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Given Maximum Acceleration calculate angle θ where Fnet=0

  1. Oct 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Given the max acceleration of a car, 7.6m/s2, calculate the angle of the slope where the cars wheels will spin out. (Fnet=0).

    2. Relevant equations
    Fnet=ma
    Ffr=μFn



    3. The attempt at a solution

    Since I'm not given μ or anything, I think I'll have to figure that out, and I guess here goes my attempt...
    I'm guessing that since there's a force of friction pulling the car back when it accelerates, there should be one as it decelerates. This should be the same coefficient, μ.
    I have no idea what to do from there, but I tried a few things.
    -> i assumed that the max accel is equal to the deceleration if the tires were to lock, so I could actually use that give acceleration.
    So:
    Fnet= Ffr =ma
    ma = μFn
    ma = μmg
    a=μg (since both sides have m, I can cancel it out)
    a/g = μ
    7.6 / 9.8 = 0.77

    Now, I would plug this into the formula for the situation involving the gravity, friction, normal force and acceleration of the car on the slope and solve for my angle θ. Where the Fnet in both Y and X directions (axis tilted to match slope) is equal to 0. (therefore tires spinning out)

    ƩFy = Fn - Fgy = 0
    Fn = Fgy = mgcosθ

    ƩFx = Fa-Ffr-Fgx = 0
    =Fa-μmgcosθ-mgsinθ
    I've continued with the algebra, but I get stuck because I cannot get rid of Fa or isolate the angle.

    (Fa)/mg = μcosθ + sinθ (I moved μmgcosθ and mgsinθ over to the other side and factored out the mg, then divded Fa by mg.

    If I divide the equation by μ, it ends up dividing sinθ... and I end up multiplying everything by μ which puts me back where I started.

    I was thinking of substituting Fa with ma...but I can't do that since I cant prove why im doing it! Even if I do it, ill end up with

    Fnet= 0 = ma -μmgcosθ-mgsinθ = a-μgcosθ-gsinθ (m's cancel)
    move μgcosθ-gsinθ over and factor out the g, then divide the equation by g

    a/g = μcosθ+sinθ hmmm... I tried to somehow get sine/cosine so I could just get tanθ, but that a/g would never allow that...

    So... any ideas? this one's been bothering me all day.
     
  2. jcsd
  3. Oct 15, 2012 #2
    Was the mass of the car not given to you?
    Is the car going up a slope or down the slope?

    I believe you solved your mu correctly, but without the mass you cannot calculate the normal force (and therefore the frictional force) of/acting on the car, and neither can you solve for the Fd in this diagram.

    incline2.jpg

    To find the angle at which Fnet = 0, you would set Fd + Ffriction to -7.6 m/s^2, or 7.6m/s^2, depending on the direction you are looking at this from. And to do this, you would need to know the mass of the car.

    Correct me if I'm wrong, and I apologize in advance for inevitably having made a mistake in my reasoning.
     
  4. Oct 15, 2012 #3
    The car is attempting to move up the slope, but the angle is such that all the forces cancel out and the car cannot move/accelerate up the slope. I'll double check and see if the mass was given to me somewhere, but I don't recall it being there. I did end up cancelling the mass out of all my attempted equations, so I dont think ill need it?
     
  5. Oct 16, 2012 #4
    I figured it out, so I guess if anybody was wondering I'll just sum up what I did.
    Since the way the car moves is because friction with the ground, on the free body diagram, the net force in the x direction is the Ffr (car trying to accelerate) - Fgx (gravity in x direction on the tilted axis) =0
    *there is no friction pulling the car back down the slope, because this is actually what the car uses to propel itself forward*
    therefor Ffr=Fgx
    Ffr is uFn, and we can use the value of u we calculated earlier. Fn is y component of Fg: Fgy
    so:
    umgcosθ = mgsinθ

    ucosθ=sinθ
    cosθ/sinθ = u
    tanθ = u
    take inverse tan of u and you get the answer!
     
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