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: Car turning on a bank (circular motion problem)

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    A racecar is turning on a bank inclined at 15%. It's speed is 225 km/h and it's mass is 1450kg.
    1. Calculate the radius of curvature
    2. Calculate the centripetal acceleration.
    3. If the car mantains circular traction, what is the magnitude of the force of static friction
    4. What is the coefficient of static friction.

    2. Relevant equations

    3. The attempt at a solution
    Is it just me or is there too many unknowns to be able to solve this question? (The answers are supposed to be actual numbers, not expressions in terms of unknowns)

    So , I will have the force of gravity pointing straight down, I will have the Normal Force pointing perpendicular to the bank, and I will have the static friction force (fs) pointing parallel to the bank towards the centre.

    So my Fy's are supposed to balance so we have:
    Ny - fy - mg = 0
    Ncosθ - μNsinθ - mg = 0
    N (cosθ - μsinθ) = mg

    And for Fx we have
    fx + Nx = mv^2/R
    μNcosθ + Nsinθ = mv^2/R

    N(μcosθ + sinθ) = mv^2/R
    plug in the N i got from the Fy equations:

    g(μcosθ + sinθ) / (cosθ - μsinθ) = v^2/R

    Now, i'm somehow supposed to provide the radius of curvature, and μ, (centripetl acceleration and magnitude of friction will then just come from those 2), but this is a single equation with 2 unknowns, how can i possibly be able to figure out both?
  2. jcsd
  3. Mar 23, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I would agree with you that more information is needed to solve this problem.

    In addition, I should point out that you are not justified in saying that the force of static friction is equal to the coefficient of static friction times the normal force. This is the case only if the two surfaces in contact are on the verge of sliding past each other. There is no language in the problem saying that this is the case.

    Finally, the direction of the force of static friction is not necessarily down slope. If the speed is low enough (lower than what is necessary for the car to make it round the curve on a friction-free road) then static friction up slope is needed to maintain the car on track.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook