A man doing a slow push-up is approximately in static equilibrium. His body is horizontal, with his weight of 750 N supported by his hands and feet, which are 1.36 m apart. One hand rests on a spring scale, which reads 268 N. If each hand bears an equal weight, how far from the shoulders is the man's center of gravity?

Question

Damon · Answer

I assume hands under shoulders x = CG to feet hands hold 268*2 = 536 N so 536 * 1.36 = 750 x x = .972 m from CG to feet 1.36 - .972 = .388 CG to hands (shoulders)