Friction
Definition Friction is just a force that tries to keep objects from moving. It occurs between materials in contact with each other when they start to slide past each other. But, until the motion begins, friction is not evident.
Types of Friction Between Solids
a) Static (starting) friction is the force preventing motionless surfaces from starting to slide past each
other.
b) If two objects are already sliding past each other, sliding (kinetic) friction is the force required to keep
their relative motion constant.
c) Rolling friction is the force required to keep one object rolling over a surface at constant speed.
Rolling friction is always less than sliding friction.
Friction Summary
1. The frictional force is parallel to the two surfaces in contact with each other and it operates in the
direction that tries to keep them from moving past each other. For example, if you try to push a chair to
the right across the floor, the static friction between the chair legs and the floor pushes to the left, back
at you, trying to stop the chair moving.
2. The amount of frictional force depends on the nature of the materials in contact with each other and
the smoothness of their surfaces. The coefficient of friction, μ, indicates the degree to which the two
surfaces interlock. Actually, if the two surfaces are extremely smooth, friction can actually increase
because so much of the two surfaces is in contact. In space, excessive frictional attraction between two
materials may create a phenomenon called “cold welding”. This may occur when flat, smooth metallic
parts are touching. With no dust layer coating the surfaces, the electrostatic attraction between atoms
in the surfaces may be strong enough to require considerable force, perhaps hammering, to separate
them.
3. Starting friction is always more than sliding friction because the motionless surfaces are interlocked.
But, once the materials are sliding, their microscopic hills and valleys have less chance to settle into
each other and lock up. Imagine a muddy field dotted with closely spaced boulders. Stepping from
boulder to boulder would be an efficient way of crossing the field – no sinking into the mud and so,
much less friction. This type of motion effectively reduces the μ by allowing the surfaces to seem
smoother.
4. The amount of friction is essentially unrelated to the area of contact between the two surfaces. You
still have the same two materials with the same surface textures and so the friction is the same. An
example that is sometimes used is using a spring scale to measure the force needed to break the grip of
friction as you pull a brick across a table. The reading on the scale will be much the same regardless of
which of the brick’s faces is in contact with the table. A bit of care is needed with this type of situation
if the object’s dimensions are very unequal. If it is quite long for example, standing it on end will
concentrate all its weight into a rather small footprint, creating a large downward pressure, i.e., a large
downward force. Then, the force needed to break the grip of friction will be more than when the object
is resting on its side. Why? Because on its side, its weight is spread out over a much larger footprint
which generates less downward pressure, less downward force, and therefore less force is needed to
break the grip of friction. Remember, the nature of the surfaces in contact with each other determines
the μ. The amount of force needed to break friction and the force jamming the two materials together
must remain in the same ratio to each other, equal to the μ.
5. As just noted above, the amount of frictional force is directly proportional to the normal force pushing
the surfaces together. The more forcefully the two surfaces are jammed together, the more energy it
takes to get them sliding past each other. It takes more force to slide a full crate across a floor than an
empty one. But remember, the μ is unchanged – the same floor and the same crate, loaded or full.
6. As motion begins, the μ decreases as the situation changes from static to sliding friction and the force
needed to keep the motion going is less than to get it started. But, once the object is under way,
increasing the speed has no further effect on the μ and so has no more effect on the frictional force.
This is true so long as the speed is not too high. Extreme speeds may cause enough frictional heating to
result in a melt zone where the two surfaces are in contact and this moves the situation into that
of liquid friction.
Some Selected μ Values
Surfaces Static Sliding
Steel on steel 0.74 0.57
Wood on wood 0.50 0.30
Rubber tire (dry road) 0.70
Rubber tire (wet road) 0.50
Friction Calculations
1. Horizontal Surfaces
Q. a) Calculate the force required to start a 30 N piece of steel moving along a horizontal steel surface.
b) Once the steel is sliding, what force is then required to keep it moving along at a constant speed?
c) What would happen to the block in b) if 25 N were applied to it?
