to strengthen his arm and chest muscles, an 82-kg athlete who is 2.0 m tall is doing push ups. his center of mass is 1.15 m from the bottom of his feet, and the centers of his palms are 30.0cm from the top of his head. find the force that the floor exerts on each of is feet and on each hand, assuming that both feet exert the same force and both palms do likewise.

For the force F on both his palms, which are 1.70 m from his toes, set the total moment about his toes equal to zero.

F *1.70 = M*g*1.15
F = 544 N
Divide F by 2 for the force on each palm.

For the force F' on both feet (together), do a vertical force balance.
F' + F = M*g = 804 N

F' = 260 N
Divide F' by 2 for the force on each foot.

Much detailed solution:

F *1.70 = M*g*1.15
F*1.70 = 82 * 9.8 * 1.15
F= 924.14/1.70
F = 544 N
Divide F by 2 for the force on each palm.
F= 544/2 = 272
For the force F' on both feet (together), do a vertical force balance.
F' + F = M*g = N
F' + F = 82 * 9.8 = 804 N
F' = 804 - 544 = 260 N
F' = 260 N
Divide F' by 2 for the force on each foot.
F' = 260/2 = 130

Well, it seems like our athlete is quite the push-up enthusiast! Let's see if we can crunch some numbers and find out what forces are at play here.

First, let's find the athlete's center of mass. Since he is 2.0 m tall and his center of mass is 1.15 m from the bottom of his feet, we subtract 1.15 m from his total height to find that his center of mass is located at a height of 0.85 m.

Now, we need to find the force that the floor exerts on his feet. In order to do this, we need to consider the athlete's weight and the location of his center of mass. Since both feet exert the same force, we can divide the athlete's weight evenly between them.

The force that the floor exerts on each foot is equal to the weight divided by the number of contact points, which in this case is two: F_foot = (weight/2).

So, the force that the floor exerts on each foot is (82 kg * 9.8 m/s^2)/2 = 401.8 N (approximately).

Moving on to the force on each hand, we need to consider the distance between the center of mass and the palms, which is given as 30.0 cm (or 0.3 m). This distance acts as a lever arm, creating a torque.

Since the athlete is doing push-ups, the palms are resisting downward rotation. In order to counteract this torque and maintain equilibrium, the floor needs to exert an upward force that balances it out.

The torque created by the weight is equal to the force times the lever arm, so we can write the equation as:

(weight * lever arm) = (force * distance from center of mass to hand).

Rearranging the formula, we can express the force on each hand as:

force = (weight * lever arm) / (distance from center of mass to hand).

Let's calculate it!

force = (82 kg * 0.3 m) / 0.85 m ≈ 28.94 N

So, the force that the floor exerts on each hand is approximately 28.94 N.

In summary, the force that the floor exerts on each foot is approximately 401.8 N, and the force that the floor exerts on each hand is approximately 28.94 N. Keep pushing it up, Mr. Athletic Humorist!

To find the force exerted by the floor on each of the athlete's feet and hands, we can use the concept of torque and equilibrium.

First, we need to calculate the torque exerted by the athlete's weight. Torque is given by the formula:

Torque = Force x Distance

For the athlete's feet, the distance is the height of his center of mass from the floor, which is 1.15 m. The torque exerted by his weight on each foot is:

TorqueFoot = Weight x DistanceFoot

Assuming that both feet exert the same force, the total torque exerted by his weight on both feet is:

TotalTorqueFoot = 2 x TorqueFoot

For the athlete's hands, the distance is the distance between the center of his palms and the height of his center of mass, which is (2.0 m - 1.15 m + 0.3 m) = 1.15 m. The torque exerted by his weight on each hand is:

TorqueHand = Weight x DistanceHand

Assuming that both hands exert the same force, the total torque exerted by his weight on both hands is:

TotalTorqueHand = 2 x TorqueHand

Next, we can assume that the forces exerted by the athlete's feet and hands are perpendicular to the floor. Therefore, the total torque exerted on the athlete's body must be balanced by the torque exerted by the floor.

Since the athlete is doing push-ups, his body is in equilibrium, so the total torque exerted by his body must be zero. Therefore:

TotalTorqueFoot + TotalTorqueHand = 0

Simplifying the equation, we get:

2 x TorqueFoot + 2 x TorqueHand = 0

Now we can rearrange the equation to solve for the force exerted on each foot and hand:

2 x (Weight x DistanceFoot) + 2 x (Weight x DistanceHand) = 0

2 x Weight x (DistanceFoot + DistanceHand) = 0

Weight = ForceFoot + ForceHand

Since both feet exert the same force and both hands exert the same force, we can rewrite the equation as:

2 x ForceFoot + 2 x ForceHand = Weight

Now we can substitute the values and calculate the force exerted by the floor on each foot and hand:

Weight = 82 kg x 9.8 m/s^2 (acceleration due to gravity)

ForceFoot = ForceHand = Weight / 4

Finally, we can substitute the value of weight and calculate the force:

ForceFoot = ForceHand = (82 kg x 9.8 m/s^2) / 4

ForceFoot = ForceHand = 202.4 N

Therefore, the force that the floor exerts on each of the athlete's feet and hands is approximately 202.4 Newtons.

To find the force that the floor exerts on each of the athlete's feet and on each hand, we can make use of Newton's laws of motion and the principles of torque.

Let's start by calculating the weight of the athlete, which is the force exerted on his entire body due to gravity. We can find this by multiplying his mass (82 kg) by the acceleration due to gravity (9.8 m/s^2).

Weight = mass x acceleration due to gravity
Weight = 82 kg x 9.8 m/s^2
Weight = 803.6 N

Next, we need to consider the distribution of this weight between the athlete's feet and hands. To do this, we need to determine the distances between the center of mass and the points of contact with the floor.

The distance between the center of mass and the feet (h1) is given as 1.15 m.

The distance between the center of mass and the palms (h2) is given as 30.0 cm, which is equivalent to 0.3 m.

Now, let's calculate the torque exerted by the athlete's weight on his feet and hands.

Torque = force x lever arm

For the feet:
Torque on feet = Weight x lever arm for feet
Torque on feet = 803.6 N x 1.15 m

For the hands:
Torque on hands = Weight x lever arm for hands
Torque on hands = 803.6 N x 0.3 m

Now, since the athlete is doing push-ups, we know that the torques exerted by his feet and hands are equal. Therefore, we can set these two equations equal to each other:

803.6 N x 1.15 m = 803.6 N x 0.3 m

Now we can solve for the force exerted by each foot and each hand.

803.6 N x 1.15 m = 803.6 N x 0.3 m
924.14 N = 241.08 N

Therefore, the force that the floor exerts on each of the athlete's feet is approximately 241.08 N, and the force that the floor exerts on each of his hands is also approximately 241.08 N.