Frictional Forces Words of the day are underlined

Frictional Forces

• Words of the day •are underlined.

Friction:

Friction: from book, ever present resistance to motion

whenever two materials are in contact with each other.

Friction: (ME) Two surfaces rubbing together and their

stickinessCauses Frictional Forces

It’s time to experience friction so rub your hands

together.

Friction: Two surfaces rubbing together and their stickiness.

Always direction opposite of motion.• frictional force: A force that slows motion, or prevents motion from starting. • Friction: frictional forces arise at the contact points between the molecules of the different bodies. Contact force that is resistance to Motion.

•ALWAYS OPPOSES MOTION, i.e, in opposite direction of motion. Friction increases as contact pressure increases, i.e., more weight means more friction. Usually dissipated as heat or usually generates heat. •SURFACE AREA DOES NOT CHANGE FRICTION.

* Friction Force is Usually lowercase “f ” usually in italics.

Frictional Forces• frictional force: A force that slows motion, or prevents motion from starting. • 2 kinds – Kinetic friction – MOTION• slows objects down. • occurs during Motion.

- Static friction – REST- prevents motion from starting. occurs before motion, when still,

not moving, or at REST.

Which is bigger?????

Frictional Forces• Which is bigger?????

Question: Does a car stop faster when the tires are locked and sliding or when rolling?

• 2 kinds – Kinetic friction – MOTION• Sliding tires

-Static friction – REST-Rolling tires

New Symbol• μ – greek letter “Mu”

• Coefficient of friction.

– μs – coefficient of static friction

– μk – coefficient of kinetic friction

• Coefficient of friction: Decimal between 0.0 and 1.0, unitless. Decimal percent of force in the normal direction. (Mu is percent of weight that turns into friction force.)

• Coefficient of friction: μ where f = μ FN

• Usually just given in the problem, depends on materials used and their surface conditions.

Static and Kinetic Frictional Forces

Kinetic Frictional Force: friction force that occurs while sliding.

The magnitude fk of the kinetic frictional force is given by:

fk = k FN

where k is the coefficient of Kinetic friction, and FN is the magnitude of the normal force. Normal Force: Force that is 90 degrees to the surfaces. (Normal is 90) degrees. Kinetic frictional forces occur when the object is moving; it acts to slow down the motion!

Frictional forces are independent of the area of contact between the surfaces! (talk about sides of block then race tires)

Instructions for demo on next slide:Place light mole box on table and slide. Place brick on table and slide.

List mu static = .4, mu k = .3

Mass of light box is 2kg.Mass of brick is 10 kg.

Find Friction resistance force when static and when in motion on board.

Find both friction resistance and Reaction Force. Same for Static but Not the same thing when in accelerating.

F = Push by fingers = 90 NFReaction = Push Back = -90 NNo exception, F always = -FR

Note: Reaction force is sometimes all friction and sometimes not.

Newton’s 3rd Law: Reaction Force Diagram

F FR

fk = Kinetic Friction = k FNF = Push Net Force = ΣF = ma ΣF = Fpush + fk = ma Can also find acceleration: a = ΣF/ m

Newton’s 2rd Law: ΣF = maFree Body Diagram in X direction

(Box is in motion)

FΣF = ma

fk

fk = Kinetic Friction = k FN

= -30N F = Push = 90N Net Force = ΣF = ma ΣF = Fpush + fk = ma ΣF = 90N - 30N = 60N Can also find accelerationa = ΣF/ m gives a = 60N / 10 kg = 6 m/s2

Newton’s 2rd Law: ΣF = maFree Body Diagram in X direction

(Box is in motion)

FΣF = ma

fk

Static Friction Forces = .4fs = s FN

Box fs = .4 * 20N = 8NBrick fs = .4 * 100N = 40NAlso known as breaking force:= the force that will begin movement

= maximum Static Friction Force = s FN

FN = weight = mg box = 2kg * 10 m/s2= 20N brick = 10kg * 10 m/s2 = 100N

Kinetic Friction Forcek = .3fk = k FN

Box fk = .3 * 20N = 6NBrick fk = .3 * 100N = 30N

Which is easier to push? A stalled car or a rolling car?Why?

fk = Kinetic Friction = k FN = -30N

F = Push = 90 N

Net Force = ΣF = ma ΣF = Fpush + fk = ma ΣF = 90N - 30N = 60N Also, a = ΣF/ m gives a = 60N / 10 kg = 6 m/s2

F = Push by fingers = 90 NFReaction = Push Back = -90 NNo exception, F always = -FR

Newton’s 3rd Law: Reaction Force Diagram

F FR

Newton’s 2rd Law: ΣFnet = manet

Free Body Diagram in X direction (with motion)

FΣF = ma

fk

Static and Kinetic Frictional Force

Example

A sled is traveling 4.00m/s along a horizontal stretch of snow. The coefficient of kinetic friction is k = 0.0500. How far does the sled go before stopping?

