Asked by sam
Four acrobats of mass 83.5 kg, 68.0 kg, 62.0 kg, and 55.0 kg form a human tower, with each acrobat standing on the shoulders of another acrobat. The 83.5-kg acrobat is at the bottom of the tower.
(a) What is the normal force acting on the 83.5-kg acrobat?
(a) What is the normal force acting on the 83.5-kg acrobat?
Answers
Answered by
Henry
Fn = m*g = (83.5+68+62+55) * 9.8
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