The drawing shows a person (weight W = 592 N, L1 = 0.841 m, L2 = 0.410 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

It is difficult to decipher the figure. Sum moments about any point, and the sum of vertical forces equals zero. Those two equations should give you a system of equations to solve.

To find the normal force exerted by the floor on each hand and each foot, we can use the principles of equilibrium.

1. Start by drawing a free-body diagram of the person in the push-up position. Label the weight W downward at the center of mass of the person and the distances L1 and L2 from the center of mass to each hand and foot, respectively.

2. The person is in equilibrium, so the sum of the forces in the horizontal direction must be zero. Since there are no horizontal forces acting on the person, this condition is automatically satisfied.

3. The sum of the forces in the vertical direction must also be zero. The normal force exerted by the floor on each hand and each foot will balance the weight of the person.

4. The total weight W can be divided between the hands and feet based on their respective distances from the center of mass. The force on each hand (Fh) can be found using the following equation:
Fh = (L2 / (L1 + L2)) * W

5. Similarly, the force on each foot (Ff) can be found using the following equation:
Ff = (L1 / (L1 + L2)) * W

6. Plug in the given values for the weight W, L1, and L2 into the equations to calculate the forces on each hand and each foot.

Let's substitute the given values:
W = 592 N
L1 = 0.841 m
L2 = 0.410 m

Fh = (0.410 / (0.841 + 0.410)) * 592 N
Ff = (0.841 / (0.841 + 0.410)) * 592 N

Now, calculate the forces:
Fh ≈ 256.93 N
Ff ≈ 335.07 N

Therefore, the normal force exerted by the floor on each hand is approximately 256.93 N, and the normal force exerted by the floor on each foot is approximately 335.07 N.