On a 15-foot seesaw, Person A weighs 100 lb and Person B weighs 150 lb. If each sits on opposite sides with their distances from the fulcrum summing to 15 feet, how far should B sit from the fulcrum to balance?

On a 15-foot seesaw, Person A weighs 100 lb and Person B weighs 150 lb. If each sits on opposite sides with their distances from the fulcrum summing to 15 feet, how far should B sit from the fulcrum to balance?

• A. 9 feet
• B. 7.5 feet
• C. 15 feet
• D. 6 feet

Answer: (D)

Torque balance is the key idea here: a seesaw stays level when the turning effect of one side equals the turning effect of the other. That turning effect is weight times distance from the fulcrum (torque). If the lighter person sits at a distance x and the heavier person sits at a distance y on the opposite side, balance requires 100 × x = 150 × y. Their distances add to 15 feet, so x + y = 15. From 100x = 150y, you get x = 1.5y. Substituting into x + y = 15 gives 1.5y + y = 15, so 2.5y = 15 and y = 6 feet. Then x = 15 − 6 = 9 feet. So the heavier person should sit 6 feet from the fulcrum (the lighter person sits 9 feet away).

Torque balance is the key idea here: a seesaw stays level when the turning effect of one side equals the turning effect of the other. That turning effect is weight times distance from the fulcrum (torque).

If the lighter person sits at a distance x and the heavier person sits at a distance y on the opposite side, balance requires 100 × x = 150 × y. Their distances add to 15 feet, so x + y = 15.

From 100x = 150y, you get x = 1.5y. Substituting into x + y = 15 gives 1.5y + y = 15, so 2.5y = 15 and y = 6 feet. Then x = 15 − 6 = 9 feet.

So the heavier person should sit 6 feet from the fulcrum (the lighter person sits 9 feet away).