To strengthen his arm and chest muscles, an 83.0 kg athlete 2.00 m tall is doing a series of push-ups. His center of mass is 1.15 m from the bottom of his feet, and the centers of his palms are 30.0 cm from the top of his head. Find the force that the floor exerts on each of his feet and on each hand, assuming that both feet exert the same force and both palms do likewise.
Draw the figure.
sum moments about thefeet.
83g*1.15=2*palm(2-.3)
That gives you each palm force. Now, 2*Palm+ 2*feet= 83g
solve for the force on each foot.
2.7
To find the force that the floor exerts on each of the athlete's feet and on each hand, we need to consider the forces acting on the athlete.
1. First, let's find the weight of the athlete. The weight can be calculated using the formula W = mg, where m is the mass and g is the acceleration due to gravity. Given that the mass of the athlete is 83.0 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight:
W = 83.0 kg * 9.8 m/s^2 = 813.4 N
2. Since the athlete is doing push-ups, he is in equilibrium and experiences no vertical acceleration. Therefore, the sum of the vertical forces acting on him is zero.
3. The vertical forces acting on the athlete include the force exerted by the floor on his feet and the force exerted by the floor on his hands. Since the athlete's center of mass is 1.15 m from the bottom of his feet, the distance between the center of mass and the floor is 1.15 m. Similarly, the distance between the center of mass and the top of his head (where his palms are) is 30.0 cm or 0.30 m.
4. According to the principle of moments, the sum of the clockwise moments is equal to the sum of the counterclockwise moments. Considering the moments about the athlete's center of mass, we can express this as:
(Wf * df) + (Wh * dh) = W * dc
where Wf is the force exerted by the floor on each foot, Wh is the force exerted by the floor on each hand, df is the distance between the center of foot contact and the center of mass (1.15 m), dh is the distance between the center of hand contact and the center of mass (0.30 m), and dc is the distance between the center of mass and the floor.
5. Since both feet exert the same force and both hands do likewise, we can simplify the equation to:
(2 * Wf * 1.15 m) + (2 * Wh * 0.30 m) = 813.4 N * (1.15 m + 0.30 m)
6. Solving for (2 * Wf) + (2 * Wh), we get:
(2 * Wf) + (2 * Wh) = 813.4 N * (1.15 m + 0.30 m) / (1.15 m * 1.15 m + 0.30 m * 0.30 m)
7. Simplifying further, we get:
(2 * Wf) + (2 * Wh) = 813.4 N * 1.15 m / (1.3525 m^2 + 0.09 m^2)
(2 * Wf) + (2 * Wh) = 936.71 N
8. Dividing both sides by 2, we find:
Wf + Wh = 936.71 N / 2
Wf + Wh = 468.36 N
9. Finally, since both feet and both hands exert the same force, we can divide the total force by 2 to find the force exerted by each foot and each hand:
Wf = Wh = 468.36 N / 2
Wf = Wh = 234.18 N
Therefore, the force that the floor exerts on each of the athlete's feet and on each hand is approximately 234.18 N.
To find the force that the floor exerts on each of the athlete's feet and on each hand, we can use the principles of equilibrium. In this case, we will use the fact that the sum of the forces in the vertical direction should equal zero.
Let's start by finding the total weight of the athlete. The weight is equal to the mass multiplied by the acceleration due to gravity:
Weight = mass × acceleration due to gravity
Weight = 83.0 kg × 9.8 m/s^2
Weight = 813.4 N
Next, let's consider the forces acting on the athlete. When doing push-ups, the athlete's hands and feet are the only points of contact with the floor. The force exerted by the floor on each foot and on each hand is equal to the force required to support the athlete's weight.
Since both feet exert the same force and both palms do likewise, let's denote the force exerted on each foot as Ff and the force exerted on each hand as Fh. Therefore, we have the following equations:
Ff + Ff + Fh + Fh = Weight
2Ff + 2Fh = 813.4 N
Now, let's consider the distances from the athlete's center of mass to each point of contact. The distance from the bottom of his feet to his center of mass is 1.15 m, and the distance from the top of his head to the center of his palms is 30.0 cm or 0.30 m.
Next, we can use the concept of torque to relate the forces and distances. Torque is equal to the force multiplied by the perpendicular distance from the point of rotation.
The torque due to the force exerted by each foot is equal to Ff multiplied by the distance from the center of mass to each foot (1.15 m). Similarly, the torque due to the force exerted by each hand is equal to Fh multiplied by the distance from the center of mass to each hand (0.30 m). These two torques should sum up to zero in equilibrium:
Torque by feet + Torque by hands = 0
Ff × 1.15 m + Ff × 1.15 m - Fh × 0.30 m - Fh × 0.30 m = 0
2.3 Ff - 0.6 Fh = 0
Now we have a system of two equations with two unknowns:
2Ff + 2Fh = 813.4 N (Equation 1)
2.3 Ff - 0.6 Fh = 0 (Equation 2)
By solving this system of equations, we can find the values of Ff and Fh, which represent the forces exerted by each foot and each hand, respectively.
Using any method of solving systems of equations, the solution will give us:
Ff ≈ 369.2 N
Fh ≈ 222.3 N
Therefore, the floor exerts a force of approximately 369.2 N on each foot and approximately 222.3 N on each hand.