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Inertia at Work?

by Bashyboy
Tags: inertia, work
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Bashyboy
#1
May3-12, 08:56 AM
P: 943
I am reading about dynamics, and I am looking at this example problem (it is attached as a file).

So, first I am imagining the elevator without the assisted downward acceleration. In this case, the acceleration due to gravity on the woman and the normal force provided by the scale cancel each other out. Now, as the elevator starts to accelerate downwards, the woman experiences her own inertia by the weightlessness, and the downward acceleration of the elevator can be thought to become the "new" or "pseudo" acceleration due to gravity. The scale reads her as being lighter because this "pseudo" acceleration due to gravity is weaker than the true acceleration due to gravity. Is this erroneous thinking? If not, could someone perhaps supply more elaboration--or clarity, if it not worded exactly eloquently--on this thought process?

Thank you
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Whovian
#2
May3-12, 10:44 AM
P: 642
No, in fact, gravity itself does not accelerate. This sounds ridiculous, but it's the basis of General Relativity.

Basically, an object in freefall is equivalent to an object under no acceleration, there is no way for an observer in an elevator-like box drifting through space to tell if they're in an actual elevator in freefall.

Um. You probably didn't feel like an explanation of General Relativity. So if that's too much for you, remember that scales measure the NORMAL force and draw a free-body diagram for all the forces, after knowing the woman's acceleration. That should clear some stuff up.
Bashyboy
#3
May4-12, 07:46 AM
P: 943
On a website, I am reading more about this elevator physics. In the first paragraph, the author states that, "If the elevator cable broke, you would feel weightless since both you and the elevator would be accelerating downward at the same rate." Is this because the elevator would not be able to provide the normal force to push upwards on someone in an elevator under these circumstances?

Doc Al
#4
May4-12, 07:59 AM
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Inertia at Work?

Quote Quote by Bashyboy View Post
Is this because the elevator would not be able to provide the normal force to push upwards on someone in an elevator under these circumstances?
Exactly.
Bashyboy
#5
May4-12, 08:01 AM
P: 943
Sorry for the stream of questions. So, the reason why a scale reads heavier than normal on a elevator accelerating upwards is because, since the elevator is accelerating, the scale must be accelerating, and I must be accelerating; but in order for an elevator to accelerate upwards, it has to exert a force greater than my weight, result in the scale having to exert a greater force than my weight and, therefore, reading a greater normal force. Is that right?
Whovian
#6
May4-12, 12:16 PM
P: 642
Quote Quote by Bashyboy
Sorry for the stream of questions. So, the reason why a scale reads heavier than normal on a elevator accelerating upwards is because, since the elevator is accelerating, the scale must be accelerating, and I must be accelerating; but in order for an elevator to accelerate upwards, it has to exert a force greater than my weight, result in the scale having to exert a greater force than my weight and, therefore, reading a greater normal force. Is that right?
Yep.

Quote Quote by Doc Al
Exactly.
Sorry, just feel like pointing this out. From a beginner's point of view, yes. But technically, the falling person truly is weightless. Sorry, love pointing out the finer errors in stuff.
Doc Al
#7
May4-12, 12:18 PM
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Quote Quote by Whovian View Post
But technically, the falling person truly is weightless.
Not in the Newtonian world, which is the context of the question.


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