A person whose weight is W = 658 N is doing push-ups. From his feet to his center of gravity he is 0.840m and from the center of gravity to his hands is 0.410m. Find the normal force exerted by the floor on (a) each hand and (b) each foot, assuming that the person holds this position.

A man whose weight is 658 N is standing on the ground. How much force does the ground exert on the man? *

To find the normal force exerted by the floor, we need to consider the forces acting on the person and the conditions for equilibrium. In this case, the person is in a stationary position, so the sum of the forces acting on them must equal zero.

Let's start with determining the forces acting on the person. We have the weight (W) acting downward, and there will be normal forces exerted by the floor at the hands and feet, denoted as N_hand and N_foot, respectively.

For equilibrium, the sum of the forces in the vertical direction must equal zero. Therefore, considering the forces at the hands and feet, we can set up the following equations:

At the hands:
N_hand - W = 0 (since the person isn't accelerating vertically)
N_hand = W

At the feet:
2N_foot - W = 0 (since there are two feet on the ground)
2N_foot = W

Now, let's substitute the given weight value:
W = 658 N

For the hand:
N_hand = 658 N

For the foot:
2N_foot = 658 N
N_foot = 658 N / 2
N_foot = 329 N

Therefore, the normal force exerted by the floor on each hand is 658 N, and the normal force exerted by the floor on each foot is 329 N.