- #1
psychochef
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Homework Statement
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).
Homework Equations
Fnet=ma
Ffr=μFn
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 can't prove why I am 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.