The drawing shows a person (weight W = 580 N, L1 = 0.849 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.
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).
To find the normal force exerted by the floor on each hand and each foot, we need to consider the forces acting on the person during push-ups.
Let's break down the forces acting on the person:
1. Weight (W) = 580N: This force acts vertically downward, pulling the person towards the ground.
2. Normal Force by the floor on each hand (N_h): This force acts vertically upward, perpendicular to the floor surface, and provides support to each hand.
3. Normal Force by the floor on each foot (N_f): This force also acts vertically upward, perpendicular to the floor surface, and provides support to each foot.
Now, let's find the normal forces:
For the hands:
Since the person is in the push-up position, the hands are the primary points of contact with the floor. Therefore, the normal force exerted by the floor on each hand is equal to the weight of the person, which is 580N.
N_h = W = 580N
For the feet:
The weight of the person is distributed between the two feet. To find the normal force exerted by the floor on each foot, we need to consider the weight distribution.
First, we calculate the total weight supported by the feet. The weight is divided between the hands and feet using the principle of moments:
W * L1 = N_f * L2
Where:
W = weight of the person = 580N
L1 = distance from the feet to the center of mass = 0.849m
L2 = distance from the hands to the center of mass = 0.410m
N_f = normal force exerted by the floor on each foot (what we need to find)
Let's solve the equation:
580N * 0.849m = N_f * 0.410m
492.02N.m = 0.41N_f
N_f = 492.02N.m / 0.410m = 1199.8N
Therefore, the normal force exerted by the floor on each foot is approximately 1199.8N.