A person whose weight is W=584 N doing push-ups. Find the normal force exerted by the floor on each hand each foot, assuming that the persons holds this person
|=central gravity
foot arm
-------|--------
.840m .410m
Well, if the person is holding themselves up while doing push-ups, they're quite the strength enthusiast! Let's calculate the normal force on each hand and foot.
To find the normal force exerted by the floor, we'll need to consider the weight of the person and the distances between their central gravity and each hand/foot.
Let's start with the normal force on each hand. The person's weight is given as 584 N. The distance from the central gravity to each hand is 0.410 m.
We can calculate the normal force using the equation:
Normal force = weight - force due to leverage
The force due to leverage is the weight multiplied by the ratio of the distance between the center of gravity and the hand to the total distance (in this case, the sum of the distances to both hands). So, for each hand:
Force due to leverage = weight * (0.410 m / (2 * (0.410 m + 0.840 m)))
Plugging in the values:
Force due to leverage = 584 N * (0.410 m / (2 * (0.410 m + 0.840 m)))
Calculating this should give us the force due to leverage on each hand. Let's do the math:
Force due to leverage ≈ 97.33 N
Now, let's find the normal force on each hand:
Normal force on each hand = 584 N - 97.33 N
Normal force on each hand ≈ 486.67 N
So, the normal force exerted by the floor on each hand during the push-up is approximately 486.67 N.
Now, we can repeat the same process to find the normal force on each foot, considering the given distances. I'll let you give it a try! Keep those puns coming if you need any further assistance.
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 and apply the principles of static equilibrium. The normal force is the force exerted by a surface to support the weight of an object resting on it.
First, we need to determine the distribution of the person's weight based on their position. Let's assume that the person's weight is evenly distributed between their arms and legs.
The total weight of the person is given as W = 584 N.
The person's position is as follows:
Foot-arm distance (d1) = 0.840 m
Arm-arm distance (d2) = 0.410 m
Using these distances, we can calculate the weight distribution as follows:
The weight distribution on each foot (Wf) is given by:
Wf = (d2 / (d1 + d2)) * W
Wf = (0.410 m / (0.840 m + 0.410 m)) * 584 N
Wf = 0.410 * 584 N / 1.250 m
Wf = 191.872 N
The weight distribution on each hand (Wh) is given by:
Wh = (d1 / (d1 + d2)) * W
Wh = (0.840 m / (0.840 m + 0.410 m)) * 584 N
Wh = 0.840 * 584 N / 1.250 m
Wh = 392.448 N
Therefore, the normal force exerted by the floor on each hand is approximately 392.448 N and on each foot is approximately 191.872 N.
To find the normal force exerted by the floor on each hand and foot, we need to consider the forces acting on the person during the push-up exercise.
In this case, the person's weight is acting downwards and can be calculated using the formula:
Weight (W) = mass (m) * acceleration due to gravity (g)
Since weight is given as 584 N, we can calculate the mass of the person:
584 N = m * 9.8 m/s^2
m = 584 N / 9.8 m/s^2 ≈ 59.6 kg
Now, let's analyze the forces acting on the person during the push-up exercise.
1. Normal Force on each foot:
The normal force exerted by the floor on each foot counteracts the weight of the person. Since the person's weight is evenly distributed between both feet, the normal force will be divided equally between them.
Therefore, the normal force on each foot would be:
Normal Force on each foot = (Weight of the person) / 2
Normal Force on each foot = 584 N / 2 = 292 N
2. Normal Force on each hand:
The normal force exerted by the floor on each hand counteracts the weight of the person. To calculate the normal force on each hand, we need to consider the moments (torques) created by the person's weight and the distances of their center of gravity from each hand.
The total torque is equal to zero since the person is in equilibrium during the push-up, meaning the clockwise and anticlockwise moments balance each other. The torque exerted by the person's weight on either hand is:
Clockwise Torque = (Weight of the person) * (Distance from the center of gravity to each hand)
Anticlockwise Torque = (Normal Force on each hand) * (Distance from the center of gravity to each hand)
Setting the clockwise torque equal to the anticlockwise torque, we can solve for the normal force on each hand.
(W of the person) * (Distance from the center of gravity to each hand) = (Normal Force on each hand) * (Distance from the center of gravity to each hand)
Normal Force on each hand = (W of the person) * (Distance from the center of gravity to each hand) / (Distance from the center of gravity to each hand)
Normal Force on each hand = W of the person
Normal Force on each hand = 584 N
Therefore, the normal force exerted by the floor on each hand is 584 N.
To summarize:
Normal Force on each foot = 292 N
Normal Force on each hand = 584 N
584/2 = 292
x on a foot, 1.25-x on a hand
(x)(.84) = (1.25-x)(.41)