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Sagot :
Answer:
Archie must sit 0.58 m relative to the center to keep the see-saw balanced
Explanation:
Given the data in the question;
Mass of Archie [tex]m_{a}[/tex] = 120 kg
Mass of clementine [tex]m_{c}[/tex] = 40 kg
total length of see-saw L = 3.5 m
as illustrated on the image below, Fulcrum is at the center,
suppose Archie sits at a distance x from center then for balancing, we will have;
[tex]m_{a}[/tex] × x = [tex]m_{c}[/tex] × ( one end = 3.5/2 = 1.75)
so we substitute
120kg × x = 40kg × 1.75m
x12okg = 70 kg.m
x = 70 kg.m / 120 kg
x = 0.58 m
Therefore, Archie must sit 0.58 m relative to the center to keep the see-saw balanced

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