free-body-diagramfrictionhomework-and-exercisesnewtonian-mechanics
I'm working on a sample problem and it asks on how steep of an incline can a car park? From what I learned the friction is in the opposite direction if there was motion in said friction-less environment. Is this the right way to think of friction?
In the case of the car; it would move down the hill and the friction would be up the hill. Why is the friction pointing down the incline and not up the incline?
I stumbled on a reference which summarizes what I think is the most correct answer to this question. You have several options, but I think we should only use terms that we have a physical reason to write. Here's the reference:
http://usna.edu/Users/physics/schneide/Buick.htm
They use a lot of unnecessary details like time between stopping that we're not interested in. I agree with their graph but not their equation. So here is my equation to explain their graph. I also limited it to flat roads (no hills).
$$F = \frac{P_0}{v} + \mu mg + c \rho A \frac{v^2}{2} $$
From the reference, known values about their car are:
I report these because there is no direct measurement available. I would then use their data to evaluate the three coefficients that determine the relative friction from each thing.
They put a constant term in the transmission factor (the 1/v term). I don't like this, because I want clean mathematics, so I'm bunching that long tail of transmission with the friction coefficient.
$$ \mu (1800 kg) (9.8 m/s^2) = 200 N + 250 N = 450 N \rightarrow \mu = 0.026 $$
$$ \frac{P_0}{15 m/s} = 500 N - 250 N = 250 N \rightarrow P_0 = 3750 W$$
$$ c (3 m^2) (1.3 kg/m^3) \frac{(31 m/s)^2}{2} = 500 N \rightarrow c = 0.27 $$
These are all consistent with what the link claimed, aside from the cases where I willfully used a different kind of definition. Another good thing that these all have physical interpretations, which those units are suggesting. I will avoid getting into the exact interpretation because I feel like there's space for quibbling.
I plotted this on Wolfram alpha. This is my altered version of that link.
This fits expectations fairly well. Take note, however, of the 1/v term. That represents fuel consumption due to constant loads (thus, units of power, of course). That might not be relevant if you're looking for a force, but it can be kind of interpreted as a force. It's a force the engine is exerting against itself (to some fraction of that number) due to idling. It is also the constant electronic loads on the battery... and the charging of the battery itself. It's not the friction of the wheels on the road of air on the car. If you're only interested in those then you might do well to just take the last two terms. If you do that, however, there is no concept of maximum gas mileage, nor should there be. What you're going to do with these terms depends on the application. I just believe this to be the best available option so I posted it.
I had fun trying to make this as intuitive as possible. I hope I've succeeded without doing the physics of the situation much injustice.
When a car is driving straight ahead, the plane in which the wheels are rotating is aligned with the direction of movement. Another way of saying this is that the rotation axis is perpendicular to the momentum vector $\vec{p}=m\vec{v}$ of the car. So the friction merely makes it harder for the car to move, which is part of the reason why you need to put your foot on the gas pedal to maintain a constant speed. At the same time, the friction is what allows you to maintain that constant speed because the rotating tires sort of grab onto the ground, which is the intuitive picture of friction. The tires grab the ground and pull/push it backwards beneath themselves, as you would do when dragging yourself over the floor (if it had handles to grab onto). Those grabbing and pulling/pushing forces are what keeps you going.
Things change when the wheels are turned. The plane in which they are rotating now is at an angle with the direction of motion. Alternatively but equivalently, we could say the rotation axis now makes an angle with the momentum vector of the car. To see how friction then makes the car turn, think again in terms of the wheels grabbing onto the ground. The fact that they now make an angle with the direction of motion, means the force the tires are exerting is also at an angle with the direction of motion - or equivalently, the momentum vector.
Now, a force is a change in momentum$^1$ and so (because the wheels are part of the rigid body that is a car) this force will change the direction of the car's momentum vector until it is aligned with the exerted force. Imagine dragging yourself forward on a straight line of handles on the floor and then suddenly grabbing hold of a handle slightly to one side instead of the one straight ahead. You'll steer yourself away from the original direction in which you were headed.
$^1$ Mathematically: $$\vec{F}=\frac{d\vec{p}}{dt}$$
Best Answer
You're right that friction points up the hill. What happens when you solve this is that you get a friction force that's negative. A negative force pointing down the hill is the same as a positive force pointing up the hill, so everything works out okay. It would have been more clear if the diagram author showed the friction vector pointing uphill to begin with, though.