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A 16ft seesaw is pivoted in the center. At what distance from the center would a 200lb person sit to balance a 120lb person on the opposite end?

1 Answer

5 votes

Answer:

9.6 ft

Step-by-step explanation:

Distance is inversely proportional to weight

distance = k / (weight), where

k is a constant

or you could say,

distance * weight = k

In this scenario,

120 * 16 = 200 * distance

On rearranging, making, distance the subject of formula, we have

Distance = 120 * 16 / 200

Distance = 1920 / 200

Distance = 9.6 ft

So the 200 pounds person should sit 9.6 feet away from the centre to balance the see saw

User Suman Kundu
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