Difficult problems, im frustrated

[chimez14]

Difficult problems, I am frustrated!

1.)A chair of mass 10.1 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force of 47.7 N that is directed at an angle of 39.6 degrees below the horizontal and the chair slides along the floor. Calculate the normal force that the floor exerts on the chair

2.)A 87.7 kg man steps off a platform 2.89 m above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.60 m before coming to rest. Treating our rigid legged friend as a particle, what is the average force his feet exert on the ground while he slows down?

Note: Assume the acceleration while he is slowing down is constant
Please also remember to try and explain as much as possible. Thanks!

 

[Nate810]

1. The normal force exerted by the floor on the chair is equal to the weight of the chair plus the component of the pushing force that is perpendicular to the floor. The weight of the chair is 10.1 kg * 9.81 m/s2 = 99.291 N. The component of the pushing force perpendicular to the floor can be calculated using trigonometry. The angle from the horizontal to the pushing force is 39.6°, so the angle from the horizontal to the perpendicular component is 90° - 39.6° = 50.4°. The magnitude of the perpendicular component is 47.7 N * cos(50.4°) = 41.1 N. Therefore, the normal force exerted on the chair is 99.291 N + 41.1 N = 140.391 N.2. The vertical acceleration of the man as he slows down is the change in velocity divided by the time it takes him to slow down. The change in velocity is the difference between his initial velocity at the moment his feet touch the ground and his final velocity when he comes to rest. His initial velocity is the velocity he had at the moment his feet touched the ground, which is the same as the velocity he had when he was 2.89 m above the ground, so his initial velocity is zero. His final velocity is also zero, since he comes to rest. Therefore, the change in velocity is zero. Since the change in velocity is zero, the vertical acceleration is also zero. The average force exerted on the ground is equal to the mass of the man multiplied by the vertical acceleration, so the average force is zero.

 

[Moetasim]

1.) The normal force is the force exerted by the floor on the chair in the direction perpendicular to the surface of the floor. To calculate the normal force, we can use Newton's Second Law, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the acceleration is in the direction of the force applied, which is at an angle of 39.6 degrees below the horizontal. We can break this force into its components, one in the horizontal direction and one in the vertical direction.

The horizontal component of the force is equal to Fcos(39.6) = 47.7 N * cos(39.6) = 38.4 N. This horizontal force does not contribute to the normal force, as it is parallel to the floor. The vertical component of the force is equal to Fsin(39.6) = 47.7 N * sin(39.6) = 30.7 N. This vertical force is responsible for the normal force, as it is perpendicular to the floor.

Therefore, the normal force exerted by the floor on the chair is equal to the vertical component of the force, which is 30.7 N.

2.) To calculate the average force exerted by the man's feet on the ground while he slows down, we can use the same equation as above, Newton's Second Law. However, in this case, we need to find the acceleration of the man as he slows down.

Since we are treating the man as a particle, we can use the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (unknown), a is the acceleration, and s is the distance traveled (0.60 m).

We can rearrange this equation to solve for the acceleration: a = (v^2 - u^2) / 2s = (0^2 - u^2) / 2(0.60) = -u^2 / 1.2

Now, we can use Newton's Second Law, F = ma, to find the average force exerted by the man's feet on the ground: F = ma = (87.7 kg)(-u^2 / 1.2) = -73.1u^2

Therefore, the average force