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For objects resting flat with $F_N = mg$, both gravity $g$ and coefficients of static friction $\mu_s$ play into calculating friction force by:

$$F_{friction} = \mu_s F_N = \mu_s mg$$

This equation implies that gravity and coefficients of static friction are equally interchangeable in determining friction force. But here is an example where I have a hard time believing that:

$\mu_s$ between ice and ice = 0.09
$\mu_s$ between rubber and dry asphalt = 0.9

Imagine two identical blocks, one sitting on two pieces of ice on the surface of Earth (gravity = $1.0g$) and the other sitting on a piece of rubber on dry asphalt on the surface of a planet with 1/10th Earth's gravity ($0.1g$). then by:

$$F_{friction\ Earth} = \mu_s F_N = (0.09)(m)(1.0g)=0.09\ mg$$ $$F_{friction\ Other Planet} = \mu_s F_N = (0.9)(m)(0.1g)= 0.09\ mg$$

Since the friction force is the same, they should behave the same way when I push them, right? But this goes against my intuition... at 1/10th the gravity of Earth, would rubber really slide around around on dry asphalt like ice? Or do the coefficients of friction need adjustment if you switch the gravity?

Coefficients of static friction from EngineeringToolbox.com

Qmechanic
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pentane
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  • -1 Why do you think rubber would not slide like ice in low $g$ environments? What is the source of your intuition about this? – sammy gerbil Jan 08 '18 at 16:07

1 Answers1

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The coefficient of friction does not depend on the origin of the normal force $\vec{F}_N$ so also not by the gravitational field that caused it. See How is frictional force dependent on normal reaction? for how a normal force can cause a friction force perpendicular to it.

Jan Bos
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