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Kinetic friction

Once the static friction maximum is exceeded, the surface involved in a problem begins to slip, and we have a problem involving kinetic friction.
If there is slipping between surfaces, then the kinetic frictional force is given by

where  is the coefficient of kinetic friction, N is the normal force, and the direction of the force is parallel to the surface in opposition to the slipping. In general  is less than  so once an object is moving, the force of friction is less than the maximum friction when the object is still.


You might have thought that, the faster an object slides, the more friction it experiences. This is not true. (It is true for air resistance, but not friction.)
Also, you might have thought that there would be more friction for an object with more surface touching (see the second picture in Figure 6-7), but again, this is not true. The friction depends only on the coefficient of friction and the normal force.

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Figure 6-7


Brad pushes a stove (100 kg) in a straight path across the level floor at constant speed 0.2 m/s. The coefficient of kinetic friction is 0.3 for the stove and the floor. What is the force that Brad must apply?


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Figure 6-8a Figure 6-8b
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Figure 6-8c Figure 6-8d


First, we DRAW A DIAGRAM (Figure 6-8a). The words “constant speed” and “straight path” should send bells off in our head. There is no acceleration, so the vertical equation becomes,


N – mg = (Fnet)y = 0


N = mg = (100 kg)(10 m/s2) = 1000 N


Equation (2) gives the friction


FkµkN = 0.3(1000 N) = 300 N


The horizontal equation becomes


FB – Fk = (Fnet)x = 0

FB = Fk = 300 N


Think about this. Brad’s pushing force is equal in magnitude to the frictional force.


“But wait a minute!” some readers will cry. “Doesn’t Brad have to overcome the force of friction for the stove to be moving? Brad’s force must be greater than the frictional force!” But that is exactly not the case. If the stove is moving at constant speed, then the forces must balance. If Brad exceeded the force of friction, the stove would be accelerating.


Perhaps it would help if we looked at the whole Brad/stove story. When Brad approaches the stove, the force diagram on the stove looks like Figure 6-8b. As Brad begins to push on the stove, the friction vector gets larger, as in Figure 6-8c. The moment Brad exceeds Fs,max, the stove budges, and the force of friction shrinks from Fs,max to Fk. Now there is a net force on the stove, and it accelerates from rest (Figure 6-8d). Once the stove is moving, it gets away (a little) from his hands, and Brad’s force decreases to become Fk. At this point the stove has attained some constant speed, which it keeps. See Figure 6-8a, where the two horizontal vectors are equal in magnitude.



A student is pushing a chalk eraser (0.1 kg) across a level desk by applying a force 0.3 Newtons at an angle directed downward, but 30˚ from the horizontal. The eraser is moving at constant speed 0.1 m/s across the desk (in a straight line). What is the coefficient of friction between the eraser and the desk?



First, we DRAW A DIAGRAM (Figure 6-9). Constant velocity tells us that (Fnet)x and (Fnet)y  are zero. The vertical equation becomes


N – Ty – mg = (Fnet)y

N = Ty + mg = Tsin 30 + mg

N = (0.3 N)sin 30 + (0.1 kg)(10 m/s2)

N = 1.15 N.

The horizontal equation becomes


Tx – Fk = (Fnet)x = 0

Fk = Tx = Tcos 30 = 0.26 N


Thus we can calculate µk


µk = Fk/N = 0.26N/1.15N = 0.23


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Figure 6-9



A car (1000 kg) is traveling downhill at 20 m/s in the rain. The grade of the road is 20%, which means that for every 100 meters of road, the vertical drop is 20 meters. The driver sees Bambi in the road and slams on the brakes. The coefficient of kinetic friction between the tires and the road is 0.5. How much time does it take the car to skid to a halt?
(Hint: If
θ is the angle between the horizontal and the road, then cosθ = 0.98 and sinθ = 0.2. Also g = 10 m/s2.)



Here we will merely sketch a solution. You should try to work out the details. Figure 6-10 shows the force diagram. Working out the vertical equation with (Fnet)y = 0 gives


N = 9800 N.

Kinetic friction is then


Fk = 4900 N.

Working out the horizontal equation (there is a net force) gives


(Fnet)x = 2900 N

ax = 2.9 m/s2

Using the acceleration, the initial velocity, and the final velocity v2x = 0, we obtain


t = 7 s.

(Bambi was unscathed, but only because he jumped off the road in time.)


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Figure 6-10


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