xaktly | Physics | Mechanics


Friction is your friend (and foe)

Friction is a necessary part of life. The world would be utterly different without it; nothing would stay in place, and given the slightest push, nothing would stop. Chaos!

We rely on friction between tires and the road to corner in and stop a vehicle. The rock climber relies on friction between shoes and rock to make upward progress →

Friction can also be something to be avoided. For example, without grease and ball-bearings to reduce friction, the axle-to-wheel joint on a car or bike would heat up and deform or break after driving/riding even a very short distance. In some crucial machines we even employ magnetic levitation to keep surfaces from rubbing and heating up while they move.

The origin of friction: Atomic Level

Friction between objects arises from the always present roughness, right down to the atomic level, of even the smoothest-looking of surfaces. Even a highly-polished silicon surface like this one has bumps because it is made up of nearly-spherical atoms stacked side by side.

In this scanning tunneling microscope (STM) image, you can see atoms, ridges where fractures occurred and other deformities that are always present on surfaces viewed at such a scale.

Friction is a direct result of this roughness.

Two flavors of friction

We divide friction into two kinds, static friction and dynamic friction. The meanings are right there in the names. Static implies a friction present when objects aren't moving — that we feel when we try to get them moving, and dynamic a friction between objects moving relative to one another.

Static = not moving

Dynamic = in motion

The origin of both is roughness of surfaces at the atomic or molecular level. In the diagram below, two hypothetical surfaces, one the bottom side of a block and the other the top surface of a table, are depicted. The spheres are atoms and the roughness is typical of any real surface.

In the static case below, the objects have been in contact for a while and high points and low points of each have fit together in a lock-and-key fashion. It will take a certain amount of force applied to the block in the direction shown to "break" these lock-and-key bonds. In this diagram, the friction force from the right is equal to the applied force from the left.

Overcoming the static friction takes a lot of force, but when the block "lifts" out of this lock-key arrangement ...

... a sustained force will keep it "skimming" along, from high-point to high-point. It generally takes more force to overcome the static friction and less to keep the block moving against the dynamic friction. Notice that the friction force is now smaller than the applied force, so the block accelerates to the right.

Notice that friction always opposes the motion. If we push an object to the right, the friction force will work in the opposite direction, to the left. A falling object experiences an upward frictional force that comes from banging into air molecules on the way down.

Friction opposes the motion

Friction always opposes any motion or intended motion. It never works in the same direction as the motion of an object.

The friction force is proportional to the squeezing force

Now it's easy to imagine that the friction force between two objects might be proportional to the force pushing them together — and it is. Imagine, for example, pushing a smooth block along a smooth table.

Now place ten pounds of weight on the block - to squeeze the surfaces closer together. It will be harder to push because of increased friction. The atomic-level basis for this is further pressing of high points and low points of surfaces together.

Quantifying friction: The coefficient of friction, μ

The trick to finding the frictional force lies in getting that squeezing force right. It's easy on a flat surface like the one illustrated above, but what about sliding friction on an inclined plane? In order to make sure that we're always talking about the same force, we use the normal force, the force that balances the component of gravitational force perpendicular to (or "normal to") the surface, like in the figure.

The friction force is the coefficient of friction μ (Greek letter "mu") multiplied by the normal force. The coefficient μ is a property of the system. It is large for systems like rubber on asphalt and small for systems like a skate on ice.

Friction force:   $F_f = \mu F_N$

The friction force

The friction force is calculated by multiplying the normal or squeezing force, FN by a coefficient of friction, μ. The coefficient of friction is measured experimentally for each situation.

Friction force:   $F_f = \mu F_N$

Example 1

A block of mass m = 0.75 Kg is pulled along a table at constant velocity. If it takes 2.0 N of force to maintain constant velocity, calculate the coefficient of friction for this system.

Solution: First draw a diagram of the situation. The 0.75 Kg mass exerts a force (F = mg) on the table, and the normal force is equal to it. I have not bothered to make these two forces of opposite sign but, strictly speaking, they are opposite vector forces.

