# Man on an Elevator -- Force Diagrams

Lugytopo
Homework Statement:
This person is riding on an elevator in a tall building. The reading on the bathroom scale when this elevator is stationary is 600 N. Draw a quantitative force diagram for each situation.

Then determine what the bathroom scale indicates for each of the three situations based on the data presented.
Relevant Equations:
F=M*A
I really need help figuring out where to start. I believe that every graph has a normal force pushing up on the person as well ag gravitational force pushing down on them. Where I am getting confused at is how the velocity plays into the scenario and if an acceleration of 9.8 m/s^2 downwards is just gravitational pull. Would that make the scale read "600 N" for the leftmost graph?

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• Phys. Force Diagram.png
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Welcome, Lugytopo! The velocity information is only to indicate direction of movement, if any.

Lugytopo
Welcome, Lugytopo! The velocity information is only to indicate direction of movement, if any.
Ok--so it has no bearing on the force diagrams or the scale reading?

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Lugytopo
What type of work would you like to see? Would you like me to fill the problem out with things I think may be correct?

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Yes, please, per PF rules, we need to see your attempt for a response.
Could you draw the balance of forces for each case?

Lugytopo
Yes, please, per PF rules, we need to see your attempt for a response.
Could you draw the balance of forces for each case?
I will try my best (although I really don't know what to do with the scale readings)

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I will try my best (although I really don't know what to do with the scale readings)
Homework? What level?

Lugytopo
Homework? What level?
I know how to draw force diagrams and such, it's just that the specific question format is confusing. I will show you what I come up with shortly.

Lugytopo
I know how to draw force diagrams and such, it's just that the specific question format is confusing. I will show you what I come up with shortly.
Also--this is High School Physics.

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how the velocity plays into the scenario
Some problem posers deliberately provide irrelevant information (a practice of which I approve).
If you seem to be able to solve without it, consider that maybe the case here.
and if an acceleration of 9.8 m/s^2 downwards is just gravitational pull.
It is the actual acceleration resulting from all the applied forces, including gravitational pull.
Would that make the scale read "600 N" for the leftmost graph?
Would what make it 600N? It is the rightmost graph that has an acceleration of 9.8 m/s^2 downwards.

In all cases, the relevant equation you quoted is the key; but remember that F is the sum of all applied forces, either as a vectorial sum or the sum of components in a given direction.
You also need to understand the relationship between the scale reading and the forces on the person. What is it?

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Thank you. When talking about forces, we should use Newtons.
Kg is only for the mass, which we never represent with a vector because is a scalar value.

For the first case, we are only concerned about the force among your feet and the scale.
You know that you feel more pressure on your feet each time an elevator that you ride in starts moving upwards.
Does that mean that the scale in our problem should indicate more or less than 600 N?

Lugytopo
Thank you. When talking about forces, we should use Newtons.
Kg is only for the mass, which we never represent with a vector because is a scalar value.

For the first case, we are only concerned about the force among your feet and the scale.
You know that you feel more pressure on your feet each time an elevator that you ride in starts moving upwards.
Does that mean that the scale in our problem should indicate more or less than 600 N?
For the first case, I guess it should be greater? So 600N*2m/s^2 which would equal 1200N? Is the force diagram correct for the first case?

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600N*2m/s^2 which would equal 1200N?
No, 600N*2m/s^2 gives 1200Nm/s2, whatever that might mean.
Multiplying a force by an acceleration is not going to give a force.

After drawing the diagram, the next step is to write the "sum of forces equals mass times resulting acceleration " equation.

Lugytopo
No, 600N*2m/s^2 gives 1200Nm/s2, whatever that might mean.
Multiplying a force by an acceleration is not going to give a force.
Right...Is the weight of the person 61.2 N or did I mess that up? I guess in that same lens is the force on the scale of the middle graph 600N?

Mentor
Ok--so it has no bearing on the force diagrams or the scale reading?
Correct

Mentor
For the first case, I guess it should be greater? So 600N*2m/s^2 which would equal 1200N? Is the force diagram correct for the first case?
Let's see your free body diagrams and force balances for the three cases.

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Right...Is the weight of the person 61.2 N
"The reading on the bathroom scale when this elevator is stationary is 600 N."
So what does the person actually weigh? What is the person's mass?

Edit: I see from your posted working you have correctly that the person's mass is 61.2kg.
So now write out the F=ma equation for the left hand picture. What forces act and what is the resulting acceleration?

Lugytopo
"The reading on the bathroom scale when this elevator is stationary is 600 N."
So what does the person actually weigh? What is the person's mass?
I guess the person actually weighs 600 N then?

