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

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray