Force Exerted on Person in an Elevator

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SUMMARY

This discussion focuses on calculating the force exerted on a person standing on a scale in an elevator under various conditions of motion. The key formula used is F = m(9.8 - a), where 'm' is the mass of the person (60 kg) and 'a' is the acceleration of the elevator. The scenarios analyzed include moving up or down at constant speeds, accelerating, and decelerating. The participants clarify that when moving upwards, the person feels heavier, while moving downwards makes them feel lighter, which directly affects the scale reading.

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  • Understanding of Newton's second law of motion (F = ma)
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Familiarity with concepts of acceleration and deceleration
  • Ability to interpret forces and vectors in physics
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  • Study the effects of acceleration on weight perception in different scenarios
  • Explore the concept of normal force in various motion contexts
  • Learn about free body diagrams to visualize forces acting on objects
  • Investigate the implications of constant versus variable acceleration in physics
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Homework Statement



A scale is placed in a normal elevator on earth. A person with a mass of 60kg steps onto the scale. What force would be exerted on the person by the scale during the following occurances?
-Moves up at constant speed
-Slows at 2.0m/s/s while moving upwards
-Speeds up at 2.0m/s/s while moving downward
-Moves downward at a constant speed
-Slows to a stop while moving downward with constant acceleration

Homework Equations


F=ma


The Attempt at a Solution


I'm somewhat stuck. I would like to answer the problem myself, but guidance towards the correct methods would be greatly appreciated. I am confused on how to decide which forces and vectors should be used. I also am confused on how to calculate the normal forces. Any help is appreciated.
 
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First, you should find the weight of the person. Then simply use the formula you posted to go through each scenario. In terms of forces, you only have the person and the scale, as far as I can tell you just have the weight and normal force, where Normal force = - Weight

Here's a few hints:

Remember that constant speed means a = 0.
When you're in an elevator and going UP, do you feel heavier/lighter?
how about when you're in an elevator and going DOWN?

Hope this helped.
 
Thanks for the Normal weight formula, can't believe I overlooked that. Up in an elevator means that you feel heavier and down means you feel lighter. So would one go about calculating the actual force that the scale reads?
 
TexasCow said:
Thanks for the Normal weight formula, can't believe I overlooked that. Up in an elevator means that you feel heavier and down means you feel lighter. So would one go about calculating the actual force that the scale reads?

Yes, you're right. So now look at your scenarios. In each scenario, something changes with your acceleration, nothing else. Your mass is constant, so the only variable here is the acceleration. In each case, just think of the direction of the acceleration and which way you're moving.

Two important things:

First, the equation is F = m a_t, where a_t is the total acceleration.

Second, you have to realize what the signs and directions mean. What I mean is, gravity is 9.8 m/s^2 DOWN, in your second case, your slowing down so you have - 2.0 m/s^2 UP. Notice how the sign changes. The sign means ACCELERATING or DECELERATING.

So you would have:

F = ma

F = (60)[9.8 - (- 2.0 ) ]

F = 708 N

Just to check if this makes sense, look at your weight while the elevator isn't moving. Like you said before, when you move up, you should be HEAVIER. So you want the new weight to be greater than the original weight, which it is. So you can check yourself that way.
 
Wow, thanks. That makes MUCH more sense. When the elevator is moving at a constant speed, is the Force the same as if they were not moving at all since the acceleration is zero?

F=ma
F=(60)[9.8-(0)]
F=588N
 
TexasCow said:
Wow, thanks. That makes MUCH more sense. When the elevator is moving at a constant speed, is the Force the same as if they were not moving at all since the acceleration is zero?

F=ma
F=(60)[9.8-(0)]
F=588N

Yup, exactly. You got it now.
 
Ok. While I do understand the two equations that I/you have done above, I am still somewhat confused on the others. Would the "Moves up at constant speed" and "Moves downward at a constant speed" have the same values? That's the way that it would seem if acceleration is the only variable. For the "Speeds up at 2.0m/s/s while moving downward", what would the sign be? I don't understand how you determind the sign of the change in acceleration.
 
TexasCow said:
... Would the "Moves up at constant speed" and "Moves downward at a constant speed" have the same values? That's the way that it would seem if acceleration is the only variable.

You're absolutely right.

TexasCow said:
For the "Speeds up at 2.0m/s/s while moving downward", what would the sign be? I don't understand how you determind the sign of the change in acceleration.

Here's the thing. Acceleration is how fast the speed changes. Speeding up is like accelerating positively at 2.0 m/s^2, while slowing down is like accelerating negatively 2.0 m/s^2. In other words, speeding up means + sign, slowing down means - sign.

If the problem was that it speeds up at 2.0 m/s^2 while moving up, the equation would be:

F = (60)[9.8 - ( 2.0 ) ]

F = 468N

which makes sense.

I think you might be confused about the equation. The equation is really:

F = m(9.8 - a)

where a is your given acceleration. Since that's the case, you just need to figure out if it's accelerating or decelerating.
 
Thanks, that cleared everything up as well. For the "It slows to a stop while moving downward with a constant acceleration, do you think it's implying that you are reading your weight while it is stopped? If not I don't see how it can be answered since an acceleration value isn't given.
 
  • #10
TexasCow said:
Thanks, that cleared everything up as well. For the "It slows to a stop while moving downward with a constant acceleration, do you think it's implying that you are reading your weight while it is stopped? If not I don't see how it can be answered since an acceleration value isn't given.

I'm a bit confused about that one. If the acceleration is constant, it means there IS a force, simply because there's acceleration. So if there's a force that you need to calculate, then you must be given an acceleration or some other information. Sorry but it seems to me that either your not given enough information and then this question has no answer, or I'm missing something.
 
  • #11
Thanks. Yeah that question is worded strangely and may be a mistype in the text. Thanks for your help!
 

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