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I'm working with the following equation:

Rs [mg/(cm^2 . day)] = K [units not given] * X [mg/cm^2] * shearstress^0.58 [dyne/cm^2]

As far as I can tell, 1 dyne = 10^-5 Newtons.

The first issue I'm having is determining the units of K. The fact that shearstress is raised to the power of 0.58 is confusing me in this regard, as I'm not sure whether to include this in the dimensional analysis.

The second thing I'm looking to do is to convert the units of the entire equation to [μg/(cm^2 . hour)]. How would I go about this conversion?

Thanks very much in advance for any advise, and please let me know if I need to provide any more information.

user142224
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1 Answers1

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You are correct 1 dyne =10^{-5} Newton. You can't raise something with units to a fractional power, so you must be intended to measure the shear stress in dyne/cm^2 and raise the value to the power $0.58$. The shear stress must really be divided by a standard value of 1 dyne/cm^2, and you should consider that term dimensionless. For dimensional analysis we can then ignore that term.

Now match up the units in the rest of the equation to find the units of K. To convert to $\mu g$, you use the fact that 1 mg=1000 $\mu$g, so terms with mg in the numerator get multiplied by $1000$. You also need to convert days to hours, so where you see day you should put 24 hours.

Ross Millikan
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  • Thanks very much for the response! So does this all mean that in order to determine the units of K, I simply ignore shear stress completely, i.e. the unit of K are therefore [1/day]?

    Now, do my measured units of shear stress HAVE to be dyne/cm^2 for this equation to work? What do I do if my measured units are, say, dyne/m^2?

    Thanks!

     – user142224 Apr 10 '14 at 13:47
  • Yes, you are correct. The units of K are 1/day. You need to convert your shear stress to dyne/cm^2 before raising to the $0.58$ power. If you had it in dyne/m^2, you would replace the m with 100 cm, giving dyne/(100 cm)^2, so 1 dyne/m^2=$10^{-4}$ dyne/cm^2 – Ross Millikan Apr 10 '14 at 15:14
  • Perfect, thanks very much for your help! – user142224 Apr 10 '14 at 15:21