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Friction

\begin{figure}\centerline{\epsfbox{dynamics/dynamics.7.eps}}
\end{figure}

Experimentally

  1. $f_{s,{\rm max}} \le \mu_s \left\vert N \right\vert$. The force exerted by static friction is less than or equal to the coefficient of static friction times the magnitude of the normal force exerted on the entire (homogeneous) surface of contact. It opposes any applied force to make the total force zero as long as it is able to do so.
  2. $f_k = \mu_k \left\vert N \right\vert$. The force exerted by kinetic friction (produced by two surfaces rubbing against each other in motion) is equal to the coefficient of kinetic friction times the magnitude of the normal force exerted on the entire (homogeneous) surface of contact. It opposes the direction of the motion of the two surfaces in such a way as to decrease the velocity along the surface of contact.
  3. $\mu_k < \mu_s$
  4. $\mu_k(v)$, but for ``slow'' speeds $\mu_k \sim$ constant.
  5. $\mu_s$ and $\mu_k$ depend on materials, but independent of contact area.

We can understand this last observation by noting that the frictional force should depend on the pressure (the normal force/area $\equiv$ N/m$^2$) times the area in contact. But then

\begin{displaymath}
f_k = \mu_k P \times A = \mu_k \frac{N}{A}\times A = \mu_k N
\end{displaymath} (2.34)

and we see that the frictional force will depend only on the total force, not the area.


next up previous contents
Next: Examples Up: Friction and Drag Forces Previous: Friction and Drag Forces   Contents
Robert G. Brown 2008-01-29