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Dimensional Analysis and constant

  • Thread starter ment4ll
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  • #1
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Homework Statement


The volume of an object is given as a function of time by
V= A + B/t + Ct^4

Find the dimension of the constant C.


Homework Equations


N/A


The Attempt at a Solution


I don't understand dimensional analysis completely .. I understand it on simpler terms, but in terms of this problem I do not.

V - A = B/t + Ct^4
V - A - B/t = Ct^4
(V - A - B/t)/t^4 = C


is that correct?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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Dimension analysis is just to find its units essentially.

When you are adding, the quantities must be the same. So if you have volume of on the left, A,B/t and Ct^4 all must have combined units of volume.

If that is the case, then what is the units of C?
 
  • #3
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also note that 2 quantities can only be added when their dimentions are equal.

what can you conclude from this?
 
  • #4
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Dimension analysis is just to find its units essentially.

When you are adding, the quantities must be the same. So if you have volume of on the left, A,B/t and Ct^4 all must have combined units of volume.

If that is the case, then what is the units of C?
I mean, yea I get that A, B/t and Ct^4 all are combined units of volume, they have to be. I don't understand how you get the units of C if it isn't the answer I posted.

Sorry if I'm noob at this, I've never taken a physics class in my life (besides this one).
 
  • #5
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also note that 2 quantities can only be added when their dimentions are equal.

what can you conclude from this?
So C has to be something measurable to volume right?
 
  • #7
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as A can be added to B, both have same dimensions... lets say [X]

now [X]/[T] = [C][T]4 = [V] = [M]3

can you find dim of C as well as A,B now?
 
  • #8
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as A can be added to B, both have same dimensions... lets say [X]

now [X]/[T] = [C][T]4 = [V] = [M]3

can you find dim of C as well as A,B now?
Sorry guy, I'm just not grasping it at this point .. wouldn't the dimensions of A+B = X then? I'm confused about where the M3 comes from ..
 
  • #9
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dimensions is just like witing units in form of 7 basic standard units
what is the unit of 5m/s + 3m/s ??

M is the dimensional symbol for mass

7 dimentions:

Length: [L]

Mass: [M]

Time: [T]

Electric current: [A]

Temperature: [K]

Luminous intensity: [Cd]

Amount of substance: [Mol]
 
Last edited:
  • #10
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dimensions is just like witing units in form of 7 basic standard units
what is the unit of 5m/s + 3m/s ??

M is the dimensional symbol for mass

7 dimentions:

Length: [L]

Mass: [M]

Time: [T]

Electric current: [A]

Temperature: [K]

Luminous intensity: [Cd]

Amount of substance: [Mol]
Okay, the unit is m/s and the dimensions is L/T, now I understand that.
 
  • #11
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GOOD!!

now that you are familiar with dimensions and that they don't change while addition or subtraction, can you now write the dimensions of A,B,C ??
 
  • #12
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Well I am assuming that C is Length while A is electric current and B is mass .. I'm trying to comprehend it man, but that is just guessing based on what you've explained :\
 
  • #13
1,137
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Well I am assuming that C is Length while A is electric current and B is mass\
Well how do you got these?
 
  • #14
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Well how do you got these?
wait, would the deminsions of A, B, and C all be L3?
 
  • #15
SammyS
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wait, would the dimensions of A, B, and C all be L3?
No, but the dimensions of A, B/t, and C·t4 are all [L]3, because each has to have the same dimension, which is the same dimension as V, which is also [L]3.
 
  • #16
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I told toy that treat dimentions just like units.

Ok what is unit of A in: (A-B)/t = 1kg ???
 
  • #17
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No, but the dimensions of A, B/t, and C·t4 are all [L]3, because each has to have the same dimension, which is the same dimension as V, which is also [L]3.
Thats what I didn't know


now I see that A=L^3, B=L^3t and C=L^3/t^4


I didn't know that you had to cancel out to get the dimension and I didn't know that V=L^3

V=L^3 because the formula for volume=lwh and those are 3 lengths multiplied together.. is that right?
 
  • #18
1,137
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Yes you are right but still A is not L3

Remember what you did ... You know Ct4 + <something> = volume

Now using table you find out that dimension of volume is [L3]

You also know that any quantity X can only be added to some quantity which has same units as X ...
(thats why 5m/s + 3m/s = 8 m/s ... but if someone says that find 5m/s + 3m/s2 ... you tell them to go home and study!!!)

................ By all this can you conclude that Ct4 has same dimentions as that of volume i.e. [L3] ...????

So Dimension of C comes out to be [L3]/[T4]


Now can you find the correct dimensions of A ???
 

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