Use the following equation: μ = F/N. The F is sometimes written as FF or FR. The FF stands for “the
frictional force to overcome”, and FR stands for “the required force to overcome friction”. Use any one of them – it just depends on the wording of the question. Also, FN is sometimes written as just N. In either case, it means the force directed straight into (normal) to the supporting surface. When the supporting surface is horizontal, the N (or FN) is just the object’s weight. Now, the solution.
a) μ = FR or FR = μ * FN = 0.74 * 30 N = 22.2 N
FN
b) μ = FR or FR = μ * FN = 0.57 * 30 N = 17.1 N Notice we use the μ for sliding friction.
FN
c) Any excess force causes the block to accelerate, after the F gets it moving first.
i) change the object’s weight to its mass using W = mg; m = W/g = 30 N / 9.8 m/s2 = 3.06 kg
ii) find the excess force; 25 N – 17.1 N = 7.9 N
iii) now, find the acceleration; F = ma or a = F / m = 7.9 N / 3.06 kg = 2.58 m/s2
2 A. Inclined Surfaces: Descending Motion, Only Gravity
Q. A 20 N wooden block just slides down a 170 wooden incline at constant speed. Find the coefficient of
friction of the two surfaces.
The thing to understand is that the object’s weight acts straight down but the geometry of the situation
creates two components of the weight. One component pushes the block straight into the surface of the
ramp so this component is the FN. The other component of the block’s weight, F∥, is directed down the
ramp, parallel to its surface. This component tries to move the block down the ramp but the friction force (F) works against it. In this problem, the F∥ is successful; no other push or pull is required to get the block moving down and so the F∥ is large enough to be called F. Of course, if the object does not move, the F∥ is too small to be called F because F is the amount of force needed to cause motion. Just because F∥ and F are both parallel to the ramp does not mean the F∥ will be always be successful in breaking the grip of friction and moving the object.
The geometry of ramp problems shows that, as the ramp angle increases, the FN decreases and the F∥ increases. Recall that the μ determines the relative sizes of the F and FN – whenever the FN lessens, so does the F. So, at steep angles both the FN and the F become less but the F∥ becomes larger. At ramp angles beyond a certain steepness, the F∥ will be larger than F and so, down slides the object. Whenever an object on a ramp slides down on its own, we know the F∥ component of the object’s weight is larger than the F trying to keep it from moving.
a) Find the weight components; FN = cos 170 * 20 N = 0.956 * 20 N = 19.12 N
F∥ = sin 170 * 20 N = 0.292 * 20 N = 5.85 N
b) The only force acting on the object to overcome friction is F∥. Because the object slides at a constant
rate, we know F∥ is just equal to the friction force. And so, we can substitute F∥ for F and write that
μ = F∥ / FN. So then, μ = 5.85 N / 19.12 N = 0.306.
2 B. Inclined Surfaces: Descending Motion, Gravity Plus Pushing
Q. A 45 N crate sits on a 100 ramp. If the μ is 0.5, what force is required to push it down the ramp?
Because the crate is not moving, we know the F∥ is not large enough to overcome F and therefore an
additional push is necessary.
a) Find the weight components; FN = cos 100 * 45 N = 0.985 * 45 N = 44.31 N
F∥ = sin 100 * 45 N = 0.174 * 45 N = 7.83 N
b) Find the friction force to be overcome; μ = F / FN or F = μ * FN = 0.5 * 44.31 N = 22.15 N
c) Find the extra push required. We know the friction force (F) keeps the crate locked on the ramp’s
surface with a grip of 22.15 N. But remember, the object’s F∥ weight component is already trying to
push it down-ramp with 7.83 N. So, we need an additional push of: 22.15 N – 7.83 N = 14.32 N.
3. Inclined Surfaces: Ascending Motion
Q. What force is required to start a 30 N wooden block up a 150 wooden incline at constant speed? (The
block is stationary and the static μ from the list is 0.5.)
