# Dimensional Analysis check

• Dec 8th 2009, 07:08 AM
OptiEng
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
• Dec 8th 2009, 07:12 AM
e^(i*pi)
Quote:

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

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

$x = L$

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

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

where M is mass, L is length and T is time
• Dec 8th 2009, 07:21 AM
Quote:

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 .
• Dec 8th 2009, 07:37 AM
OptiEng
Quote:

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?
• Dec 8th 2009, 07:43 AM
e^(i*pi)
Quote:

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
• Dec 8th 2009, 01:02 PM
OptiEng
and I also need to lose one of the 'L's too?
• Dec 8th 2009, 01:12 PM
e^(i*pi)
is V meant to be cm^3 or cm^3/cm = cm^2?

Note that acceleration has dimensions $L \cdot T^{-2}$
• Dec 9th 2009, 01:36 AM
OptiEng
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.