two students sit on a see-saw. archie is a hulking football player with a mass of 120 kg. clementine is a dainty cheerleader with a mass of 40 kg. the see-saw is 3.5 m in total length with the fulcrum at the center. if clementine sits at the end on one side, where must archie sit relative to the center to keep the see-saw balanced

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 = 120 kg

Mass of clementine = 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;

× x = × ( 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

Force to stretch the wire is 250 N

Explanation:

As we know that modulus of elasticity will remain the same for the wire if the applied stretch to the wire is within elastic limit

So we will have

now we have

so we can write it as