Apparent weight loss 2nd Newtons Law elevator problem

In summary, a 65-kg woman is descending in an elevator that accelerates at 0.20g downward. During this acceleration, her weight is still 640N, but the scale only needs to exert a force of 0.80mg and will display her weight as 52kg. This is because the scale reads in kg and interprets force as mass. The acceleration allows for a net downward force and the scale only needs to balance out this force.
  • #1
Frankenstein19
56
0

Homework Statement


A 65-kg woman descends in an elevator that briefly acclerates at 0.20g downward when leaving a floor. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight and what does the scale read?

Homework Equations


ƩF = ma

The Attempt at a Solution


I understand that mg - Fn = m(0.20g) happens and then solving for Fn = 0.80mg.

I also understand that F'n is -0.80mg.

What I don't understand is why the "scale only needs to exert a force of only 0.80mg and that it will only give a reading of 0.80m"; why did the g get eliminated?
 
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  • #2
Because you want the net force to be downward so it can actually accelerate downward and n is just the force that the scale will push you up with. It also equal to the apparent weight because the springs in the scale gets compressed for example. Draw a FBD and try to figure what I am saying
 
  • #3
I still don't understand. My book says the F'n=0.80mg downward. Her weight is still mg=(65kg)(9.8m/s^2)=640N. But the scale, needing to exert a force of only 0.80mg and that it will only give a reading of 0.80m=52kg". I don't understand why it's only giving a reading of 0.80m, instead of 0.80mg. WHy was g dropped?
 
  • #4
Biker said:
Because you want the net force to be downward so it can actually accelerate downward and n is just the force that the scale will push you up with. It also equal to the apparent weight because the springs in the scale gets compressed for example. Draw a FBD and try to figure what I am saying

I still don't understand. My book says the F'n=0.80mg downward. Her weight is still mg=(65kg)(9.8m/s^2)=640N. But the scale, needing to exert a force of only 0.80mg and that it will only give a reading of 0.80m=52kg". I don't understand why it's only giving a reading of 0.80m, instead of 0.80mg. WHy was g dropped?
 
  • #5
Frankenstein19 said:
I still don't understand. My book says the F'n=0.80mg downward. Her weight is still mg=(65kg)(9.8m/s^2)=640N. But the scale, needing to exert a force of only 0.80mg and that it will only give a reading of 0.80m=52kg". I don't understand why it's only giving a reading of 0.80m, instead of 0.80mg. WHy was g dropped?
The scale reads in kg, not Newtons. But the spring inside it is sensitive to force, not mass. So the scale interprets a force of mg as a mass m, and displays it as such.
 

1. What is Apparent Weight Loss in the context of 2nd Newton's Law elevator problem?

Apparent weight loss refers to the change in weight that a person experiences when riding in an elevator that is accelerating or decelerating. This change in weight is due to the application of Newton's Second Law of Motion, which states that the net force on an object is equal to its mass multiplied by its acceleration.

2. How does Newton's Second Law apply to elevator problems?

In the context of elevators, Newton's Second Law explains that when the elevator is accelerating or decelerating, there is a net force acting on the person inside. This force is equal to the person's mass multiplied by the acceleration of the elevator. This results in a change in the person's apparent weight.

3. What factors affect the apparent weight loss in an elevator?

The apparent weight loss in an elevator is affected by the acceleration of the elevator and the mass of the person. The greater the acceleration, the greater the apparent weight loss. Similarly, the heavier the person, the greater the apparent weight loss.

4. Why do we experience apparent weight loss in an elevator?

We experience apparent weight loss in an elevator because of the normal force exerted by the elevator floor on our feet. When the elevator is accelerating or decelerating, the normal force changes to counteract the person's weight and maintain equilibrium. This results in a change in the person's apparent weight.

5. How can I calculate the apparent weight loss in an elevator?

To calculate the apparent weight loss in an elevator, you can use the formula F = m(a + g), where F is the net force, m is the mass of the person, a is the acceleration of the elevator, and g is the acceleration due to gravity (9.8 m/s^2). The apparent weight loss is equal to the difference between the person's actual weight and the net force calculated using this formula.

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