In these cases of constant velocity, there is no acceleration, so all forces must be balanced. Therefore it doesn't matter how fast the block is moving, only that it is moving at constant velocity. Then the friction force Ff is equal to the pulling force of 2.0 N (but again of opposite sign, which I'll ignore).

Now it's a simple matter of rearranging the friction equation and using the normal and friction forces to calculate the coefficient of friction. Notice that in calculating μ, the units cancel.

$$F_f = \mu \cdot F_N \; \color{#E90F89}{\longrightarrow} \; \mu = \frac{F_f}{F_N} \; \color{#E90F89}{\longrightarrow} \; \mu = 0.272$$

(The coefficient of friction is unitless; notice that we're dividing two like quantities, in terms of units.)

Example 2

A 4.5 Kg mass is pulled up a 30˚ ramp at constant velocity by a falling 8.0 Kg mass, as shown. Calculate the coefficient of friction.

Solution: Here's the diagram.

All of the relevant calculations are shown above. Notice that this is another of those problems that makes μ easy to find because the force up the ramp (22.0 N) is equal to the frictional force. The normal force of 38.2 N gives the result:

$$F_f = \mu \cdot F_N \; \color{#E90F89}{\longrightarrow} \; \mu = \frac{F_f}{F_N} = \frac{22.0 \; N}{38.2 \; N} \; \color{#E90F89}{\longrightarrow} \; \mu = 0.576$$

Example 3

Here's one of my favorite examples because the answer is hidden beneath quite a few layers of interesting physics.

A 1200 Kg cannon situated on a cliff 22 m above flat ground is fired on a level trajectory, as shown. The 12 Kg cannonball travels 300 horizontal feet before hitting the ground. If the cannon rolls back 0.8 meters in recoil after the shot, calculate the coefficient of friction between the cannon and the ground.

Solution: First, here's the diagram (not to scale) →

Strategy: The basic strategy goes like this. Use the distance fired and the height to determine the initial velocity of the cannon ball. Calculate the momentum of the ball, which has the same magnitude as the momentum of the cannon, but in the opposite direction at recoil. Use the momentum of the cannon to calculate its initial velocity, then use the recoil distance to calculate the average velocity during the deceleration period, and the acceleration (deceleration) itself. The friction force is what causes the cannon to decelerate; calculate that. Now it's easy to find μ.

The freefall equation is rearranged to give the time it takes the cannonball (or any object, neglecting air friction) to fall 22 m.

The horizontal velocity of the cannonball is then

$$ d = \frac{1}{2} gt^2 \; \color{#E90F89}{\longleftarrow \; \text{freefall equation}}$$

$$ \require{cancel} t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 \cdot 22 \, \cancel{m}}{9.8 \, \cancel{m}s^{-2}}} = 4.49 \, s$$

$$v = \frac{d}{t} = \frac{300 \, m}{4.49 \, s} = 66.82 \frac{m}{s}$$

The momentum of the cannonball is

$$ \begin{align} p &= mv \\[4pt] &= (66.82 \, \frac{m}{s}(12 \, Kg) \\[4pt] &= 801.2 \; Kg \frac{m}{s} \end{align}$$

The cannon has a backward momentum of the same magnitude.

We can then calculate the initial velocity of the cannon at the moment of recoil from the momentum:

$$ \require{cancel} \begin{align} v &= \frac{p}{m} \\[5pt] &= \frac{801.2 \, \cancel{Kg} \frac{m}{s}}{1200 \cancel{Kg}} \\[5pt] &= 0.668 \, \frac{m}{s}. \end{align}$$

