Friction

Friction Friction is the force that opposes the motion between two surfaces that touch. Friction forces are parallel to the contact surface and opposite the direction of motion It allows you to walk, turn a corner on your bike, warm your hands in the winter.

What Causes Friction? The surface of any object is rough. Even an object that feels smooth is covered with tiny hills and valleys. The contact between the hills and valleys of two surfaces causes them to stick, resulting in friction. The surface area in contact has negligible effect on friction!

What Causes Friction? The amount of friction depends on: Roughness of the surfaces Force pushing the surfaces together

Types of Friction Kinetic friction occurs when force is applied to an object and the object moves.

Types of Friction Static friction occurs when force applied to an object does not cause the object to move.

Coefficients of Friction Static coefficient … s. Kinetic coefficient … k. Both depend on the materials in contact. Small for steel on ice or scrambled egg on Teflon frying pan Large for rubber on concrete or cardboard box on carpeting The bigger the coefficient of friction, the bigger the frictional force.

Static Friction Force Ffs  s FN Ffs, max = s FN normal force static frictional force coefficient of static friction Ffs, max = s FN maximum force of static friction Fs, max is the force you must exceed in order to budge a resting object.

Static friction force varies Ffs, max is a constant in a given problem, but Fs varies. Ffs matches the applied force until FA exceeds Ffs, max. Example: If s for a 10 kg wooden crate on a tile floor is 0.6, what is Fs,max? What does this mean? Ffs, max = 0.6 (10 ) (10) = 60 N The box finally budges when FA surpasses 60N. Then kinetic friction acts on the box.

kinetic frictional force Ffk = k FN normal force kinetic frictional force coefficient of kinetic friction Once an object starts to move, use k instead. Ffk is constant so long as the materials involved don’t change.

 values Typically, 0 < k < s < 1. This is why it’s harder to budge an object than to keep it moving. If k > 1, it would be easier to lift an object and carry it than to slide across the floor. Dimensionless (’s have no units, as is apparent from Ff =  FN).

Friction Example 1 You push a giant barrel o’ monkeys sitting on a table with a constant force of 63 N. If k = 0.35 and s =0.58, when will the barrel have moved 15 m? Never, since this force won’t even budge it! 63 < 0.58 (14.7) (10)  83.6 N answer: Barrel o’ Monkeys 14.7 kg

Friction Example 2 Same as the last problem except with a bigger FA: You push the barrel o’ monkeys with a constant force of 281 N. k = 0.35 and s =0.58, same as before. When will the barrel have moved 15 m? Forget Fs and calculate Fk: fk = 0.35 (14.7) (10) = 51 N Barrel o’ Monkeys 14.7 kg (continued on next slide)

Friction Example 2 (continued) step 4: Free body diagram while sliding: N FA fk mg step 5: Fnet = FA – fk = 281 - 50.421 = 230.579 N Note: To avoid compounding of error, do not round until the end of the problem. step 6: a = Fnet / m = 230.579 / 14.7 = 15.68564 m/s2 step 7: Kinematics: x = +15 m, v0 = 0, a = +15.68564 m/s2, t = ? x = v0 t + ½ a t 2  t = 2 x / a  1.38 s

Friction Conceptual Example Two cars, identical in mass and initial speed are traveling along two different roads. The road the first car is on is dry and the road the second car is on is wet. Why does it take longer for the 2nd car to come to a stop? The wet road has a lower coefficient of friction and since the two cars have the same mass, they will have the same normal force so Ff = µFN will be lower for the second car than for the first and since the force of friction is lower, there is less force slowing the car down and therefore less negative acceleration (by Newton’s second law) and it will take longer to stop

Friction as the net force example A runner is trying to steal second base. He’s running at a speed of 8 m/s; his mass is 70 kg. The coefficient of kinetic friction between his uniform and the base is 0.6. How far from second base should he begin his slide in order to come to a stop right at the base? KEY: Once the slide begins, there is no applied force. Since N and mg cancel out, fk is the net force. So Newton’s 2nd Law tells us: fk = ma. But the friction force is also given by fk =  N =  m g. FN Ff Fg