To find the normal force exerted by the floor on each hand and each foot, we need to analyze the forces acting on the person during the push-up exercise.
In this scenario, there are four contact points between the person and the floor: two hands and two feet.
Let's start by considering the forces acting on one hand:
1. Weight force (W = 580 N): This force acts vertically downwards from the center of mass of the person.
2. Normal force (N1) exerted by the floor on the hand: This force acts vertically upwards and balances the weight force.
Since the person is holding this position, there is no vertical acceleration, which means that the sum of the vertical forces must be zero:
ΣFy = N1 - W = 0
This equation tells us that the normal force on the hand (N1) is equal to the weight force (W):
N1 = W = 580 N
Similarly, the normal force exerted by the floor on the other hand (N2) will also be equal to the weight force:
N2 = W = 580 N
Now, let's consider the forces acting on one foot:
1. Weight force (W = 580 N): This force acts vertically downwards from the center of mass of the person.
2. Normal force (N3) exerted by the floor on the foot: This force acts vertically upwards and balances the weight force.
Similarly to the hand, the sum of the vertical forces acting on the foot must be zero:
ΣFy = N3 - W = 0
From this equation, we can deduce that the normal force on the foot (N3) is equal to the weight force (W):
N3 = W = 580 N
And finally, the normal force exerted by the floor on the other foot (N4) will also be equal to the weight force:
N4 = W = 580 N
To summarize:
The normal force exerted by the floor on each hand is 580 N.
The normal force exerted by the floor on each foot is also 580 N.
Note: In this analysis, we have assumed that the person's arms and legs are fully extended and that there are no horizontal forces (since the person is not moving horizontally).