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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?

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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.

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  • $\begingroup$ A variation on joshphysic's expression which might be useful (I've no idea whether it is, because I haven't ever thought about doing problems like this in quite these terms before) is $\vec{f}_k = \mu_k \vec{F}_n\wedge \frac{\vec{F}_n \wedge \vec{v}}{F_n\,v}$. $\endgroup$ – WetSavannaAnimal Sep 16 '13 at 1:11
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That is a great observation regarding the vector nature of friction that is not treated in the text; however; the notation in the book is strictly correct. Since the equation in the book does not include harpoons above either force, you can accept the equation as providing only the magnitude of the force of friction, based on the magnitude of the normal force. This is somewhat common; one suggestion given by many physics teachers and texts is to use scalars when possible in order to reduce the complexity of having to solve vector equations.

As we know, the direction is simply opposite the direction of motion (for kinetic friction; it is opposite the direction the object would be moving for static friction). If you are working on a free body diagram, you don't need to complicate the equation to include the direction since you draw the arrow to indicate the direction. If you have 1-D motion, your need to indicate direction is minimal; it will either be positive or negative. In general, if you need all mathematical rigor in 2- or 3-D motion, you can simply put $-\hat{d}$ on it.

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