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I'm currently studying Friction $f_a$ and Normal force $N$, and I read that those two forces are nothing but the parallel and perpendicular components of a "general" force called Contact Force $C$.

In this sense, we can write

$$\vec{C} = \vec{N} + \vec{f_a}$$

So, since I never heard of a similar explanation, I started wondering about their modulus.

We know that the basic expression for the friction is

$$f_a = \mu N$$

Since $f_a$ and $N$ are the parallel and perpendicular components, I can use vector sum to define the modulus of the contact force:

$$|\vec{C}| = \sqrt{f_a^2 + N^2} = \sqrt{\mu^2 N^2 + N^2} = N\sqrt{\mu^2 + 1}$$

In the simplest case in which $N = mg$ (on a horizontal plane), we get

$$|\vec{C}| = mg\sqrt{\mu^2 + 1}$$

First question: is this meaningful or is just a wrong derivation (why?)

Second question: is there some interpretation for the square root term $\sqrt{\mu^2 + 1}$? Just asking because it looks strange.

Thank you!

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  • $\begingroup$ Note that frictional force is generally $f_a \leq \mu N$ and only in the case of 100% sliding the two are equal. $\endgroup$ Commented May 5, 2022 at 16:53
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    $\begingroup$ What happened to the $μ^2$? $\endgroup$
    – R.W. Bird
    Commented May 5, 2022 at 18:29
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    $\begingroup$ Indeed, there's an error in expression, should be $N\sqrt{\mu^{2} + 1}$ $\endgroup$ Commented May 5, 2022 at 18:31

2 Answers 2

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We know that the basic expression for the friction is

$$f_a = \mu N$$

That's the basic equation for kinetic friction where $f_{k}=\mu_kN$ and $\mu_k$ is the coefficient of kinetic friction.

In the case of static friction is is the expression for the maximum possible static friction where motion is impending and $\mu=\mu_s$ the coefficient of static friction. Up until the maximum possible static friction force is reached, static friction matches the applied force. That is, for static friction,

$$f_{s}\le \mu_{s}N$$

First question: is this meaningful or is just a wrong derivation (why?)

Since it is only the magnitude of the sum of two vector quantities, there is nothing wrong with it mathematically. However, if fails to distinguish between kinetic and static friction. For kinetic friction it is

$$|\vec{C}| = N\sqrt{\mu_{k}^{2} + 1}$$

But for static friction it is

$$|\vec{C}| \le N\sqrt{\mu_{s}^{2} + 1}$$

Hope this helps.

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  • $\begingroup$ Should be $\mu^2$ $\endgroup$
    – Eli
    Commented May 5, 2022 at 19:13
  • $\begingroup$ @Eli Thanks. I didn't check the OPs math. $\endgroup$
    – Bob D
    Commented May 5, 2022 at 19:24
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Yes, since friction is limited by $| \vec{f}_a | \leq \mu | \vec{N}|$ you can write as a condition for stiction the following expression

$$ | \vec{C} | = | \vec{N} + \vec{f}_a | \leq \sqrt{1+\mu^2}\; | \vec{N} | $$

This is interpreted geometrically as the contact force $\vec{C}$ vector must lie inside a cone of half-angle $\theta$, such that $\mu = \tan \theta$. The axis of the cone is along $\vec{N}$.

Using a cosine to define the angle between vectors from their dot product, we have following relationship

$$ \begin{aligned}\vec{C}\cdot\vec{N} & \leq\cos\theta\,|\vec{C}||\vec{N}|\\ |\vec{N}|^{2}+\vec{f}_{a}\cdot\vec{N} & \leq\cos\theta\,|\vec{C}||\vec{N}|\\ |\vec{N}| & \leq\cos\theta\,|\vec{C}| \end{aligned}$$

If you assert that $\mu=\tan\theta$, then $\cos\theta=\frac{1}{\sqrt{1+\mu^{2}}}$ (* a lot of trig skipped here) and the above relation becomes

$$\sqrt{1+\mu^{2}}\,|\vec{N}|\leq|\vec{C}|$$

which is the same as the first equation here.


The way to use this is

  1. Find out the friction force $\vec{f}_s$ needed for two bodies together to stick together.
  2. Check if $| \vec{N} + \vec{f}_s | \leq \sqrt{1+\mu^2} | \vec{N} |$
  3. If not then the contact is slipping. Specify $|\vec{f}_a| = \mu | \vec{N} |$ and apply it in the direction that opposes the relative motion.
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  • $\begingroup$ Insighful! Thank you! $\endgroup$
    – Heidegger
    Commented May 26, 2022 at 18:21

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