# Homework Help: Car on an incline, friction related question

1. Mar 21, 2009

### nugget

1. The problem statement, all variables and given/known data
Car is driving up a slope, 16o to the horizontal, trying to accelerate as much as possible. Static friction coefficient = 0.9. Kinetic Friction coefficient = 0.8. Find the magnitude of the maximum possible acceleration (assuming a powerful enough engine).

2. Relevant equations

F=ma
F(friction) = (static friction coefficient) x Normal force

Normal force = mgcos(16)

3. The attempt at a solution

I drew a force diagram and worked out the Engine force required to overcome the gravity and friction forces and keep the car in constant velocity:

No mass is given so i canceled out the masses and found acc. to be 11.2 m/s-2. I used the static friction coefficient because I was told that the wheels have to be skidding for me to use the kinetic friction coefficient.

Can anybody help me with this question? I have a value for acceleration but I think I am doing something wrong, otherwise the question wouldn't ask for a maximum value.

2. Mar 21, 2009

### Delphi51

You get the maximum acceleration when the wheel is just at the point of slipping. Try to get more and the coefficient of friction slips to the kinetic value - and you get less.
My older cousin taught me that when I couldn't drive my Mom's car up the icey driveway. He put it in second gear (manual transmission) where it had less power and it crawled up the driveway.

I get a smaller value for acceleration. Perhaps you should share your calculation so we can see how it worked out. Did you subtract the component of the weight down the ramp from the friction force?

3. Mar 21, 2009

### nugget

Initially i thought:

Force from the cars engine has to be equal to the force down the ramp due to its own gravity + the force down the ramp due to friction.

So I wrote: Fengine = Ffriction + Fcomponent of weight downhill

ma = μsN + mgsin(16)
therefore, ma = 0.9x(mgcos(16)) + mgsin(16)

but using this i get a = 11.18 m/s^2

4. Mar 21, 2009

### Delphi51

Close. You must think of the friction force in a slightly different way for this question. The friction force is the maximum grip the car has on the road. The car pushes downhill on the road with this force, and the road pushes the car uphill with the same force (a pair of Newton's 3rd Law forces). So the force up the hill is μs*N. That mg*sin(16) pulls the car down the hill.
ma = μs*N - mg*sin(16). (Here g is a positive 9.81.)