Static and Kinetic Frictional Force

Known Variables k = 0.0500vo = 4.00 m/svf = 0.00 m/s

Unknown Variables ax = x =

Formula 1) FN= mg = weight

2) ΣF= fk , fk = k FN

3) ΣF= max = k FN

4) max = k mg

5) (/m=>) ax= k g

6) ax = .05 (9.80m/s2)

7)vf2=vo

2+2axx. Find x

Static and Kinetic Frictional Force

Known Variables k = 0.0500vo = 4.00 m/svf = 0.00 m/s

Unknown Variables ax = 0.49 m/s2

x =16.3m

Formula 1) FN= mg = weight

2) ΣF= fk , fk = k FN

3) ΣF= max = k FN

4) max = k mg

5) (/m=>) ax= k g

6) ax = .05 (9.80m/s2)

7)vf2=vo

2+2axx. Find x

Static and Kinetic Frictional Force

Did we need to know the mass of the sleder? No. Why? It cancels out in the ax equation. Real Life Ap: This applies to car tires in accidents. By measuring the length of skid marks, they can calculate the speed a car was going before an accident. k of a tire is the same for all cars since it does not depend on car mass or surface area of the tires.

Formula 1) FN= mg = weight

2) ΣF= fk , fk = k FN

3) ΣF= max = k FN

4) max = k mg

5) (/m=>) ax= k g

6) ax = .05 (9.80m/s2)

7)vf2=vo

2+2axx. Find x

Static and Kinetic Frictional ForcesStatic Frictional Force:

Reaction force to anything trying to

start motion.

•Equal and opposite to applied force.

•DOES NOT EXCEED THE

APPLIED FORCE,

but is equal to it.

Static and Kinetic Frictional ForcesStatic Frictional Force:

Reaction force to anything trying to

start motion.

Equal and opposite to applied force,

until reaches maximum value and motion

starts.

friction “breaks” when F is great enough

and motion begins.

Static and Kinetic Frictional ForcesStatic Frictional Force Breaks at a certain value:

fs = s FN

fs = force of static friction

s = coefficient of static friction

FN = Normal force

Static and Kinetic Frictional ForcesStatic Frictional Force Breaks at a certain value:

fs = s FN

fs = force of static friction

s = coefficient of static friction

FN = Normal force

s is a given value. It depends on the object and

the surface.

Static and Kinetic Frictional ForcesStatic Frictional Force:

fs = force of static friction

s = coefficient of static friction

FN = Normal force (usually weight)

Normal force is usually just the

weight of the object.

FN = Mass* 9.80 m/s2

IMPORTANT!!!!!

If the surface is not horizontal use trig.

Multiply by cos of the angle of incline.

Notes on friction

Almost always:

μs > μk

It is easier to keep an object moving than it is to start from rest. Think about pushing a car.

Both are almost always less than 1. If it was greater than one, it would be easier to pick the object up and carry it than it would be to push it across the flat surface (something like velcro)

Problem

Box:

How much force is needed to “budge” this box?

If we keep pushing that hard, what will the acceleration be?

ProblemBox:

fs = s FN = .4 (10kg) ( 9.80m/s2)

fs = 39 N (Breaking Force)

ProblemBox:

fk = k FN = .2 (10kg) ( 9.80m/s2)

fk = 19.5 N (Kinetic Force)

Net Force = Pushing Force – Kinetic Friction Force

Net Force = 39N – 19.5N = 19.5N

a = F / m = 19.5N / 10kg = 1.95 m/s2 = 2 m/s2

The Tension Force

The Tension Force

Tension is the force balanced by a rope, cable or wire.

A “simple pulley” changes direction without affecting tension.

Tension is the same at every point in a single rope.

Equilibrium Applications ofNewton’s Laws of Motion

An object is in equilibrium when it has zero acceleration

“equilibrium” refers to a lack of change, but in the sense that the velocity of an object isn’t changing, i.e, there is no acceleration. Equilibrium: Constant Speed and Direction. Fx = 0 and Fy = 0, ax = 0 m/s2 and ay = 0 m/s2

ax = 0 m/s2 and ay = 0 m/s2

Fx = 0 and Fy = 0

Reasoning Strategy

If F = 0, then Fx = 0 and Fy = 0.