Again, the block’s weight must be resolved into the FN and F∥ components. Also, since the block is being moved up the incline, two forces must be overcome – friction (F) and the block’s F∥ weight component. Why two forces to overcome? Friction must always be overcome. And, in this problem, the block is being dragged uphill so its own F∥ acts against it. The F∥ is always downslope – last time it helped because the object’s motion was downslope. But, this time it hinders because the object’s motion is upslope ands so it acts against it.
a) Find the weight components; FN = cos 150 * 30 N = 0.966 * 30 N = 28.98 N
F∥ = sin 150 * 30 N = 0.259 * 30 N = 7.77 N
b) Find the friction force to be overcome; μ = F / FN or F = μ * FN = 0.5 * 28.98 N = 11.49 N
c) Find the combined forces to overcome, F∥ + F; 7.77 N + 11.49 N = 19.26 N
4. Horizontal Surface: Pulling Up on Handle
Q. A 27 N sled is pulled across the snow (μ = 0.2) by a child who pulls up on its tow rope with 50 N. If
the rope angles up at 380 and the child’s force is just sufficient to move the sled, what weight can the
child drag on the it?
The child must pull on the rope with more force than if all her efforts were directed horizontally. Because
of the geometry of the situation, only part of her energy, only the FH component, is available to try to move the sled. The upward component of her pulling force, the FV component, lessens the sled’s weight so more “cargo” can be pulled. This last point is the reason you can pull more weight than you can push.
a) Vertical force component (FVERTICAL or FV) = sin 380 * 50 N = 0.616 * 50 N = 30.8 N
Horizontal force component (FHORIZONTAL or FH) = cos 380 * 50 N = 0.788 * 50 N = 39.4 N
b) Because we are told the child can just pull the sled, we know the FH component of her pulling force
will be enough to overcome friction and so, FH = F = 39.4 N.
c) To find the total weight the child’s FH can move, we find the FN. This value is the total of the sled and
its cargo. FN = F / μ = 39.4 N / 0.2 = 197 N. By the way, notice that because the snow is slick, a large
FN (197 N) can be moved with a small FH (39.4 N.)
d) The sled weighs only 27 N but 197 N can be moved. So, to find the cargo, we subtract:
197 N – 27 N = 170 N of cargo. But, recall that the child’s FV reduces the sled’s weight by 30.8 N.
So, another 30.8 N can be loaded onto the sled. Now, the total cargo is 170 N + 30.8 N = 200.8 N.
5. Horizontal Surface: Pushing Down on Handle
Q. A gardener moves a 2 N hand plow across a patch of ground (μ = 1.3) by pushing down its handle
with 37 N. If the handle dips at 240 and the gardener’s force is just sufficient to move the plow, what
weight of soil (cargo) can it carry?
Because the force is at an angle, the person must push with more force than if her efforts were all
horizontal because only the FH component of her effort is moving the plow. The downward component of her pushing force, the FV component, increases the effective weight of the plow and so the amount of “cargo” her FH force component can move is reduced. In contrast to pulling, pushing lessens how much we can move.
a) Vertical force component (FVERTICAL or FV) = sin 240 * 37 N = 0.407 * 37 N = 15.06 N
Horizontal force component (FHORIZONTAL or FH) = cos 240 * 37 N = 0.914 * 37 N = 33.82 N
b) Because the gardener’s force is just sufficient to move the plow, we know the FH of her pushing force
will equal friction and so FH = F = 33.82 N. To find the total weight that can be moved, use
FN = F / μ = 3.82 N / 1.3 = 26.02 N. See that because the soil is rough and friction is large, it takes a
large FH (33.82 N) to move a smaller weight (26.02 N).
d) Although 26.01 N can be moved by the girl’s FH component, the plow weighs just 2 N and so the
amount of soil that can be moved is: 26.01 N – 2 N = 24.02 N. But, recall the FV component of the
girl’s pushing force increases the plow’s weight by another 15.06 N. This further reduces the amount
of soil that can be moved: 24.02 N – 15.06 N = 8.96 N. Note: If the soil were rougher or the girl’s
pushing angle were steeper, there would be so much friction that her FV would not be enough to move
even the plow itself! In such a case, this last calculation, the amount of soil that could be moved,
would yield a negative value.