# Dimensional Analysis

1. Oct 7, 2013

### Pajamas

1. The problem statement, all variables and given/known data
A=B^3C^1/2 where A has the dimensions L/M and C has dimensions L/T. What are the dimensions of B?

2. Relevant equations

3. The attempt at a solution
When I worked the problem I got B=M/T but it is wrong. I'm not sure how to approach the question.

2. Oct 7, 2013

### Simon Bridge

You substitute the dimensions into the equation and work out what dimenstions B has to have to make the LHS match the RHS.

3. Oct 7, 2013

### Pajamas

So far I have B^3=TL/M^2

L/M=B^3(L/T)^1/2
with L/T then in the square root, square both sides and get T/L*L/M^2=B^3, cancel the L on bottom and one on top to get the above answer.

4. Oct 7, 2013

### Pajamas

This doesn't make a lot of sense to me since B is still B^3. Am I supposed to have it look similar to the other side with TLM?

5. Oct 7, 2013

### nasu

You have several errors in this.
Squaring will produce B^6. And L^2 in L/M.
Using parentheses will make the things clearer. For you as well as for the people reading your posts.
You don't need to square. Just solve for B and put the dimensions.

6. Oct 7, 2013

### Simon Bridge

T/L*L/M^2=T/L^2/M^2=T/(L^2M^2) ...

Yike... you need to use brackets more to group your terms.
Use square brackets to represent when you mean "dimensions of"

[B^3]=(T/L)(L/M^2)=T/(M^2)

OK - but you need ... you've not finished.
(And - check your arithmetic.)

Last edited: Oct 7, 2013