• Draw a free-body diagram the object. Be sure to include only the forces that act on the object; do not include forces that the object exerts on its environment.

Example

A jet plane is flying with a constant speed along a straight line at an angle of 30.0o above the horizontal. The plane has a weight W whose magnitude is W=86,500 N and its engine provide a forward thrust T of magnitude T=103,000 N. In addition, the lift force L (directed perpendicular to the wings) and the force R of air resistance (directed opposite to the motion) act on the plane. Find L and R.

List our Forces:

Weight – 86,500 @ 270Thrust - 103,000 @ 30 Lift - ??? @ 120Drag - ??? @ 210(drag is a friction)

List our Forces:

Weight – 86,500 @ 270Thrust - 103,000 @ 30 Lift - ??? @ 120Drag - ??? @ 210(drag is a friction)

This looks really complicated.

Shortcut:

Since 3 of our forces are Perpendicular, lets change The axes...

Now:

Weight – 86,500 @ 240Thrust - 103,000 @ 0Lift - ??? @ 90Drag - ??? @ 180(drag is a friction)

3 nice angles are better than 1.

x component W: 86,500 cos 240

L: 0 (L cos 90) T: +103,000N R: -R (R cos 180)

y component W: 86,500 sin 240

L: +L T: 0 R: 0

Fx = W cos240.0o + T -R = 0 Fy = -Wcos30.0o + L= 0

R=59,800N and L = 74,900 N

Nonequilibriuium Applications of Newton’s Laws of Motion

Non-equilibrium conditions occur when the object is accelerating and the forces acting on it are not balanced so the net force is not zero.Non-equilibrium: Fx = max and Fy = may

Fx = max and Fy = may

Example

A supertanker of mass m = 1.50 X 108 kg is being towed by two tugboats. The tensions in the towing cables apply the forces T1 and T2 at equal angles of 30.0o with respect to the tanker’s axis. In addition, the tanker’s engines produce a forward drive force D whose magnitude is D = 75.0 X 103 N. Moreover, the water applies an opposing force R, whose magnitude is R = 40.0 X 103 N. The tanker moves forward with an acceleration that points along the tanker’s axis and has a magnitude of 2.00 X10-3 m/s2. Find the magnitudes of T1 and T2.

x component T1: +T1cos30.0o

T2: +T2cos30.0o D: +D = 75.0 X103 N R: -R=-40.0 N

y component T1: +T1sin30.0o

T2: -T2sin30.0o D: 0 R: 0

Fy = +T1cos30.0o – T2sin30.0o = 0 Fx = +T1cos30.0o + T2cos30.0o +D -R = max

T=1.53 X 105 N

Example

The figure shows a water skier at four different moments: a) The skier is floating motionless in the water b) The skier is being pulled out of the water and up onto the skis c) The skier is moving at a constant speed along a straight line d) The skier has let go of the tow rope and is slowing down For each moment, explain whether the net force acting on the skier is positive, negative, or zero.

Example

A flatbed is carrying a crate up a 10.0o hill. the coefficient of static friction between the truck bed and the crate is s = 0.350. Find the maximum acceleration that the truck can attain before the crate begins to slip backward relative to the track. (p. 114)

x component givensW: Fwx = -mgsin10.0o (gravity pulls backwards at10.0o )s = 0.350 f s = s FN =.35 mgcos10.0o

Ftruck = max = Fx

y component givensW: -mgsin10.0o

FN: FN = mgsin10.0o

Equations

Fx = -Gravity + Friction = Truck Engine Accel

Fx = Fwx + sFN = max

Fx = -mgsin10.0o + .35 mgcos10.0o= max

(/m out)= -9.80sin10.0o +.35(9.8)cos10.0o =ax

y comp equations Fy = -mgsin10.0o + FN = 0

ax = 1.68 m/s2

Who uses this info? Not the driver. The engineers use it to figure out if they need to add more tie downs to the truck bed design.

Free Body Diagram for X comp

Gravity

Friction

Truck Engine

10o

Stop Here.

Example

Block 1 (mass m1 = 8.00 kg) is moving n a frictionless 30.0o incline. This block is connected to block 2 (mass m2 = 22.0 kg) by a cord that passes over a massless and frictionless pulley, Find the acceleration of each block and the tension in the cord. (p. 115)

x component W1: -W1sin30.0o

T: T

y component W2: -W2

T: T

Fx = -W1sin30.0o + T = m1a Fy = T – W2 = m2(-a)