Dimensional Analysis Help (Algebra Based)

In summary: Don't forget that that the m's canceled out.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 :).In summary, the first problem involves finding the consistency of units in an equation involving time, length, and acceleration. The 2pi in the equation has no role in unit consistency. In the second problem, the force depends on momentum and time, and the units of momentum and time can be equated with the base units of force to find the values of a and b.
  • #1
Dig
13
0

Homework Statement


Problem #1
http://img230.imageshack.us/img230/705/hwpicvf5.jpg

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

Homework Equations





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. :)
 
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  • #2
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

[tex]\sqrt{\frac{m}{ms^{-2}}[/tex]
 
  • #3
rock.freak667 said:
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

[tex]\sqrt{\frac{m}{ms^{-2}}[/tex]

Thank you for your quick reply :).

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
Dig said:
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?



Acceleration is m/s^2, another way to write that is ms^-2 . You can treat them like indices.
 
  • #5
rock.freak667 said:
Acceleration is m/s^2, another way to write that is ms^-2 . You can treat them like indices.

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
Dig said:
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?

Very much so.
 
  • #7
rock.freak667 said:
Very much so.

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
Dig said:
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.

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
rock.freak667 said:
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.

I thought that s^-2 became 1/s^2.
 
  • #10
Dig said:
I thought that s^-2 became 1/s^2.

[tex]\frac{m}{ms^{-2}}=\frac{1}{s^{-2}}[/tex]


Like that. Don't forget that that the m's canceled out.
 
  • #11
rock.freak667 said:
[tex]\frac{m}{ms^{-2}}=\frac{1}{s^{-2}}[/tex]


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

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
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.
 

FAQ: Dimensional Analysis Help (Algebra Based)

What is dimensional analysis?

Dimensional analysis is a problem-solving technique used in science and mathematics to convert between different units of measurement. It involves using conversion factors and cancelling out units to get the desired unit for a given quantity.

Why is dimensional analysis important?

Dimensional analysis is important because it allows scientists to accurately and efficiently convert between units of measurement and ensure that their calculations are correct. It also helps to identify and prevent errors in calculations due to incorrect units.

How do I use dimensional analysis to solve a problem?

To use dimensional analysis, you need to identify the starting unit, the desired unit, and any conversion factors that relate the two. Then, you can set up a conversion factor ratio and use it to cancel out units until you are left with the desired unit.

What are some common conversion factors used in dimensional analysis?

Some common conversion factors used in dimensional analysis are those relating units of length (such as inches to centimeters), units of mass (such as pounds to kilograms), and units of time (such as minutes to seconds). It is important to use accurate and precise conversion factors to ensure the correct answer.

Can dimensional analysis be used in other areas of study besides science and math?

Yes, dimensional analysis can be used in other areas of study such as engineering, pharmacy, and economics. It is a versatile problem-solving technique that can be applied to any situation that involves converting between units of measurement.

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