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Some people said that moving a car on the moon would make the tires spin a bit more compared to Earth's. But then I started thinking that if static friction and gravity are perpendicular to each other, how can gravity affect the 'more' spinning on the wheels? Since the forces act perpendicularly, shouldn't they be independent of each other no matter what?

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    $\begingroup$ It would probably be better if you wrote your question and multiple sentences, instead of the single run on sentence it is now, (and capitalised your "i" s!) Also, I wasn't the guy who downvoted: this is a good question, it just needs to be formatted a bit nicer. $\endgroup$ Oct 13, 2018 at 9:17

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The force of friction depends on the force that acts perpendicular to the surface. This is called the normal force $F_N$, and it basically just describes how strongly the car is being pushed into the inclined plane by its own weight. The force of friction $F_F$ is directly related to the normal force: $$F_F = \mu F_N$$

$\mu$ here is the coefficient of friction, it's a number between 0 (very strong friction, body doesn't move at all) and 1 (body moves without friction).

How do we calculate the normal force? If you have a body sitting on a flat surface, the force of gravity $F_G = mg$ acts in the same direction as the normal force - perpendicular to the surface, i.e. straight downwards. In such a case, $F_N = F_G$.

On an inclined plane, the force of gravity still goes downwards, but it is not in the same direction as the normal of the plane. Instead, $F_N$ and $F_G$ form an angle (the same angle as the inclination $\alpha$). In such a case we can write: $$F_N = cos(\alpha) \cdot F_G$$

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So, as you can see, if the normal force is lower, also the friction decreases. Under what circumstances is the normal force lower?

For once, we can increase the angle $\alpha$: we make the plane more inclined, which in turn means that $\cos(\alpha)$ approaches 0 as soon as we get closer to 90°. In the limiting case of 90°, the normal force is 0 and we get frictionless...falling.

The other way is to decrease the force of gravity: $F_G = mg$. If we keep the mass $m$ constant, the only other option is to decrease $g$, the acceleration due to gravity. On Earth this is also fixed at $9.81 m/s^2$, but on the moon it has a much lower value of $1.63 m/s^2$, or about a sixth of Earth's. With this new value, at the same angle and mass, the normal force is now lower, and so is the force of friction.

$$F_F = \mu \cos(\alpha) m \cdot g_{moon} < \mu \cos(\alpha) m \cdot g_{Earth}$$


Now what about the tires? In the case of non-slipping tires, the contact point where the wheel touches the ground should be stationary momentarily. If it did move, we would see that the point is being dragged across the ground.

How fast does the wheel need to spin then to not slip? The rotation of the tire needs to be matched to the forward speed of the car. During one rotation the center of the wheel moves $$x_{center} = 2\pi r = x_{car}$$ where $r$ is the radius of the tire. We can divide both sides by the time of one rotation to get expressions for the velocity. $$v_{center} = \omega r$$ where $\omega$ is the angular velocity of the tire. Finally, we reach an expression (through differentiation with respect to time) involving accelerations: $$a_{center} = a_{car} = \alpha r$$

What forces act on the wheel? We have a force $F$ acting forwards on the wheel itself (at the center), and a force of friction $F_F$ acting at the contact point. This force of friction also depends on the coefficient of friction and the weight of the car, which is where gravity comes in again.

With gravity being lower, the force of friction at the contact point is reduced, while the force forward is still the same. This means that the wheel does not "feel" enough friction at its contact point to gain the necessary torque (it doesn't turn fast enough), and hence slips.


TL;DR

The force of friction depends directly on the normal force and consequently on the force of gravity. A lower force of friction increases slipping of the tires (all other parameters being equal).

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