[SOLVED] Friction Problem


by MellowOne
Tags: friction, solved
MellowOne
MellowOne is offline
#1
Jan12-08, 09:13 PM
P: 44
1. The problem statement, all variables and given/known data
A bicyclist can coast down a 5 hill at a constant 8.0 km/h. Assume the force of friction (air resistance) is proportional to the speed v so that Ffr = cv.

(a) Calculate the value of the constant c.
________ kg/s

(b) What is the average force that must be applied in order to descend the hill at 18 km/h? The mass of the cyclist plus bicycle is 85 kg.
________N


2. Relevant equations
Sigma F = ma
Ffr = cv

3. The attempt at a solution
First i converted my 8 km/hr into 2.22 m/s. Then i went into the parallel forces which are Ff and Fwx. The sum of their forces is equal to 0 because its acceleration is 0 so Ff is equal to Fwx and Fwx is equal to sin5 times mg which equlaed .854m. Then i set that equal to 2.22c because Ff also equal that. Now I'm stuck because I don't have mass. I tried writing out all units and canceling out so that on kg/s was left, but that didn't work out.

Any ideas?
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dynamicsolo
dynamicsolo is offline
#2
Jan12-08, 10:27 PM
HW Helper
P: 1,664
You correctly found that cv = mg sin 5 . Go ahead and use the mass given for the cyclist in part (b); I think it's not intended to be used there exclusively. The "resistance" coefficient (largely due to air drag) will depend on the mass of the rider and bicycle; a different rider would have a different value for c.
MellowOne
MellowOne is offline
#3
Jan12-08, 10:43 PM
P: 44
When using the mass given in part b, the answer came out to be approx 1 kg/s which wasn't correct. Is there something I missed or another way to approach this problem.

dynamicsolo
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#4
Jan12-08, 11:24 PM
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P: 1,664

[SOLVED] Friction Problem


Quote Quote by MellowOne View Post
When using the mass given in part b, the answer came out to be approx 1 kg/s which wasn't correct. Is there something I missed or another way to approach this problem.
I used what you set up and found

c = (85 kg) (9.81 m/sec^2) (sin 5) / (2.22 m/sec) .

Did you omit something?
MellowOne
MellowOne is offline
#5
Jan13-08, 12:15 AM
P: 44
Yeah, I must have forgotten something cause I did it again and I got the right answer. Thanks a lot.


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