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!
Best Answer
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,
$$f_{s}\le \mu_{s}N$$
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.