# Using couette flow/viscosity to find force

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1. Sep 15, 2014

### reed2100

1. It is a rainy day in Houston. You are a poor student (or TA or professor) with an old car that has bald tires. You are traveling at 60 miles/hour and you slam on the brakes (i.e., your wheels are no
longer turning. Assume there is a layer of pure water 0.1 mm thick between your tires and the
road, i.e., the rubber is not meeting the road. Assume each of your tires contacts the water layer
over a rectangle 10 cm wide and 15 cm long.

2. Relevant equations
τ=μ(∂v/∂y) ?
τ=F/A

3. The attempt at a solution

y = .1mm = .01cm = .0001m
A= 10cm(15cm) = 150cm^2 = .015m^2
v = 60mph = 2.682 m/s
μ= 1 cp = .001 N*s/m^2

My professor posted the notes online and I looked at them again, but it's confusing me because he mentioned integration and differential equations which I know how to do but don't see where to begin on, and there was also (∂v/∂y) = vmax/ymax. I've been doing ODE's in my engineering math class but we're always given a general solution or an ODE to work with; with this I don't see a function to integrate or anything.

I tried this:

F/.015m^2 = (.001 N*s/m^2) * ([2.682m/s]/.0001m)
F = .402 N * (.225 lb/N)
F = .09 lb

I have no idea if it's right because these aren't problems out of a book with the solutions or online with feedback, I'll only know once it's graded. I don't understand how to do these couette flow/viscosity problems, so any help would be appreciated. Given the context of the problem I feel like .09 lb of braking force is probably extremely wrong.

2. Sep 9, 2015

### lahudson

First, your velocity is off by a factor of 10.. it should be 60 mph = 26.83 m/s, which would then make your force PER TIRE F=0.9 lb. The answer would then be 4F to be the total force because you have 4 tires which gives the answer Ftot=3.6lbs