1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Friction bicycle homework Problem

  1. Jan 12, 2008 #1
    [SOLVED] Friction Problem

    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.

    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?
  2. jcsd
  3. Jan 12, 2008 #2


    User Avatar
    Homework Helper

    You correctly found that cv = mg · sin 5º . 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.
  4. Jan 12, 2008 #3
    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.
  5. Jan 12, 2008 #4


    User Avatar
    Homework Helper

    I used what you set up and found

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

    Did you omit something?
  6. Jan 13, 2008 #5
    Yeah, I must have forgotten something cause I did it again and I got the right answer. Thanks a lot.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook