A man of mass 75.0 kg is doing push-ups. He stops moving. What is the normal for

Question

A man of mass 75.0 kg is doing push-ups. He stops moving. What is the normal force on each hand and each foot? The distance between his hands and his feet is 1.35m. His center of mass is 0.950m from his feet. Assume all angles are 90.0 degrees. Please show all work and explain the equations used.

Explanation / Answer

since the man is symmetric ( his body) the reaction on both his hands is same and also the reaction on both his feet let the reactions be Rf and Rh => 2*(Rf+ Rh) = m*g now balancing the moments about the feet we have => 2*Rh*1.35 = m*g*.95 => Rh = 258.875 N => the reaction on each hand is Rh = 258.875 N while that on his foot is 2*Rf = m*g -2*Rh => Rf = 109 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.