1. ## Dimensional Analysis check

Hello all,

Just want to check what other people get as an answer for this dimensional analysis.

V=(kLx)/p

Units are:

'k' is dimensionless
'x' is length (m)
'p' is (grammes/cm^2)

Any help and workings shown would be much appreciated.

Thanks

2. Originally Posted by OptiEng
Hello all,

Just want to check what other people get as an answer for this dimensional analysis.

V=(kLx)/p

Units are:

'k' is dimensionless
'x' is length (m)
'p' is (grammes/cm^2)

Any help and workings shown would be much appreciated.

Thanks
$\displaystyle N = M \cdot L \cdot T^{-2}$

$\displaystyle x = L$

$\displaystyle p = M \cdot L^{-2}$

$\displaystyle \frac{M \cdot L \cdot T^{-2} \cdot L}{M \cdot L^{-2}} = L^4 \cdot T^{-2}$

where M is mass, L is length and T is time

3. Originally Posted by OptiEng
Hello all,

Just want to check what other people get as an answer for this dimensional analysis.

V=(kLx)/p

Units are:

'k' is dimensionless
'x' is length (m)
'p' is (grammes/cm^2)

Any help and workings shown would be much appreciated.

Thanks
HI

What is V ? You need to know what v is so that you can verify whether the equation is dimensionally homogenous .

HI

What is V ? You need to know what v is so that you can verify whether the equation is dimensionally homogenous .
V is supposed to be Cm^3/cm, and I obtained the same answer as "e^.." so that means the equations can't hold true?

5. Originally Posted by OptiEng
V is supposed to be Cm^3/cm, and I obtained the same answer as "e^.." so that means the equations can't hold true?
As it stands yes, you'd need to get rid of time

6. and I also need to lose one of the 'L's too?

7. is V meant to be cm^3 or cm^3/cm = cm^2?

Note that acceleration has dimensions $\displaystyle L \cdot T^{-2}$

8. Hi,

V is supposed to be a specific rate of change of volume per cm. It's a strange unit I know. That's why I am questioning it. It is from the Archard's equation of wear used in engineering since the 1950's.