Coefficient of friction

[KingNothing]

In my physics class, when we involved friction...the book kept making it a point of "why the coefficient of friction has no units"...the answer was always because it was a ratio.

Well.., if F_{friction}=umg then u=\frac{F_{friction}}{mg}, right? So why can't we say that the coefficient of friction is "Newtons per kilogram meters per second per second"?


edit: fix latex
Integral

[garytse86]

F = uR, both LHS and RHS have units N, so u (friction coefficient) must have no units (homogenous)

[Doc Al]

Well.., if F_{friction}=umg then u= \frac{F_{friction}}{mg}, right? So why can't we say that the coefficient of friction is "Newtons per kilogram meters per second per second"?
Because 1 Newton = 1 Kg/m^2. The units cancel.

[KingNothing]

I see, so in essence I was saying that the units should be \frac{ma}{ma}. I see how I was wrong. I still don't think "It's a ratio" is a good way of saying it. Almost everything is a ratio aside from distance and time.

[Chi Meson]

You are correct. The better way of saying it is that "the coefficient of friction is a ratio of two forces" (the normal force over the frictional force); this implies that units will cancel.

[jdavel]

Decker said: "I still don't think "It's a ratio" is a good way of saying it. Almost everything is a ratio aside from distance and time."

In some systems of units even distance or time come out as a ratio!

But you're right, just saying "It's a ratio" doesn't explain the coefficient of friction very well.

Try this.

Many (maybe even most) of the questions physics tries to answer boil down to this: Given the location and velocity of an object at a certain time, what will its location and velocity be at any later time? Once Newton defined the force acting on objects as F=ma, the answer to these questions became: What is the force acting on the object at any location? (The simple fact that this is usually an easier question to answer, is why Newton is considered the greatest physicist whoever lived!)

In most cases the force can be expressed as the product of 3 factors:

F = T*M*f(x)

where T is a value that's already been measured and Tabulated, because it can be applied in many similar situations, M is a value that you can Measure or determine for your particular situation, and f(x) is a universal law of physics that shows how the force on the object will depend on the object's location.

Some examples:

1) The force on a charged object near another charged object

F = Em*(q1*q2)*(1/x^2)

Em depends on the the particular material in the space between the particles, and there are tables of these values for many different materials

q1 and q2 are the charge on the particular objects in question

The last factor says the force decreases as the square of the distance between the objects.

2) The force that a bungee cord exerts on somebody when they jump off a bridge:

F = Y*(pi*R^2/L)*(-x)

Y is the Young's modulus of the material the cord is made of, again values have been tabulated for lots of materials.

The second is calculated from the dimensions (R & L) of the cord

-x shows that the force is in the opposite direction that the cord is stretched (very important for the jumper!) and that it gets greater the more the cord is stretched (also good for the jumper).

3) Friction works the same way, but it's particularly simple

F = Cf*W*1

Cf is the coefficient of friction between your object and the surface it slides on (you can look it up)

W is the weight of the object (you can measure it)

1 is how F depends on x, but since 1 is a constant, F doesn't depend on x.


In all these cases, the units of the 3 factors have to combine to be Newtons. That's why Cf has no units! The other 2 factors are weight(Newtons) and 1(no units).

So Cf is the conversion factor from one force (weight, which is easily measured) to another force, friction, which is part of the total force used in F=ma to find where the object will be and how fast it will be going at any time in the future.

Well that's a lot more than "It's a ratio"! Hope it helped a little.