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Flofloflo
Member
Joined: 07 Nov 2007
Posts: 120
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 Posted: 31 Oct 2008 04:27:42 pm Post subject: |
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Hello,
I had Physics today, and I we where working with some formulas, namely:
F = C * U and
F = M * A.
Basicly, if you put both in their 'units' so to speak, you get this:
[N] = [N/M] * [M]
[N] = ([Kg] * [M])/[Seconds^2)
Which basically means [N/M] = [Kg]/[Seconds]^2
Does that mean anything whatsoever; I mean, is there some way to explain that C (from F = C * U) can be expressed as Kg/S^2?? |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 31 Oct 2008 08:56:05 pm Post subject: |
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Hi,
First, what does "F=c*u" mean? I have never seen 'c' and 'u' used in any way to express force. For 'u' all I can come up with is it's either speed (which I don't think can be right), potential energy, or a generalized lagrangian coordinate (which I highly doubt  ), and for 'c' either a constant or capacitance and/or capacity (which I both doubt even more).
Now for your question (the part I do know, I think). I think what you're getting at is dimensional analysis, or using the units of known variables to figure out the dimensions of some other quantity. One thing to keep in mind is that even if you have something that can be expressed in units such as kg/s/s, it doesn't mean that it is calculated as mass divided by the square of the time.
Since I'm not exactly sure what the significance of F=c*u is I'll just explain dimensional analysis with another example (and if I'm on the wrong track just ignore me )
Take the gravitational attraction between two objects, of masses m1 and m2 (in kg) a distance of r meters away. The force between them (in Newtons) is [font="courier"]F=G*m1*m2/r^2 where G is some constant. The left side is in N, or kg-m/s/s. The units on the left and right sides must be equal (because there's no way 3 meters could equal something like 3 kg), so overall the units of G*m1*m2 / r^2 has to be kg-m/s/s. On the right side we have [font="courier"](something)-kg^2/m^2 and on the left we have [font="courier"]kg-m/s/s. One of the kg's cancel so we have [font="courier"](something)-kg/m^2 = m/s/s and 'solving' for (something) yields [font="courier"](something) = m^3/(kg-s^2) which sounds really wacky, but the gravitational constant G must have units of cubic meter per kg per second per second for Newton's law to be right.
Hope this helps :biggrin:
PS: Sorry if I confused you/am way off from what you meant. (It's kind of sad that a physics major's never seen something that's in an introductory physics course ) |
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thornahawk
μολών λαβέ
Active Member
Joined: 27 Mar 2005
Posts: 569
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| Posted: 31 Oct 2008 09:12:25 pm Post subject: |
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You got me at the "F == c*u" as well. Matter of fact, the only time I've seen the N mֿ¹ unit anywhere was when I was dealing with surface tension, and even then, it was (usually) too big to be convenient.
But simplethinker is right; a way of proving the equivalence of two seemingly different units is to decompose them in terms of the basic units.
So, going with your units example: N == Kg*(m sֿ²) == (N/m)*m , and canceling the meter unit in the numerators of the second equation gives Kg*sֿ² == N/m . That can also be rearranged to give N == Kg*m*sֿ² which is exactly how the newton unit is defined.
thornahawk |
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Flofloflo
Member
Joined: 07 Nov 2007
Posts: 120
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| Posted: 01 Nov 2008 09:26:22 am Post subject: |
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Aah! Sorry, you see F = C * U is the Dutch version of Hooke's law.
F=-kx
Apears to be the English version where, k can be expressed in Nm^-1, or, Kg/s^2
Y'all catch my drift?  |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 01 Nov 2008 03:06:03 pm Post subject: |
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Flofloflo wrote: Aah! Sorry, you see F = C * U is the Dutch version of Hooke's law.
F=-kx [post="128372"]<{POST_SNAPBACK}>[/post]
Ah! I had forgotten about Hooke's law...
To answer your initial question (the real one), N/m is the right unit for the constant for Hooke's law. The restoring force from a spring is proportional to the distance it is stretched, so the constant is in "newtons per meter" (or N/m) because it's 'c' Newtons per meter stretched. The definition of N/m in base units (kg, m, s) is kg/s/s, but I don't think that would have any intuitive physical meaning for you. |
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Flofloflo
Member
Joined: 07 Nov 2007
Posts: 120
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| Posted: 01 Nov 2008 04:32:28 pm Post subject: |
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| Ok, thanks! I think it's quite surprising that it's possible to end up with these really weird dimensions! But, are you saying they are truly meaningless or would it somehow be possible to actually explain their meaning? (Btw, if there's an explanation for them, you probably shouldn't bother trying to explain them, I don't think I speak English well enough to understand all scientific terms... ). |
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simplethinker
snjwffl
Active Member
Joined: 25 Jul 2006
Posts: 700
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| Posted: 01 Nov 2008 05:10:31 pm Post subject: |
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Flofloflo wrote: But, are you saying they are truly meaningless or would it somehow be possible to actually explain their meaning?[post="128380"]<{POST_SNAPBACK}>[/post]
They all do have a meaning, but depending on what they're being used for the different ways to write them may not make sense (or even be valid). Here are two examples:
1) Temperature is defined as the average kinetic energy of something, but we don't measure temperature in the units that are used for energy (like joules).
2) Energy is measured in joules (J) and 1J is defined as 1 N-m. Torque (you'll get to later in your class) is measured in N-m, but even though that's technically one joule we don't say "the torque is __ joules" since it has a different meaning and application. |
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