Finding Dimensional Homogeneity

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In summary, the third equation is not dimensionally homogeneous because the left hand side is not simply F, but has time with it.
  • #1
Physics_Grunt
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Homework Statement


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)

Homework 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)

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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  • #2
Physics_Grunt said:

Homework Statement


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)


Homework 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)


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.
 
  • #3
Physics_Grunt said:
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. :wink:
 
  • #4
collinsmark said:
Try that one again. It seems you forgot to square something or forgot to divide by something. One of the two. :wink:

Uh, V^2 dimensions are m^2/s^2. You try it again.
 
  • #5
PeterO said:
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" [Broken]) 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?
 
Last edited by a moderator:
  • #6
Physics_Grunt said:
Okay, so further googling turned up this(http://physics.nist.gov/cuu/Units/units.html" [Broken]) 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.
 
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  • #7
Dick said:
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?
 
  • #8
Physics_Grunt said:
Okay, so further googling turned up this(http://physics.nist.gov/cuu/Units/units.html" [Broken]) 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.
 
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  • #9
Physics_Grunt said:
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.
 
  • #10
Thank you all. Your explanations made it click in my head and I think I have it now.
 

What is dimensional homogeneity?

Dimensional homogeneity is a concept in dimensional analysis that states all terms in an equation must have the same units in order for the equation to be valid. This means that the dimensions of all physical quantities on both sides of the equation must be the same.

Why is it important to find dimensional homogeneity?

Finding dimensional homogeneity is important because it ensures the validity of an equation. If an equation is not dimensionally homogeneous, it means there is an error in the equation or in the units used. This can lead to incorrect results and conclusions.

How do you check for dimensional homogeneity?

To check for dimensional homogeneity, you must analyze the dimensions of each term in the equation. This involves identifying the fundamental dimensions (such as length, mass, and time) for each physical quantity and ensuring they are the same on both sides of the equation.

What are some common mistakes in finding dimensional homogeneity?

Common mistakes in finding dimensional homogeneity include forgetting to convert between units, using incorrect fundamental dimensions for a physical quantity, and forgetting to include all terms in the equation. It is important to double check all dimensions and units to avoid these errors.

How can dimensional homogeneity be used in problem solving?

Dimensional homogeneity can be used in problem solving to help check the validity of equations and to convert between different units. It can also be used to identify relationships between physical quantities and to determine the appropriate units for a given quantity.

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