FRICTION

Friction- a force that resists the motion of an object Acts parallel to the surface and opposite the direction of the motion. Dependent on the type of contact surfaces. Independent of contact area. Equal to applied force when object is at rest or traveling at a constant velocity. Fnet

TYPES OF FRICTION Static friction(starting friction)- Friction that is produced by surface projections and bonding. Kinetic friction(sliding friction)- friction an object has while in motion. Rolling friction-one object rolls over another Air resistance- friction produced by the air pushing against something in free fall.

Ff = µ FN FORMULA Ff = frictional force (N) FN = normal force (N) µ-Coefficient of friction *(µ) Has No Units* *See reference table*

ExA. A block of wood rests on a wood desktop ExA. A block of wood rests on a wood desktop. If it has a mass of 10kg, what is the minimum force required to move it? FN = Fg = mg FN = 10kg(9.81m/s2) FN = 98.1N Ff = µ FN = .42(98.1N) = 41.2N FN= 98.1N Ff= 41.2N F= 41.2N Fg= 98.1N All forces are balanced if the object is at rest or traveling at a constant velocity. The value for static friction is used because its at rest.

exB-If a 60N force is applied to the block, what will be its acceleration? Fnet = 60N – 41.2N Fnet = 18.8N a = Fnet/m = 18.8N/10kg a = 1.88 m/s2 Ff= 41.2N F = 60N

FORCE VECTOR DIAGRAMS STATIONARY OBJECT: Fg- weight Fg = mg Fn Fn – normal force- force pushing surfaces together Fn = Fg Fn Fg

Applied Parallel force Fn Fp - Parallel force Ff – frictional force Fn – normal force Fg- weight- Fn Ff Fp Fg At constant velocity Fp = Ff

Example A cross country skier has a mass of 67kg and is sliding on waxed ski’s across a flat snow covered surface. If they have a constant velocity, a) what is their normal force? B) what is the frictional force? FN=Fg=mg FN= 67kg(9.81m/s2) FN= 657.27N Ff=µFN Ff= .05(657.27N) Ff= 32.82N

Applied force at an angle Fa – applied force Fp – parallel force Fp = Fa cosΘ = Ff Fy = FasinΘ Fn = Fg - Fy Fn Fa Fy Fp Ff Fg = mg

A worker pulls a 40kg crate across the floor at a constant velocity A worker pulls a 40kg crate across the floor at a constant velocity. If he applies 214N of force along a rope held at 30 degrees with the ground, find Fg , Fp, Fy , Fn , and μ. Fp = Fa cosθ = Ff Fp = 214N cos 30 Fp = 185.3N Fg = mg F = 40kg(9.81m/s2) Fg =392.4N Fy = Fa sin θ Fy = 214N sin 30 Fy = 107N

Fn = Fg – Fy Fn = 392.4N – 107N Fn = 285.4N μ = Ff / Fn μ = 185.3N/ 285.4N μ = 0.65

Sliding object on an incline Fg = mg (always perpendicular to the ground) Ff Fn Fn = Fg cosθ (perpendicular to the surface) Ѳ Fp = Ff ( both are parallel to the surface) Fp = Fg sinθ Fp Fn Fg

Ex: A 20kg wooden box is sliding down an incline of 30 degrees at a constant velocity. Find Fg , Fp , Fn and μ. Fg = m g Fg = 20kg(9.81m/s2) Fg = 196N θ Fg Fn Fp Fn = Fg cosѲ Fn= 196N cos 30 Fn = 170N Fp = Fg sin Ѳ Fp = 196N sin 30 Fp= 98N µ = Ff / Fn = 98N/170N = .576

Pulling an object up an incline Fnet = Fp – Ff So Fp = Fnet + Ff Fparallel Ffriction Fp ≠Ff

ExA: A net force of 35N is required to move a 10kg crate up an incline of 20 degrees at a constant velocity. What is the parallel force applied to the block? Ff = Fgsinϴ Ff=10kg(9.81m/s2)sin 20° Ff =33.5N Fp=Fnet+ Ff Fp= 35N + 33.5N Fp = 68.5N

exB: What is the coefficient of sliding friction between the crate and the incline? µ = Ff /Fn Fn= Fgcosϴ Fg=mg µ = Ff = 33.5N mg cosϴ 10kg(9.81m/s2)cos 20° µ = 33.5N / 92.18N µ = .36

COOL MATH TRICK Sin Ѳ = opp/hyp cos Ѳ= adj/hyp µ = Ff = Fgsin Ѳ = opp/hyp = opp/adj Fn Fg cos Ѳ adj/hyp µ = tan Ѳ *Only works for objects sliding down an incline*