Question

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?

Answer 1

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)