weight is W = 506 N doing push-ups. Find the normal force exerted by the floor on (a) each hand and (b) each foot

Question

         Physics HW 

Answer 1

New Answer with Diagram:

 

The normal forces at the hands and feet are different because the center of mass is closer to the hands.

 

F_NF + F_NH = W to balance the forces in equilibrium.

 

F_NH(0.840 + 0.410) = W(0.840) to balance the torques in equilibrium about the feet.

 

F_NH = W(0.840/1.250) = 506(0.840/1.250) = 340 N

 

So, F_NF = 506 - 340 = 166 N.

 

 

Therefore, each hand will support 170 N and each foot will support 83 N.

 

 

Old Answer without Diagram:

 

Assuming that the center of mass is equidistant from each appendage, then the normal force on each hand and foot would be 1/4 the weight of the person doing pushups if the person is not in motion.

 

When the person is going up, the normal forces have to be bigger to accelerate the body up.  When the person is going down, the normal forces are less than the weight.