Coefficient of Friction The force of friction is directly proportional to the normal force. When we increase the mass or acceleration of an object we also increase the normal force. The coefficient of friction (µ) is a ratio between the force of friction and the normal force. µ is a constant and has no units. Fs = µsFN Fs – force of static friction (N)

µs – coefficient of static friction (no units)FN – normal force (N)

Fk = µkFN Fk – force of kinetic friction (N)

µk – coefficient of kinetic friction (no units)FN – normal force (N)

Generally, the force needed to start an object moving is always greater than the force needed to keep it moving.

µs ≥ µk

We gather the coefficient of friction from the tables. (pg. 101)

Example A car skids on wet concrete (µk = 0.97) for 50m. The mass of the car is 2000 kg and the initial velocity is 50 m/s. Calculate:a) the kinetic frictionb) the acceleration (deceleration)c) the final velocity (a) Find the normal force m = 2000 kg

g = 9.81 m/s2

FN = ?FN = mg = (2000)(9.81) = 19 620 N Use the coefficient of kinetic friction

µ k = 0.97Fk = ukFN FN = 19620 = (0.97)(19 620) Fk = ? = 19 031 N The kinetic friction is 19 031 N.

(b) Fnet = FA + Fk

FA = 0 Fnet = Fk

Fnet = ma Fk is slowing down the car. (deceleration)

Fk = -19 031 Nm = 2000 kga = ?

Fnet = ma -19 031 = 2000a a = -9.5 m/s2

The car is decelerating at -9.5 m/s2

(c) final velocity v1 = 50 m/s

a = -9.5 m/s2

d = 50 mv2 = ?

v22 = v1

2 + 2ad = (50)2 + 2(-9.5)(50) = 1550

v2 = = 39.0 The final velocity is 39.0 m/s

Friction Applet

3.4 Prac. #1-4, 6-8 UC #1-6