When a seesaw is perfectly balanced each persons distance d

When a seesaw is perfectly balanced, each person\'s distance, \"d, \" from the fulcrum varies inversely as his weight, \"w.\" If a 120-pound person sits 5 feet from the seesaw fulcrum, where must a 150-pound person sit to balance the seesaw?

Solution

Dear Student

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Given that distance from fulcrum of seesaw varies inversely from person\'s weight

Therefore d = K / w ( Where K is proportionality constant)

From above we cn deduce the relation between two people sitting at opposite ends of see saw

d1/dd2 = w2/w1

Given

d1 = 5 ft w1 = 120 pounds

d2 = ? w2 = 150 pounds

=>

5 / d2 = 150 / 120

d2 = 4 ft

Solution