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Dimensions help

  • Thread starter Beholder
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8
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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 refering to is: http://www.physics.uoguelph.ca/tutorials/dimanaly/dimanaly_ans1a.html
 

brewnog

Science Advisor
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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...
 

Integral

Staff Emeritus
Science Advisor
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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 refering to is: http://www.physics.uoguelph.ca/tutorials/dimanaly/dimanaly_ans1a.html
 
8
0
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.

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

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