6
$\begingroup$

Possible Duplicate:
Does the force of kinetic friction increase with the relative speed of the objects involved? If not, why not?

Layman alert...I last did physics in high school, and am just trying to understand something my son is working on.

To keep a box moving across the floor at a constant velocity I need to apply a force equal and opposite to that of the kinetic friction force working on it.

If I start applying a force greater than the kinetic friction force, the box will speed up. Will it keep speeding up indefinitely, or will the kinetic friction force I need to overcome increase as the box gets faster?

$\endgroup$
2
  • $\begingroup$ Well, the exposition of friction that is usually done at the high school level is the simplest and most analytically tractable case and provides one answer to this problem (no dependence), but it may not apply to the whole regime you are asking about. $\endgroup$ Commented Jan 7, 2013 at 15:51
  • $\begingroup$ The power increases the faster you go, not force. Power is work per time. Work is force * distance. So force * velocity = power. $\endgroup$
    – atlex2
    Commented Jun 23, 2019 at 15:24

2 Answers 2

7
$\begingroup$

No for "dry", yes for "wet".

For "dry friction", such as a box on a floor, it is relatively constant. Why is this? Most objects are microscopically rough with "peaks" that move against each-other. As more pressing force is applied, the peaks deform more and the true contact area is increases proportionally. The surfaces adhere forming a bond that will take a certain amount of shear force to break. Since the molecules are moving much faster ~300m/s than the box (due to thermal vibrations) velocity will not affect how many molecules adhere (with the exception of "static friction"). However, static friction is sometimes be higher, in one explanation because the peaks have time to settle and interlock with each-other. Neglecting static friction, force is constant.

The simplest case in wet friction is two objects separated by a film of water. In this case there is zero static friction, as the thermal energy is sufficient to disrupt any static, shear-bearing water molecule structure. However, water molecules still push and pull on each-other, transferring momentum from the top to the bottom. The rate of momentum transfer i.e. "friction" grows in proportion to how much momentum is available, which in turn grows with velocity. Thus, force is linear with velocity.

However, interesting things happen when the bulk mass of the water gets important. In this case, bumps, etc on the surface push on the water creating currents that can ram into bumps on the other surface. If you double the velocity, your bumps will push twice as much water twice as fast for 4 times the force; force is quadratic to velocity. You can plug in formulas for the linear case (which depends on viscosity) and quadratic case (which depends on density) to see which one "wins" (this is roughly the Reynolds number), if there is no clear winner the answer is complex (see the Moody diagram).

Nevertheless these are approximations and the real answer could fail to follow these "rules".

$\endgroup$
3
  • $\begingroup$ "Since the molecules are moving much faster ~300m/s than the box they have plenty of time to adhere". I couldn't understand this part. Did you mean slower here? $\endgroup$ Commented Jul 10, 2014 at 6:27
  • $\begingroup$ I meant to say the box's motion does not matter, since it is much slower than the particals. $\endgroup$ Commented Jul 10, 2014 at 15:19
  • $\begingroup$ Also, I think it might be cubic to velocity for a rotating shaft when bulk mass is significant. If you think about a cam shaft that rotates- dispensing oil to lubricate its bearing (like an automobile engine)- then the rate of oil flow is probably proportionate to a centrifugal force, $f_c = w^2*r$. Then using $f_{friction}=\frac{d m}{dt}*wr$ for the force of a cam shaft oil momentum transfer, you end up with $f_{friction} \approx w^3*r^2$ $\endgroup$
    – atlex2
    Commented Jun 23, 2019 at 14:39
0
$\begingroup$

For "low" speeds the kinetic friction can be considered constant [1]. I am however unsure what effects dominate at higher speeds. If there's air around it then there's for example drag which causes a force proportional to $v^2$[2].

[1] http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html#kin

[2] http://en.wikipedia.org/wiki/Drag_(physics)#Drag_at_high_velocity

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.