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I am trying to understand the nature and direction of friction force acting on from the ground a car driving up an incline.

I was thinking about a simple situation where a car of mass $ m $ is driving up an $ \alpha = 10 ^{\circ} $ with constant speed $ v $. Let's neglect the air resistance to simplify things.

All right, so down the incline we should have: $mg \cdot \text{sin}(\alpha)$ and friction force $ \mu \cdot mg \cdot \text{cos}(\alpha) $.

The thrust force that acts on the car from the ground (supposedly friction as well?) should have magnitude $T=mg \cdot \text{sin}(\alpha) + \mu \cdot mg \cdot \text{cos}(\alpha) $.

Here's my problem. If the thrust force acting on the car has friction-like nature, then shouldn't $T=\mu \cdot mg \cdot \text{cos}(\alpha)$ ? Which gives $mg \cdot \text{sin}(\alpha)=0$ and that is of course nonsense.

Please tell me, what are my misconceptions here? Nudge me in the right direction :)

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Mistake one: The friction force is only $\mu N$ ( aka $\mu mg \: \text{cos}(\theta)$ for a ramp) if the friction is kinetic (wheels are spinning out) or if you are at your maximum value of static friction (wheels are $\textit{about}$ to start sliding out). So for a car driving up an incline the best you can say is the friction is some value $F$. $$ $$ Mistake two: The friction force of the ground on the tires $\textit{is}$ the force that propels the car up the ramp! The drive shaft attempts to rotate the tires against the ground, but the ground and tire interface resists slipping, so the friction force actually points up the incline and provides the upward force driving the motion (sometimes this is called "traction" as opposed to friction to let you know what role it has in the mechanics, but it is fundamentally the same force). So, the friction force $F$ is some value that depends on how hard the driveshaft is trying to spin the wheels, up until the point when the static friction maxes out and it becomes kinetic friction (wheel slipping on the ground).

$$ $$ Clarifying remark: There is definitely going to be some rolling friction that points down the incline, but it has its own coefficient/ modeling that is different from the sliding friction.

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  • $\begingroup$ Much appreciated Mike :) I organized everything in my head. So in my example traction force would be just $T=mg \cdot \text{sin} (\alpha) + F$ assuming rolling without slipping, right? And $0 <F \le \mu_{static} \cdot mg \cdot cos(\alpha) $. I think in many problem books authors create problems with cars speeding up or slowing down without specifying the details and thus my misconception arose. $\endgroup$
    – Lukasz Skowron
    Commented Jun 9 at 0:20
  • $\begingroup$ If we have rolling without slipping, then yes, the maximum value of the required friction is $\mu mg \: \text{cos}(\theta)$, and for the car to move up the ramp without slowing down and coming to a stop the traction must at minimum be equal to $mg \: \text{sin}(\theta)$, plus any downward rolling friction or air resistance. Mechanics can have some slippery concepts, especially when wheels and active motion using friction to propel bodies! $\endgroup$
    – mike1994
    Commented Jun 9 at 1:33
  • $\begingroup$ slippery concepts haha! $\endgroup$
    – Lukasz Skowron
    Commented Jun 9 at 15:32

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