A rubber eraser and an ice cube slide across the same desk, pressed down equally.
The eraser stops almost at once. The ice cube glides on and on.
The desk is the same and the normal forces are equal, yet the friction differs hugely.
The last lesson tied friction's size to the normal force. The missing piece is a number for the surface pair itself.
Here is a measurement run on one pair: a wooden crate sliding on a concrete floor. At a normal force of 100 N, the friction is 40 N. At 200 N, it is 80 N. At 300 N, it is 120 N.
Each time, the friction is the same fraction of the normal force: 40 ÷ 100 = 0.40, and 80 ÷ 200 = 0.40, and 120 ÷ 300 = 0.40.
That is the pattern for any given pair of surfaces: the friction force on a sliding object is a fixed fraction of the normal force.
The fixed fraction is called the coefficient of friction, written μ (the Greek letter mu).
For the crate on this floor, μ = 0.40.
For one pair of surfaces, what does the coefficient of friction tell you?
μ is a plain number with no unit, because it is one force divided by another force, and the newtons cancel.
The two materials set μ together. Rubber on dry concrete has a large μ, about 0.7. Steel on ice has a tiny μ, about 0.05. Wood on wood sits between, about 0.3.
One surface alone never has a μ. The pair does: rubber on concrete is one pair, and rubber on ice is a different pair with a different μ.
Beth says: 'The same tire has one coefficient of friction on dry pavement and a different one on ice, because the coefficient belongs to the pair of surfaces.' Is she correct?
For a ski-on-snow pair, μ = 0.10. What does that number say about a sliding ski?
μ is the fixed fraction linking friction to the normal force for the pair.
So the friction on the sliding ski is 0.10 of the normal force, whatever the normal force happens to be.
The friction on the sliding ski is one tenth of the normal force, at every normal force.
Pair A has μ = 0.60. Pair B has μ = 0.15. Under equal normal forces, which slider feels more friction, and by what factor?
Equal normal forces: the friction compares as the coefficients compare.
0.60 is four times 0.15.
Pair A's slider feels four times the friction of pair B's.
For a given pair of surfaces, the friction on a sliding object is a fixed fraction of the normal force. Type the name of that fixed fraction.
Which statement about the coefficient of friction μ is correct?
The normal force on a slider doubles. What happens to μ for the pair?
Of these surface pairs, which should have the largest coefficient of friction?