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Dimensional consistency

GreenGoblin

Member
Feb 22, 2012
68
Which of the following are dimensionally consistent:

[TEX]U = \frac{EA\delta^{2}}{2\ell}, F = \frac{EA\delta}{l} + mg\ell[/TEX].

Right so. I get the concept, but the thing is I don't have (nor do I know where to find) expressions for each of these components. A is area so that is [TEX]L^{2}[/TEX]. E is a 'pressure' which I have as [TEX]ML^{-1}T^{-2}[/TEX]. This is never having done any mechanicsy type stuff before. l is just length. But thing is U is a 'potential energy', which I have no expretion for so I don't know the LHS of the first which is kind of the point of what I need to do. I also don't know what to do with a scalar multiple (the 1/2)? if this affects it at all? And the delta, I have no indication what it represents.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
1.) $\displaystyle U=\frac{EA\delta^2}{2L}$

Since you state $U$ is energy, which is $\displaystyle \frac{\text{mass}\times\text{length}^2}{\text{time}^2}$, let's see what $\delta$ must be if there is dimensional consistency.

Pressure is force per area, so $EA$ has units of force or $\displaystyle \frac{\text{mass}\times\text{length}}{\text{time}^2}$

Scalar constants may be ignored as they are dimensionless.

So we are left with:

$\displaystyle \text{length}=\frac{\delta^2}{\text{length}}$

So, if the units of $\delta$ are length, then it is consistent.

Can you try the second one?
 

GreenGoblin

Member
Feb 22, 2012
68
1.) $\displaystyle U=\frac{EA\delta^2}{2L}$

Since you state $U$ is energy, which is $\displaystyle \frac{\text{mass}\times\text{length}^2}{\text{time}^2}$, let's see what $\delta$ must be if there is dimensional consistency.

Pressure is force per area, so $EA$ has units of force or $\displaystyle \frac{\text{mass}\times\text{length}}{\text{time}^2}$

Scalar constants may be ignored as they are dimensionless.

So we are left with:

$\displaystyle \text{length}=\frac{\delta^2}{\text{length}}$

So, if the units of $\delta$ are length, then it is consistent.

Can you try the second one?
Hi, thanks a lot. In the second, we have two terms. I take it both would have to be consistent? I.e. then the RHS would be 2 x (whatever) which has dimension of (whatever). I get that the first term is consistent, but the 2nd is not.. if g has the dimension of acceleration which I believe is correct? Also, is delta being length likely to make sense? I have no indication of what delta should represent.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I suspect that $\delta$ does have units of length, and so the first is consistent.

You are correct on the second one, both terms on the right have to be consistent with the left side. The first term is consistent as you discovered, but unless the factor $\ell$ is dimensionless, then this second term is not consistent. By Newton's second law, we know $mg$ is a force.
 

GreenGoblin

Member
Feb 22, 2012
68
I suspect that $\delta$ does have units of length, and so the first is consistent.

You are correct on the second one, both terms on the right have to be consistent with the left side. The first term is consistent as you discovered, but unless the factor $\ell$ is dimensionless, then this second term is not consistent. By Newton's second law, we know $mg$ is a force.
Yes I noticed that the 2nd term is just ma = F but multiplied by an additional l which makes it inconsistent. I thank you for your help and I am confident with this question now. I decided to accept delta as a length, it makes more sense that way. Gracias.