I read a text on mechanics and in the chapter on friction, there written that the kinetic friction is in the form
$$f_k = \mu_k F_N$$
where $f_k$ is the kinetic friction, $\mu_k$ is the kinetic friction coefficient, $F_N=mg$ is the normal force due to the weight of the object. In another chapter, there mentioned the vector form of the force. But it is confusing that if $f_k = \mu_k F_N$, should the vector form written as $\vec{f}_k = \mu_k\vec{F}_N$? It doesn't look right to me since $\vec{F}_N$ is downward but $\vec{f}_k$ is along horizontal. So what's the right way to write the vector for, of kinetic friction? How does people know that $f_k = \mu_k F_N$? From experiment?
I am guessing that if the object is moving at the direction $\hat{\mathcal{d}}$, so the vector form of kinetic friction should be
$$
\vec{f}_k = -\mu_k(\vec{F}_n\cdot\hat{\mathcal{d}})\hat{\mathcal{d}}
$$
Is that correct?
Best Answer
You're very close. The vector giving the friction force has magnitude $\mu_k F_N$ and is opposite the direction of travel, so it can be written as the product of $\mu_k F_N$ with a unit vector pointing in the direction opposite the direction of travel. Since the velocity $\vec v$ is in the direction of travel, the unit vector $-\vec v/v$, where $v$ is the object's speed, points opposite to the direction of travel, so the force of friction can be written as \begin{align} \vec f_k = -\mu_k F_N \frac{\vec v}{v} \end{align} The issue with your last expression is that $\vec F_N\cdot \hat d = 0$ since the normal force is perpendicular to the surface, and the direction of travel is parallel to the surface.