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Frictional force on an incline

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A 1,970-kg car is moving down a road with a slope (grade) of 14% while slowing down at a rate of 3 m/s^2. What is the direction and magnitude of the frictional force? (define positive in the forward direction, i.e., down the slope)?


    2. Relevant equations
    ∫=μ*m*g*cosθ
    Fn = m*g*cosθ
    F= m*a


    3. The attempt at a solution
    i know m = 1970kg
    a = -3m/s^2
    g=-9.8m/s^2
    So i converted the grade using arctan(14/100) = ~8°
    then did tan(8°) = .14 = μ

    Fn = (1970)(9.8)cos(8) = 19118N
    19118*.14 = 2676.5N = ∫

    thats as far as i could get but i know i missed a step somewhere. I did a similar problem and got it right but that was with a constant velocity. not sure where to throw that in here. would m*a be the normal force in this case?


    thanks in advance ive been reading these threads for a while and the community seems very helpful
     
  2. jcsd
  3. Oct 23, 2012 #2


    Why did you assume [itex] tan(\theta) [/itex] is the friction coefficient? That would mean that the friction coefficient is always just the grade, independent of what causes the friction. Obviously, that can't be the case.

    To solve this problem, we don't even need to work with the normal force or friction coefficient, since they are both contained within [itex] F_{friction} [/itex] and we know all of the other necessary variables.

    Here's how I worked it, where [itex] F_{s} [/itex] is the force "down the slope," or parallel to the slope surface:

    [itex] F_{s} = mgsin(\theta) = (1970)(9.8)sin(8) = 2686.88 N[/itex]

    It is only accelerating parallel to the plane's surface, so:

    [itex] F_{s} - F_{f} = F_{net} = ma [/itex]

    [itex] 2686.88 - F_{f} = 1970(-3) = -5910 [/itex]

    [itex] F_{f} = 8596.88 N [/itex]

    Then if you did want the friction coefficient, you would just divide that answer by the normal force. You would get:
    [itex] \mu \approx .45 [/itex]

    Edit: technically, I should have written [itex] F_{s} + F_{f} = F_{net} = ma [/itex], so that [itex] F_{f} [/itex] would have a negative sign since I defined "up the slope" as the negative direction, but you get the idea.
     
    Last edited: Oct 23, 2012
  4. Oct 23, 2012 #3
    Hint:
    Force = mass * acceleration
    (Be careful what value you take for the acceleration. Remember acceleration is a vector)
     
  5. Oct 23, 2012 #4

    PhanthomJay

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    Homework Helper
    Gold Member

    Welcome to PF!
    No, this method of calculating μ only applies when the object is moving at constant velocity.
    You have the correct equation for the normal force, but you don't need it in this problem. Apply newton's 2nd law parallel to the incline after first identifying all the forces acting on the car. The friction force is one of the forces acting up the incline, what is the other force acting on the car down the incline? To solve for the friction force, you don't need to calculate the friction coefficient first...you can solve for it later if the problem asks for it. Watch signage.
     
  6. Oct 23, 2012 #5
    sorry i have difficulty explaining thought processes.

    i did (m*g*sin)/(m*g*cosθ) after eliminating the m and the g from both sides im left with sinθ/cosθ. which is the same as tanθ


    Fs+Ff=Fnet=ma

    where Fs is the force of x so Fs = (1970)(-9.8)sinθ =-2686.88 (got something different here)
    then Fnet = 1970(-3) = -5910N

    5910- (-2686.88) so i got 8596.88? i dont even know anymore. going on 6 hours trying to wrap my head around this. starting to get delirious i will be back with fresh eyes in the early morning
     
  7. Oct 23, 2012 #6
    (m*g*sinθ)/(m*g*cosθ) is simply the ratio of force parallel to the slope to force perpendicular to it. It makes sense that this would be equal to the grade(i.e. also the ratio of the height/length of the slope) but it doesn't really do anything for you in this problem since you already know the angle of the slope.

    You are actually correct. That was my mistake, I boneheadedly used the grade instead of the angle. I guess you aren't the only one who needs some sleep here. :wink:

    If you check back, my previous reply will be corrected.
     
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