Finding Dimensional Homogeneity


by Physics_Grunt
Tags: dimensional, homogeneity
Physics_Grunt
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#1
Aug29-11, 11:06 PM
P: 7
1. The problem statement, all variables and given/known data
Which one of the following equations is dimensionally homogeneous?

Where:

F= force (N)
m= mass (kg)
a= acceleration (m/s2)
V= velocity (m/s)
R= radius (m)
t= time (s)


2. Relevant equations

1. F=ma
2. F=m(V2/R)
3. F(t2-t1)=m(V2-V1)
4. F=mV
5. F=m(V2-V1)/(t2-t1)


3. The attempt at a solution

Through what I can gather from my textbook and the internet, I started by entering what I know. So:
F=ma
becomes:
N=(kg)(m/s2)

From here, I'm not really sure where to go.

F=m(V2)/(R)
becomes:
N=(kg)((m/s2)/(m))

And again, I plug everything in but in my textbook at this point is where they determine if it is or isn't dimensionally homogeneous.

I would really appreciate any guidance on this, I realize it's a super basic question, but you've got to start somewhere! Thank you.
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PeterO
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#2
Aug29-11, 11:18 PM
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Quote Quote by Physics_Grunt View Post
1. The problem statement, all variables and given/known data
Which one of the following equations is dimensionally homogeneous?

Where:

F= force (N)
m= mass (kg)
a= acceleration (m/s2)
V= velocity (m/s)
R= radius (m)
t= time (s)


2. Relevant equations

1. F=ma
2. F=m(V2/R)
3. F(t2-t1)=m(V2-V1)
4. F=mV
5. F=m(V2-V1)/(t2-t1)


3. The attempt at a solution

Through what I can gather from my textbook and the internet, I started by entering what I know. So:
F=ma
becomes:
N=(kg)(m/s2)

From here, I'm not really sure where to go.

F=m(V2)/(R)
becomes:
N=(kg)((m/s2)/(m))

And again, I plug everything in but in my textbook at this point is where they determine if it is or isn't dimensionally homogeneous.

I would really appreciate any guidance on this, I realize it's a super basic question, but you've got to start somewhere! Thank you.
Homogeneous usually means "made up of the same types" or words to that effect.

If you express each quantity in base units - m , kg , s - you could compare left to right.

hint: there are two formulas which look very similar, but with one variable different. I would suspect one of those.
collinsmark
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#3
Aug29-11, 11:31 PM
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Quote Quote by Physics_Grunt View Post

F=m(V2)/(R)
becomes:
N=(kg)((m/s2)/(m))
Try that one again. It seems you forgot to square something or forgot to divide by something. One of the two.

Dick
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#4
Aug29-11, 11:35 PM
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Finding Dimensional Homogeneity


Quote Quote by collinsmark View Post
Try that one again. It seems you forgot to square something or forgot to divide by something. One of the two.
Uh, V^2 dimensions are m^2/s^2. You try it again.
Physics_Grunt
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#5
Aug29-11, 11:42 PM
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Quote Quote by PeterO View Post
Homogeneous usually means "made up of the same types" or words to that effect.

If you express each quantity in base units - m , kg , s - you could compare left to right.

hint: there are two formulas which look very similar, but with one variable different. I would suspect one of those.
Okay, so further googling turned up this(http://physics.nist.gov/cuu/Units/units.html) handy chart.

Am I to understand correctly that 1N= 1kg(m/s2) And then using that I can compare left to right? And if the right side doesn't come out to kg(m/s2) it is NOT dimensionally homogeneous?
Dick
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#6
Aug29-11, 11:44 PM
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Quote Quote by Physics_Grunt View Post
Okay, so further googling turned up this(http://physics.nist.gov/cuu/Units/units.html) handy chart.

Am I to understand correctly that 1N= 1kg(m/s2) And then using that I can compare left to right? And if the right side doesn't come out to kg(m/s2) it is NOT dimensionally homogeneous?
Yes. And I don't there is just one that is homogeneous.
Physics_Grunt
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#7
Aug29-11, 11:55 PM
P: 7
Quote Quote by Dick View Post
Yes. And I don't there is just one that is homogeneous.
I think the question may have been mis-typed on my handout. I think it should read "which one of the following is NOT dimensionally homogeneous?"

When I do the substitutions for F=mV it becomes:

N=(kg)(m/s) or further:

(kg)(m/s2)=(kg)(m/s)

Because the (m/s) on the right is not squared as it is on the left, would this be a correct example of an equation that is NOT dimensionally homogeneous?
PeterO
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#8
Aug30-11, 12:00 AM
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Quote Quote by Physics_Grunt View Post
Okay, so further googling turned up this(http://physics.nist.gov/cuu/Units/units.html) handy chart.

Am I to understand correctly that 1N= 1kg(m/s2) And then using that I can compare left to right? And if the right side doesn't come out to kg(m/s2) it is NOT dimensionally homogeneous?
That is the task for all but the third one, where the left hand side is not simply F, but has time with it.

I would have the units of F as kgms-2

Perhaps made clearer as kg m s-2 or kg.m.s-2

Often these are actually written "dimensionally" using [M] for mass, [T] for time and [L] for length.

then we would have [M][L][T]-2

That certainly takes care of countries that use pounds instead of kg, and feet instead of metres.

Oh and rest assured - only one of the examples is not homogeneous - perhaps you left the word not out of your original post.
PeterO
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#9
Aug30-11, 12:04 AM
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Quote Quote by Physics_Grunt View Post
I think the question may have been mis-typed on my handout. I think it should read "which one of the following is NOT dimensionally homogeneous?"

When I do the substitutions for F=mV it becomes:

N=(kg)(m/s) or further:

(kg)(m/s2)=(kg)(m/s)

Because the (m/s) on the right is not squared as it is on the left, would this be a correct example of an equation that is NOT dimensionally homogeneous?
certainly would.
Physics_Grunt
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#10
Aug30-11, 12:12 AM
P: 7
Thank you all. Your explanations made it click in my head and I think I have it now.


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