# Dimensional Analysis Help (Algebra Based)

1. Sep 3, 2008

### Dig

1. The problem statement, all variables and given/known data
Problem #1
http://img230.imageshack.us/img230/705/hwpicvf5.jpg [Broken]

Problem #2
http://img149.imageshack.us/img149/3196/hwpic2fb4.jpg [Broken]

2. Relevant equations

3. The attempt at a solution

Problem #1)
I do not know where to proceed from here

I have.
T = 2pi * square root of L/g
where T = time, g = acceleration
L = length
I know that acceleration = Length/Time^2
so I get
T = 2pi * square root of L/(L/T^2)
I also do not understand how 2pi eventually plays a part in this.

Problem #2)
I honestly do not know where to start with this problem. If I understood the first part, the second part wouldn't be a problem. What does kg.m^2/s^2 mean?

I know that these are very basic questions, but my book assumes a lot and takes leaps and I unfortunately need things spelled out for me from time to time. Any help given is more than appreciated. :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 3, 2017
2. Sep 3, 2008

### rock.freak667

For the first one, to show that it is consistent, you need to show that the units on the right=units on the left.

The unit of the period is seconds,s. The 2pi has no role when dealing with the units since, it has none!
length=m
g=ms-2

now just simplify this

$$\sqrt{\frac{m}{ms^{-2}}$$

3. Sep 3, 2008

### Dig

I was under the impression that g(acceleration) was equal to m(length) divided by s(time) squared. Am I wrong about this or is there some sort of exception I am not understanding here?

Thanks.

4. Sep 3, 2008

### rock.freak667

Acceleration is m/s^2, another way to write that is ms^-2 . You can treat them like indices.

5. Sep 3, 2008

### Dig

Ok, I'm going to take a shot at what you left me with.

If I multiply it by ^1/2, the m's cancel out, and I am left with (s^2)^1/2 which leaves me with s. So, that would make the original equation correct as s=s.

Am I correct here?

I understand what you meant by acceleration being ms^-2 now. Thanks :)

With your method of ms^-2, the m's cancel out and you are left with the square root of s^-2. This would be the square root of 1/s^2. This would be s * square root of 1; which is s. Am I correct here? It's been a while since college algebra lol.

6. Sep 3, 2008

### rock.freak667

Very much so.

7. Sep 3, 2008

### Dig

I edited my message and did it your way at the bottom. Can you tell me if I applied that correctly?

Also, why doesn't the pi count? I hate to be difficult, but it's just so confusing to see that 2pi out in front and it not mean anything lol.

8. Sep 3, 2008

### rock.freak667

you'd get the sq.root of 1/s-2 which is the same as s2

Well 2 and pi are just constants with no units.

9. Sep 3, 2008

### Dig

I thought that s^-2 became 1/s^2.

10. Sep 3, 2008

### rock.freak667

$$\frac{m}{ms^{-2}}=\frac{1}{s^{-2}}$$

Like that. Don't forget that that the m's cancelled out.

11. Sep 3, 2008

### Dig

Noted!

I appreciate your help a lot man.

Any chance you could take a look at the second problem?
I mostly need to understand the first part and the measurement and stuff.

Any help is appreciated :).

12. Sep 3, 2008

### rock.freak667

Well the force F depends on momentum and time right?

So you could simplify the equation as this:

F=k patb

k is just a constant with no units. a and b are numbers that you need to find.

just write the units of momentum and time in the formula. Assuming that you have the base units of Force, just equate the terms.

e.g. kgm=kgxm

then x=1. Like that.