Lugytopo
"The reading on the bathroom scale when this elevator is stationary is 600 N."
So what does the person actually weigh? What is the person's mass?
I guess the person actually weighs 600 N then?

Lugytopo
I guess the person actually weighs 600 N then?
Nevermind.

Lugytopo
"The reading on the bathroom scale when this elevator is stationary is 600 N."
So what does the person actually weigh? What is the person's mass?

Edit: I see from your posted working you have correctly that the person's mass is 61.2kg.
So now write out the F=ma equation for the left hand picture. What forces act and what is the resulting acceleration?
I know that there is the force of the Earth on the person and the elevator on the person with a net upwards acceleration of 2 m/s^2. If my force diagram is correct (which it may not be) the sum of the accelerations is 21.6. So would I take 21.6 * 61 kg to get 1321.9 N for the first graph?

Mentor
Sure, here is an updated version.
The diagram and equations (hen scratchings) are unreadable. Treat the man's body as a free body. Representing it as a rectangle shape would be adequate. Please show the forces acting on it only (There are only two).

Lugytopo
The diagram and equations (hen scratchings) are unreadable. Treat the man's body as a free body. Representing it as a rectangle shape would be adequate. Please show the forces acting on it only (There are only two).
There is a pre-drawn dot in the middle of the diagrams that represent the person. I also thought I showed the forces acting on the person--I only showed 2.

Mentor
There is a pre-drawn dot in the middle of the diagrams that represent the person. I also thought I showed the forces acting on the person--I only showed 2.
I can see that now. Thanks.

The same force balance equation applies algebraically to all three diagrams. What is that equation (please type it out or, preferably, use LaTex)?

Lugytopo
I can see that now. Thanks.

The same force balance equation applies algebraically to all three diagrams. What is that equation (please type it out or, preferably, use LaTex)?
I believe that the equation is F=M*A (I don't know how to use LaTeX yet). Do I have the force diagrams right? If the diagrams are not right I'm pretty sure that I will mess up everything.

Mentor
I believe that the equation is F=M*A (I don't know how to use LaTeX yet). Do I have the force diagrams right? If the diagrams are not right I'm pretty sure that I will mess up everything.
And, algebraically, in terms of the two forces, F is equal to ?

Lugytopo
And, algebraically, in terms of the two forces, F is equal to ?
I'm not 100% understanding what you are saying. The two forces are the elevator on the man and the Earth on the man. The mass is 61.2 Kg. A little bit more direction would be helpful.

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For the first case, I guess it should be greater?
Correct!
Before the elevator starts moving, there is a pair of vertical forces acting up and down among feet and scale.
Both have the same magnitude and opposite directions; therefore, it is like there is no force at all.
There is the mass of the man, but no net force and no resulting acceleration.

##a=F_{net}/m##

The elevator's door closes and the elevator pushes the scale up harder than before, which pushes the man up harder than before.
The scale "feels" that the man is now heavier while it is pushing him upwards in an accelerated manner.
If indicated weight, which is a force, is proportional only to the mass and to the acceleration of the man, which one of those has changed?

Lugytopo
Correct!
Before the elevator starts moving, there is a pair of vertical forces acting up and down among feet and scale.
Both have the same magnitude and opposite directions; therefore, it is like there is no force at all.
There is the mass of the man, but no net force and no resulting acceleration.

##a=F_{net}/m##

The elevator's door closes and the elevator pushes the scale up harder than before, which pushes the man up harder than before.
The scale "feels" that the man is now heavier while it is pushing him upwards in an accelerated manner.
If indicated weight, which is a force, is proportional only to the mass and to the acceleration of the man, which one of those has changed?
Only the acceleration of the man changed.

Lugytopo
Only the acceleration of the man changed.
Would the acceleration used in the F=M*A equation be a sum of the upward and downward accelerations?

Mentor
The force balance on the man is $$N-mg=ma$$where N is the upward force that the scale exerts on the man, mg is the downward force of the Earth on the man, and a is the upward acceleration. Does that make sense?

• Lnewqban
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Would the acceleration used in the F=M*A equation be a sum of the upward and downward accelerations?
The thing is that, when the elevator is not moving, the man is being accelerated upwards by the scale at exactly the same rate as the gravity is trying to accelerate him downwards (like in a free-fall), resulting in no net acceleration respect to Earth in any direction.
Action force of feet against scale equals reaction force of scale against feet and there is no change of state (repose or constant up or down velocity).

That balance is lost once the elevator starts pushing the scale, and the scale pushes the man upwards with a force greater than his normal weight (when in repose).