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Newton's laws of motion...

by Nervous
Tags: laws, motion, newton
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Nervous
#1
Mar28-12, 03:11 PM
P: 18
Newton's third law says that every force X exerts onto Y, Y exerts and equal and opposite force onto X. Ergo, if I'm standing on my linoleum floor, I'm exerting a force equal to my weight (mass x force of gravity) onto the floor and it's exerting an equal and opposite force on to me.

However, Newton also told us F=mA and that would mean that the force I am exerting on the floor, my weight, is equal to 0. (Assuming my downward acceleration is zero.)

Though, as I mentioned, I am exerting my weight, w=mG and F=mA; Ergo if w=F then mG=mA and G=A.

Can gravity really just be thought of as acceleration?
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Doc Al
#2
Mar28-12, 03:19 PM
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Quote Quote by Nervous View Post
However, Newton also told us F=mA and that would mean that the force I am exerting on the floor, my weight, is equal to 0. (Assuming my downward acceleration is zero.)
No. Since your acceleration is zero, Newton's 2nd law just tells you that the net force on you must be zero. There are two forces acting on you: The earth pulling you down and the floor pushing you up. They sum to zero.
Nervous
#3
Mar28-12, 03:39 PM
P: 18
I thought my weight was cancelled out by the force exerted on the floor, but not non-existent.;
w= mG & F=mA ->
w+F=0
w=-F
mG=-(mA)
G=-A

And since negative acceleration is simply called acceleration...
G=A


I must have phrased part of my question wrong, but what I'm concerned with is the ultimate implication I just showed in these two posts, can gravity be thought of as acceleration?

Doc Al
#4
Mar28-12, 04:05 PM
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Newton's laws of motion...

Quote Quote by Nervous View Post
I thought my weight was cancelled out by the force exerted on the floor, but not non-existent.;
w= mG & F=mA ->
w+F=0
w=-F
mG=-(mA)
G=-A

And since negative acceleration is simply called acceleration...
G=A
I don't quite understand what you're doing here. Try this:
ƩF = ma
The forces on you are your weight (acting down) and the normal force from the floor (acting up). They are equal and opposite:
-mg + N = 0 (since you are not accelerating)

If you were in free fall, with the only force acting on you being gravity, then you would have a downward acceleration equal to g = 9.8 m/s^2.

I must have phrased part of my question wrong, but what I'm concerned with is the ultimate implication I just showed in these two posts, can gravity be thought of as acceleration?
Not sure what you mean.
OldEngr63
#5
Mar28-12, 04:21 PM
P: 343
I think a part of the problem comes from wanting to apply the second law, which really only applies to dynamics, in a statics problem where the acceleration is zero.

The other part of the difficult is the failure to write the force SUMMATION, writing instead simply F. It is extremely important to recognize that it is the sum of all of all forces that is equal to m*a, not any one force.

With a few simple oversights, it is easy to get confused.
Nessdude14
#6
Mar28-12, 04:27 PM
P: 172
Quote Quote by Nervous View Post
I thought my weight was cancelled out by the force exerted on the floor, but not non-existent.;
w= mG & F=mA ->
w+F=0
w=-F
mG=-(mA)
G=-A

And since negative acceleration is simply called acceleration...
G=A
Newton's second law is only applicable to the net force on an object, which is the total of all of the forces on the object. The variable F in your equations above is in fact the normal force (the force from the ground pushing against gravity). Newton's second law cannot be applied to this force which you labeled F, since it is not the net force.

If you're standing stationary on the ground, your net force is the vector sum of your weight and the normal force (which equals 0). This net force is the only force that Newton's second law applies to.
Nessdude14
#7
Mar28-12, 04:29 PM
P: 172
Quote Quote by OldEngr63 View Post
I think a part of the problem comes from wanting to apply the second law, which really only applies to dynamics, in a statics problem where the acceleration is zero.
The other part of the difficult is the failure to write the force SUMMATION, writing instead simply F. It is extremely important to recognize that it is the sum of all of all forces that is equal to m*a, not any one force.
With a few simple oversights, it is easy to get confused.
Newton's second law is just as applicable to a statics problem as in dynamics. The problem is that Newton's second law is no good for solving most statics problems, because you'll just get an answer of 0 acceleration.
Nervous
#8
Mar28-12, 04:30 PM
P: 18
OK, thanks for that.
I knew I was too dumb to be onto something. :P
OldEngr63
#9
Mar28-12, 05:05 PM
P: 343
Quote Quote by Nessdude14 View Post
Newton's second law is just as applicable to a statics problem as in dynamics. The problem is that Newton's second law is no good for solving a statics problem, because you'll just get an answer of 0 acceleration.
Wow! Did you read what I wrote?
Nessdude14
#10
Mar28-12, 05:33 PM
P: 172
Quote Quote by OldEngr63 View Post
Wow! Did you read what I wrote?
Of course; you said this:
Quote Quote by OldEngr63 View Post
the second law, which really only applies to dynamics
And that statement is wrong, because Newton's second law does not only apply to dynamics. Newton's second law holds true in all statics problems as well as dynamics. In fact, there are statics problems where use of Newton's second law is required to come up with an answer.
OldEngr63
#11
Mar28-12, 07:21 PM
P: 343
Split many hairs? If you really want to get picky, we could say that statics is only a subset of dynamics, and that the first law is only a special case of the second law.
Nessdude14
#12
Mar28-12, 09:44 PM
P: 172
Quote Quote by OldEngr63 View Post
Split many hairs? If you really want to get picky, we could say that statics is only a subset of dynamics, and that the first law is only a special case of the second law.
You're the one who made the distinction between statics and dynamics when you said that Newton's second law applied to one of the cases and not the other. I was just correcting your statement.
K^2
#13
Mar28-12, 09:55 PM
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Quote Quote by OldEngr63 View Post
Split many hairs? If you really want to get picky, we could say that statics is only a subset of dynamics, and that the first law is only a special case of the second law.
Statics is a special case of dynamics. Any rule that applies to dynamics applies to statics. E.g., Newton's Laws. But statics problems also have some of their own rules that don't apply to dynamic problems in general.
AlephZero
#14
Mar29-12, 08:39 AM
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Quote Quote by K^2 View Post
But statics problems also have some of their own rules that don't apply to dynamic problems in general.
Really? Whether the velocity of something is zero or nonzero depends what coordinate system you measure it in. Are you saying the laws of physics are different in different coordinate systems?
shadi_s10
#15
May10-12, 06:20 AM
P: 82
Dear All,
I have a question about newton's third law.
As you know , by pushing your hand through the wall the force F is done on the wall. But a small amount of this force is converted to heat.
So let's say the force which the wall is experiencing is F-e.
Does the wall exert F on your hand or F-e?
Doc Al
#16
May10-12, 06:28 AM
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Quote Quote by shadi_s10 View Post
But a small amount of this force is converted to heat.
Force doesn't 'convert' to heat.

Whatever force you end up exerting on the wall will be equal and opposite to the force that the wall exerts on you.
DrewD
#17
May10-12, 08:17 AM
P: 459
Can gravity really just be thought of as acceleration?
I knew I was too dumb to be onto something. :P
You are onto something. It is something very important, but your "proof" is wrong. Looking into the laws of Newtonian Mechanics will probably not help you think about it.

So, you were wrong, but there are profound relationships between gravity and acceleration.
shadi_s10
#18
May10-12, 08:19 AM
P: 82
Quote Quote by Doc Al View Post
Force doesn't 'convert' to heat.

Whatever force you end up exerting on the wall will be equal and opposite to the force that the wall exerts on you.
But when you touch the wall there is heat because of friction force.


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