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Through some driver's ed class i encountered a claim stating that a car traveling at 30 kmph has a minimal braking distance of 8 meters (The specific number is not important). My question is whether there is indeed a maximum force a body can have acted upon it by friction?

The naive model I know states that given a friction coefficient $\mu$ and a body of mass $m$, the friction force would be $m\cdot g\cdot \mu$. Is there a very strict upper bound on values of $\mu$?

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    $\begingroup$ You seem to mix up friction and breaking distance. Any given number of breaking distance is necessarily dependent on the body's properties, mass being among them. So the quote from your driver's class IMHO must refer to some "average" car. Plus, when referring to breaking distance, it is the rolling friction, you're interested in (see e.g. byjus.com/physics/types-of-friction), at least in a driving school. This differs e.g. from sliding friction, only very talented drivers use on purpose ;-) $\endgroup$
    – StefG
    Commented Jan 28, 2020 at 11:38
  • $\begingroup$ There was an offhand remark stating that this braking distance can not be significantly decreased, which I suspect that it is not true. I think one can improve this by saying making tires out of materials which can have greater friction while braking. $\endgroup$
    – Keen-ameteur
    Commented Jan 28, 2020 at 13:55
  • $\begingroup$ You could make a pawl and ratchet system rather than a rubber tire, and could generate much larger forces. But outside of exotic materials, you can assume you're not going to find production tires with coefficients in excess of 1.0. Even high-performance racing tires (like motorcycle and F1) won't be in excess of 2.0. $\endgroup$
    – BowlOfRed
    Commented Jan 28, 2020 at 22:20

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There is an upper bound to the amount of static friction force that can be achieved without skidding. It is given by $f_{s}=μ_{s}N$ where $μ_{s}$ is the coefficient of static friction between the surfaces, and $N$ is the normal (perpendicular) force acting on the surface (=$mg$ for a mass $m$ on a horizontal surface).

When braking the maximum force that can slow the vehicle is the static friction force, above which skidding will occur. Then you will be dealing with a kinetic friction force, which given by $f_{k}=μ_{k}N$ where $μ_{k}$ is the coefficient of kinetic friction. The coefficient of kinetic friction, $μ_{k}$ ,is generally less than $μ_{s}$ making the kinetic friction force bringing the vehicle to a stop less than the static friction force.

Hope this helps

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  • $\begingroup$ I was wondering whether one can conceivably find materials to make tires out of, which will continually better by a non-negligible amount the deceleration. $\endgroup$
    – Keen-ameteur
    Commented Jan 28, 2020 at 13:52
  • $\begingroup$ It's not just a matter of tire materials, but a host of other tire parameters that influence static friction relied on for deceleration and acceleration. Tire designers have spend many years trying to optimize traction taking into other design considerations that are needed as well. In any event, the maximum possible theoretical deceleration and acceleration equals $g$, the acceleration due to gravity. Thats because $F_{max} =ma_{max} = μ_{s}mg$, therefore $a_{max}=g$. $\endgroup$
    – Bob D
    Commented Jan 28, 2020 at 14:10
  • $\begingroup$ You seem to assume that $\mu_s \leq 1$? That is not necessarily true. $\endgroup$
    – user137289
    Commented Jan 28, 2020 at 23:05

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