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

I post this eailer and O'm still not getting it. I read my book and i know we would be using the center og gravity equation. I also know L2 is the hand force and L1 is the force of each foot

L1 and L2 are not forces; they describe locations where forces are applied. Between what points are the lengths L measured? Where is the person's center of mass?

I can visualize the problem but not the notation you are using.
The sum of the forces on arms and legs equals the person's weight. The distances of hands and feet from the center of mass determines how the forces are distributed between hands and feet. Set the moment about hands. feet or the center of mass equal to zero for the second equation you need.

WAT IS PHYSICS

To find the normal force exerted by the floor on each hand and each foot, we need to consider the equilibrium condition. In this case, the sum of the vertical forces should be zero since there is no vertical acceleration.

To start, let's define our known values:
Weight of the person, W = 592 N
Distance from the hands to the center of gravity (L2) = 0.410 m
Distance from each foot to the center of gravity (L1) = 0.841 m

First, we need to determine the location of the center of gravity (COG). The center of gravity is the point where the total weight of an object can be considered to act. Since the person is doing push-ups and assuming a symmetrical body, we can consider the COG to be at the midpoint between the hands and feet.

Let's calculate the position of the COG:
COG = (L1*W1 + L2*W2) / (W1 + W2)
Here, W1 and W2 represent the weights of the respective parts of the body.

Since the person is holding the position, the total weight W is distributed equally between the hands and feet:
W1 = W2 = W / 2 = 592 N / 2 = 296 N

COG = (0.841 m * 296 N + 0.410 m * 296 N) / (296 N + 296 N)
COG = 493.736 N*m / 592 N
COG = 0.833 m

Now, with the COG position determined, we can analyze the forces acting on each hand and each foot.

Let's consider the forces acting on one hand:
Sum of vertical forces on the hand = Normal Force - Weight of the hand part
Since the person is holding the position, the normal force and weight of each hand part should be equal.
Normal force on each hand = Weight of each hand part = W1 / 2 = 296 N / 2 = 148 N

Similarly, let's consider the forces acting on one foot:
Sum of vertical forces on the foot = Normal Force - Weight of the foot part
Normal force on each foot = Weight of each foot part = W2 / 2 = 296 N / 2 = 148 N

Therefore, the normal force exerted by the floor on each hand is 148 N, and the normal force exerted by the floor on each foot is also 148 N.