Basic fluid mechanics of sled question

AI Thread Summary
The problem involves determining the thickness of a water layer under sled runners, given specific parameters such as speed, force, area, and viscosity. The calculations show that the thickness is 11.7(10^-5) ft, but there is confusion as the textbook states 11.7(10^-4) ft. The discussion highlights the importance of correctly interpreting scientific notation and verifying calculations. A suggestion is made to double-check the textbook for possible errors in notation. The conversation emphasizes the need for accuracy in fluid mechanics calculations.
stinlin
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Homework Statement



A sled slides along on the snow on a thin horizontal layer of water between the ice and the runners. The horizontal force that the water puts on the runners is equal to 1.2 lb when the sled's speed is 50 ft/s. The total area of both runners in contact with the water is 0.08 ft^2, and the viscosity of the water is 3.5(10^-5) lb*s/ft^2. Determine the thickness of the water layer under the runners. Assume a linear velocity distribution in the water layer.

Homework Equations



Shear stress = viscosity * (du/dy)

du/dy is the rate of strain.


The Attempt at a Solution



Well, because it has a linear velocity distribution, du/dy = Umax / h, where Umax is the maximum velocity and h is the height or depth of the fluid.

Umax = 50 ft/s
Max shear stress = P/A = 1.2 lb / 0.08 ft^2 = 15 lb/ft^2

So rearranging with substitutions, the equation looks like this:

h = (viscosity)*Umax/h

With numbers:

h = (3.5(10^-5) lb*s/ft^2) * 50 ft/s / 15 lb/ft^2 = 0.00175/15 ft

h = 11.7(10^-5) ft

That seems legit, no? But my book says 11.7(10^-4) ft. Now I know I shouldn't ALWAYS trust the book, but I just have a feeling I missed something if I'm off by only one order of magnitude. Thoughts?
 
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stinlin said:
my book says 11.7(10^-4) ft.
That is an unusual use of scientific notation. Are you sure it doesn’t say 1.17(10^-4) ft?
 
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