and balance equation. The 8-foot board weighs 36 pounds, so each foot of board weighs 4.5 pounds. That means on the 2-foot side of the fulcrum, the board weighs 9 pounds; on the 6-foot side, it weighs 27 pounds. The fact that 9 and 27 add up to 36 is good for solving the puzzle.

Now it really helps here to understand something of the physics of seesaws. A seesaw balances when the torque on both sides of the fulcrum is equal. Consider the torque on the 6-foot side of the fulcrum. The center of the mass of the board is at 3 feet (half the length). To calculate the torque of a seesaw, multiply the weight of the board by the feet from the center of mass. In this case, on the right side of the fulcrum it is 27 pounds 3 feet = 81 pound feet. And that’s without an additional weight on it.

Now consider the other side. On the 2-foot side of the fulcrum, there is a 2-foot section that weighs 9 pounds, but that 2-foot section has a center of mass that is 1 foot from the fulcrum. So that piece of board by itself has a torque of 9 1, or 9 pound feet. Now the block weighs 48 pounds. Its center of gravity is a foot and a half from the fulcrum. So it’s 48 times a foot and a half, which comes out to be 72 pound feet. Adding the weight of the block (72 pound feet) plus the weight of the board (9 pound feet) makes 81 pound feet, exactly the torque of the other side. There is no need to add any weight. The seesaw is perfectly balanced as it is.