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.

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.

FORMULA

Ff = µ FNFf = 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. 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

• All forces are balanced if at rest or traveling at a constant velocity.

• The value for static friction is used because its at rest.

Ff= 41.2N FN= 98.1N

F= 41.2N

Fg= 98.1N

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 DIAGRAMSSTATIONARY OBJECT:

Fg- weight Fg = mg

Fn – normal force- force

pushing surfaces together Fn = Fg

Fg

Fn

Applied Parallel force

Fp - Parallel force

Ff – frictional force

Fn – normal force

Fg- weight- Fn

At constant velocity Fp = Ff

FpFf

Fg

Fn

Applied force at an angle

Fa – applied force

Fp – parallel force

Fp = Fa cosΘ = Ff

Fy = FasinΘ

Fn = Fg - Fy

Fa

Fp

Fg = mg

Ff

Fy

Fn

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.

Fg = mgF = 40kg(9.81m/s2)Fg =392.4N

Fp = Fa cosθ = Ff

Fp = 214N cos 30Fp = 185.3N

Fy = Fa sin θFy = 214N sin 30Fy = 107N

Fn = Fg – Fy

Fn = 392.4N – 107NFn = 285.4N

μ = Ff / Fn

μ = 185.3N/ 285.4Nμ = 0.65

Sliding object on an incline

Ѳ

Fn

Fn

Ff

Fg

Fp

Fg = mg (always perpendicular to the ground)

Fn = Fg cosθ (perpendicular to the surface)

Fp = Ff ( both are parallel to the surface)

Fp = Fg sinθ

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 gFg = 20kg(9.81m/s2)Fg = 196N

Fp = Fg sin ѲFp = 196N sin 30Fp= 98N

Fn = Fg cosѲFn= 196N cos 30Fn = 170N

µ = Ff / Fn = 98N/170N = .576

FnFg

Fp

θ

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*