Question

The drawing shows a person (weight W = 587 N, L₁ = 0.853 m, L₂ = 0.398 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position.

force on each hand	N
force on each foot	N

Answer 1

If the person holds the position, then there is no acceleration nor rotation, so we have vertical and rotational equilibrium.

Vertical equilibrium tells us all vertical forces are balanced, which means

2F+2H=587 where F is he force of the floor on each foot and H is the force of the floor on each hand; dividing by 2:

F+H=293.5

the sum of torques is also zero; if we sum torques around the feet, we have

sum of torques = 587 L1 + 2H (L1+L2) =0

the force F exerts no torque around the feet since F has no lever arm around the feet
substituting numbers, we have
587x0.853+2H(0.853+0.398)=0
H=200.12N

since H+F=297N, F=93.4N