Now the cannon will decelerate from an initial velocity of 0.668 m/s to a final velocity of 0 m/s over a distance of 0.8 m. We want the acceleration of the cannon, so we'll need to know how long that slowdown takes. Assuming a smooth deceleration, the average velocity over the 0.8 meters is

$$v = \frac{v_f - f_i}{2} = \frac{0 - 0.668}{2},$$

where $(v_f - v_i)/2$ is just the average of the starting and ending velocities. Then the time it takes for the cannon to stop is

$$t = \frac{d}{v} = \frac{0.8 \, m}{0.334 \, m\cdot s^{-1}} = 2.39 \, s$$

Now we can calculate the acceleration,

$$a = \frac{\Delta v}{\Delta T} = \frac{0.334 \, m\cdot s^{-1}}{2.39 \, s} = 0.140 \, m\cdot s^{-2}$$

and the backward force on the cannon:

$$ \begin{align} F = ma &= (1200 \, Kg)(0.140 \, m\cdot s^{-2}) \\[5pt] &= 168 \, N \end{align}$$

That 168 N is the only force slowing the cannon down, and it's the friction force, Ff. The normal force on the cannon is

$$ \begin{align} mg &= (1200 \, Kg)(9.8 \, m·s^{-2}) \\[5pt] &= 11,760 \, N \end{align}$$

so we can finally calculate the coefficient of friction:

$$ \begin{align} F_f = \mu \cdot F_N \; \color{#E90F89}{\longrightarrow} \; \mu &= \frac{F_1}{F_N} \\[5pt] \mu &= \frac{168 \, N}{11,760 \, N} \\[5pt] \mu &= \bf 0.014 \end{align}$$


Example 4 - Terminal velocity

A falling object experiences the friction of "air resistance." Sometimes this is called drag force, but on the atomic level it's just a simple matter of things bumping into one another. A falling object bumps into the atoms and molecules that make up the air below it. Each collision slows the object somewhat, and that slowing has to be proportional to the speed at which the collision happens.

Generally, there are drag forces (Fd) proportional to the velocity v and to its square, v2. We generally set the air resistance of small objects proportional to v and of larger objects, like a falling person, proportional to v2

$$F_d = kv^2,$$

where k is a constant that is a property of the system.

The illustration shows the forces at work on a falling baseball. Actually, a spinning baseball has other kinds of forces on it that lead to curve balls, fast balls and so on in the game, but we'll just assume for now that our baseball is smooth and not spinning.*

The gravitational force is Fg = mg and the drag force is Fd = kv2. The constant k will be smaller for a baseball than for a larger object — like a skydiver. When these forces are balanced, the ball has reached what we call "terminal velocity." The forces from collisions with atoms and molecules of the air below it are in balance with the gravitational force at this point, thus the net force is zero, so there can be no acceleration.

If we denote terminal velocity as vterm, we can write the equation

$$v_{term}^2 = \frac{mg}{k}$$

Taking a root, we find the terminal velocity:

$$v_{term} = \sqrt{\frac{mg}{k}}$$

* The book "The Physics of Baseball", by Robert Adair, is a great little book about all of the physics aspects of the game.

Note: This is a simplistic view of air resistance. There's a lot that goes into the constant k, which is different for every falling object. This is really a problem in fluid mechanics (air is a "fluid" from the physics perspective, because it can flow. Fluid problems can get pretty complicated.)

Practice problems

1. It takes 350 N of horizontal force to set a 50 Kg box in motion across a floor. Once the box is in motion, a 325 N horizontal force keeps it moving at a constant velocity. Calculate the coefficient of dynamic friction between the box and the floor.
2. The coefficient of friction between a 168 Kg box and a carpeted floor is μ = 0.900.
  1. With how much force would you have to push on the chair to move it across the carpet at a constant speed?
  2. With how much force would you have to push on the chair to accelerate it at 0.5 m/s2?
3. A 160g hockey puck is accelerated to a velocity of 5.25 m/s by a stick, then released on the ice. If the coefficient of dynamic friction between the puck and the ice is 0.029, how far will the puck travel before it comes to rest?
4. A golf cart of mass 600 Kg and traveling at 5 m/s runs out of energy and stops after traveling 24 m. Calculate the stopping friction between the cart and the grass.
5. After being pushed, a 50 Kg box slides across a rough surface, causing it to slow down with an acceleration (deceleration) of -7.35 m/s2. Calculate the coefficient of friction between the box and the surface.

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