Unraveling the Mystery of Dimensional Analysis

In summary, Dimensional Analysis is a mathematical method used to convert units of measurement and solve problems involving different quantities. It involves setting up a conversion factor and using the canceling units method to arrive at the desired unit. This method is commonly used in science and engineering fields, as well as in everyday life to ensure accuracy and consistency in measurements. By understanding the principles of dimensional analysis, one can easily convert between different units and solve complex problems involving multiple quantities.
  • #1
Beholder
8
0
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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  • #2
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...
 
  • #3
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
 
  • #4
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
 

Related to Unraveling the Mystery of Dimensional Analysis

1. What is dimensional analysis?

Dimensional analysis is a mathematical tool used to analyze and convert physical quantities and units. It involves identifying the dimensions (such as length, time, mass) of a given quantity and using this information to convert it into different units or to solve problems involving multiple physical quantities.

2. How is dimensional analysis useful?

Dimensional analysis is useful in many scientific fields, particularly in physics and engineering. It allows for easy conversion between units and helps to identify and correct errors in calculations. It also helps in understanding the relationships between different physical quantities and can be used to solve complex problems involving multiple variables.

3. Can dimensional analysis be used for all types of units?

Yes, dimensional analysis can be used for any type of unit as long as the units can be expressed as combinations of the base units (such as meters, seconds, kilograms). This includes both SI units and imperial units.

4. How do you perform dimensional analysis?

To perform dimensional analysis, you first need to identify the dimensions of the given quantity. Then, using conversion factors and the rules of algebra, you can manipulate the units to convert them into the desired units or to solve for a specific variable. It is important to keep track of units and cancel out any that are not needed in the final answer.

5. Why is dimensional analysis important in research and experiments?

Dimensional analysis is important in research and experiments because it helps to ensure accuracy and consistency in measurements and calculations. It also allows for easy comparisons between different experiments and helps to identify any potential errors or inconsistencies in the data. Additionally, dimensional analysis can be used to derive new equations and understand the relationships between different physical quantities, furthering scientific understanding in various fields.

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