Dimensional Analysis Question

In summary: G as kg*m/s^2. I wasnt actually putting G back in. my bad! In summary, the student is a new college physics student that has been out of the science/math arena for about 2.5 years. They are a Software Engineer by day, but are going back to get their mechanical engineering degree. Unfortunately, they have not used physics or calculus in their business systems designs in the past, and are struggling with a problem from a physics book they are using. They have tried different methods to solve the problem, but have not been successful. A coincidence or not, the same question was posted here the day before.
  • #1
Grave9
8
0
First off, I'm glad I found a forum like this! Thank jeebus!

I'm a new college physics student that's been out of the science/math arena for about 2.5 years. I'm a Software Engineer by day but I"m going back to get my mechanical engineering degree. Unfortunately I build business systems all day and physics nor calculus is used much in these applications that I build, hence why I'm kind of off the rocker on this stuff now.

I've been pouring over these problems ALL DAY LONG and I've beat my self up (mentally) because I cannot figure them out.

Just so you know, I'm not looking for the answer, I'm looking for the direction to go down to get hte answer. I want to learn this stuff as best as I can, but I need to know how it works!

With the introduction to the side, here I go...

In my physics book, I have a quesion that says:

"Newtons law of universal graviatation is represented by

F = GMm/r^2

Here F is the magnitude of the gravitational force exrted by one small object on another. M a m are the masses of the objects and r is a distance. Force has the SI units kg * m/s^2.

What are the SI units of the propotionally constant G? "

Here's what I've tried..

kg * m/s^2 = GMm/s^2

=> kg * m/L^2 = GMm/s^2

This is where my math skills are a little rusty. I know the dividing by a fraction is a little weird, but I think i remember how to do it. (I took Calc I and Calc II a couple years back, but like i said, I'm just rusty. As soon as I figure it out, I remember everything about it).

I'm not sure if it goes like this (which is how I've done it)

=> kg * s^2 = GMmm/r^2 => Gm^3/r^2

=> kg * s^2 = Gm^3/r^2

=> (kg * s^2)/G = m^3/r^2

=> 1/G = m^3 * (kg * s^2) / r^2 * 1

But this answer is COMPLETELY WRONG. I MEAN, WAY OFF... The answer is in the back of the book as:

m^3 / kg * s^2

I've tried moving around The G, replacing r^2 with L (length) and setting s^2 = L (lenth) and I've moved around fractions, numerators, demonminators, etc etc and I CANNOT get close to this answer at all.

What might I be doing wrong? TO tell you the truth, I"m not really sure what this question is asking. Is it asking me to solve for G or what is it asking me to do? The question is under the section of dimensional analysis so I'm assuming that I need to make sure both sides are dimensionally correct. My answers never seem to add up!

I've tried thinking of all possibilities, solving for other variables, and I cannot wrap my head around this. The entire chapter never touches on anything like this at all.

Any help would be appreciated!

Thanks!
 
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  • #2
Coincidence or not, but the same question was posted here yesterday.

As I said in the other topic, I'd work out for G first like this:

[tex]F = G\frac{{m_1 \cdot m_2 }}{{r^2 }} \Leftrightarrow G = \frac{{F \cdot r^2 }}{{m_1 \cdot m_2 }}[/tex]

Now just replace everything on the RHS by their dimensions and simplify a bit.
 
  • #3
We have, [tex]F = G\frac{Mm}{r^2}[/tex], and we want to solve for G.

[tex]G = \frac{Fr^2}{Mm}[/tex]

The SI unit for mass is kg (kilogram) and for length it's m (meter). We know that:
[tex] {F = \frac{kg*m}{s^2}[/tex]

With all this in mind we have that:

[tex]G = \frac{\frac{kg*m}{s^2}*m^2}{kg*kg}[/tex]

Now all you have to do is to simplify.
 
  • #4
TD said:
Coincidence or not, but the same question was posted here yesterday.

As I said in the other topic, I'd work out for G first like this:

[tex]F = G\frac{{m_1 \cdot m_2 }}{{r^2 }} \Leftrightarrow G = \frac{{F \cdot r^2 }}{{m_1 \cdot m_2 }}[/tex]

Now just replace everything on the RHS by their dimensions and simplify a bit.


Thats weird, but assuming that this is the first semester of this year, its understanable that someone would have the same book and possibly the same class. But in all honesty, I don't know who this is.

I'm going to Solve for G here. These images/equations you posted pointed out a very big math flaw that I had, which makes A TON of difference in this case.
 
  • #5
iNCREDiBLE said:
We have, [tex]F = G\frac{Mm}{r^2}[/tex], and we want to solve for G.

[tex]G = \frac{Fr^2}{Mm}[/tex]

The SI unit for mass is kg (kilogram) and for length it's m (meter). We know that:
[tex] {F = \frac{kg*m}{s^2}[/tex]

With all this in mind we have that:

[tex]G = \frac{\frac{kg*m}{s^2}*m^2}{kg*kg}[/tex]

Now all you have to do is to simplify.


AHHHHHHHHHHH ! it makes sense! I wasnt putting G back into the eqaution. I was solving for G and then stopping. I was only doing half of the problem! Oh man, I feel like an idiot now!

Thanks for the help guys, I appreciate it.

Also, where do you find the codes to post these images?

D
 
  • #6
I keep track of which "m" stands for the variable, "mass" ,
and which "m" stands for the unit "meter" (which you goofed up)
by always putting units inside brackets:
[m] means meter, which can never cancel a mass m .
 
  • #7
Grave9 said:
...

Also, where do you find the codes to post these images?

D
Have a look here.
 

1. What is dimensional analysis and why is it important?

Dimensional analysis is a mathematical technique used to convert units from one system to another. It is important because it allows scientists to accurately compare and analyze data without being limited by different unit systems.

2. How do I perform dimensional analysis?

To perform dimensional analysis, you must first identify the units you want to convert from and to. Then, use conversion factors or unit equivalencies to cancel out the unwanted units and end up with the desired units.

3. Can dimensional analysis be used for any type of measurement?

Yes, dimensional analysis can be used for any type of measurement as long as the units are known and the conversion factors or equivalencies are available.

4. Is dimensional analysis only used in the field of science?

No, dimensional analysis can also be applied in other fields such as engineering, economics, and finance, where unit conversions are necessary for accurate calculations and comparisons.

5. Are there any limitations to dimensional analysis?

Yes, dimensional analysis may not be effective for complex or non-linear relationships between units. It also assumes that the units being converted are consistent and compatible with each other.

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