# Man on an Elevator -- Force Diagrams

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.

Chestermiller
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)?

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.

Chestermiller
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 ????

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.

Lnewqban
Gold Member
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?

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.

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?

Chestermiller
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
Lnewqban
Gold Member
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).

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?
Well, let's see. We can change that equation to N= ma + mg.
ma =(61.2*11.8)= 722.2
mg= (61.2*9.8)= 599.8

So N=1322 N--am I understanding this correctly?

Chestermiller
Mentor
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).
This seems very confusing to me, especially the part about the man being accelerated upwards by the scale when the elevator is not moving.

Or would the force just be 600 + (61.2 * 2)= 722.4?

Chestermiller
Mentor
Or would the force just be 600 + (61.2 * 2)= 722.4?
That is correct

That is correct
Alright, thanks for the help.