Unraveling the Mystery of Dimensional Analysis

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Homework Help Overview

The discussion revolves around the concept of dimensional analysis in physics, specifically focusing on the dimensions of various physical quantities such as volume, acceleration, density, force, and charge. Participants are exploring how these dimensions are derived and the underlying principles of dimensional analysis.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the derivation of dimensions for physical quantities, expressing confusion about the relationships between units and their representations. They are attempting to understand how dimensions like length cubed for volume and length per time squared for acceleration are established.

Discussion Status

Some participants have offered explanations regarding the breakdown of units into their simplest forms and how dimensional analysis works. There is an ongoing exploration of the definitions and relationships between different physical quantities, with some clarity emerging around the concept of acceleration and its dimensional representation.

Contextual Notes

Participants are grappling with the definitions and assumptions related to dimensional analysis, noting that certain dimensions may only hold true under specific conditions, such as the volume of a perfect cube. There is also a mention of the need for a clearer understanding of the fundamental units involved.

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Hello I've been learning the basics of physics over the past few weeks and thought I knew what they meant when they would refer to the dimensions of something for instance if you asked me what are the dimensions of acceleration I'd say 'change in velocity/time'' or 'v2-v1/T' but I came across a page on a topic I'm not familar with 'Dimensional analysis' and they had a little quiz and said the following

1. volume = Length cubed
2. acceleration (velocity/time) = Length/Time squared
3. density (mass/volume) = Mass / Length cubed
4. force (mass × acceleration) = Mass x Length/Time squared
5. charge (current × tiime) = Current x Time

Where did they get these 'dimensions from'?

For the first one its only true for the volume of a perfect cube.

The second one it seems they are substituting length for velocity and I don't know where they got the squared from.

The third one again uses a perfect cube.

Number four I though would just be the well known F=MA I don't know where the length or T squared come in

and number five makes sense...

Can anyone clear this up, I'm baffled... thanks
BTW the link to the page I'm referring to is: http://www.physics.uoguelph.ca/tutorials/dimanaly/dimanaly_ans1a.html
 
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Let's see.

Dimensional analysis works by breaking down the units of whatever you're measuring into their simplest forms. When you do this, you find that all units are made up from just four basic units; these are length (L), mass (M), time (T) and force (F). (Temperature is a fifth, but you shouldn't be needing this for now).

When you're being asked for the dimension of something, you're really being asked for its units, in their most basic form.

So for (1), as you know, the units of volume are cubic metres; m^3. This one's simple, - the only units are those of length. Because it's volume, it's cubed.

I take it you're with me so far.

So, similarly, the units of 'acceleration' are m/s^2 (or "metres per second per second"). In dimensional terms, this breaks down to (L/T)/T, - units of length (metres), divided twice by units of time (seconds). The (L/T) part of this is in fact velocity, - it's a distance (or a length, L) divided by a time T. You know that acceleration is simply velocity/time. The squared comes from the fact that you've divided the length by time twice.

For (4), you're right that the force F comes from F=Ma. Substitute what you got for the dimensions of acceleration in (2) into this equation.


Sorry if I've fudged through this a bit, I didn't really understand it when I came across it first either.

I really must get round to learning Latex one day too...
 
Beholder said:
Hello I've been learning the basics of physics over the past few weeks and thought I knew what they meant when they would refer to the dimensions of something for instance if you asked me what are the dimensions of acceleration I'd say 'change in velocity/time'' or 'v2-v1/T' but I came across a page on a topic I'm not familar with 'Dimensional analysis' and they had a little quiz and said the following

1. volume = Length cubed
2. acceleration (velocity/time) = Length/Time squared
3. density (mass/volume) = Mass / Length cubed
4. force (mass × acceleration) = Mass x Length/Time squared
5. charge (current × tiime) = Current x Time

Where did they get these 'dimensions from'?

For the first one its only true for the volume of a perfect cube.
Let us say that T represents time and L length.
Any time you are dealing with a volume the units will be in terms of L3. For example the volume of a sphere is [tex]\frac 4 3 \pi r^3[/tex] note that the radius is cubed therefore if the radius is given in meters the volume will be in m3. Any volume will have dimensions of L3.

The second one it seems they are substituting length for velocity and I don't know where they got the squared from.
What are the dimensions of velocity? [itex]\fr ac L T[/itex] so if you divide by time one more time you get for acceleration:
[tex]\frac V T = \frac {\frac L T} T = \frac L { T^2}[/tex]
The third one again uses a perfect cube.
Read the the first one again.

Number four I though would just be the well known F=MA I don't know where the length or T squared come in
Read the acceleration explanation again.
and number five makes sense...

Can anyone clear this up, I'm baffled... thanks
BTW the link to the page I'm referring to is: http://www.physics.uoguelph.ca/tutorials/dimanaly/dimanaly_ans1a.html
 
Dimensions

Integral said:
Let us say that T represents time and L length.
Any time you are dealing with a volume the units will be in terms of L3. For example the volume of a sphere is [tex]\frac 4 3 \pi r^3[/tex] note that the radius is cubed therefore if the radius is given in meters the volume will be in m3. Any volume will have dimensions of L3.


What are the dimensions of velocity? [itex]\fr ac L T[/itex] so if you divide by time one more time you get for acceleration:
[tex]\frac V T = \frac {\frac L T} T = \frac L { T^2}[/tex]

Read the the first one again.


Read the acceleration explanation again.


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Ah very clear, I understand now, just breaking them down. Thanks
 

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