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 stickiness Causes Frictional Forces It’s time to experience friction so rub your hands together. 

Friction: Two surfaces rubbing together and their stickiness 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 Static friction – REST occurs during Motion. 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.

Newton’s 3rd Law: Reaction Force Diagram F FR F = Push by fingers = 90 N FReaction = Push Back = -90 N No exception, F always = -FR Note: Reaction force is sometimes all friction and sometimes not.

Free Body Diagram in X direction Newton’s 2rd Law: ΣF = ma Free Body Diagram in X direction (Box is in motion) ΣF = ma F fk fk = Kinetic Friction = k FN F = Push Net Force = ΣF = ma ΣF = Fpush + fk = ma Can also find acceleration: a = ΣF/ m

Free Body Diagram in X direction Newton’s 2rd Law: ΣF = ma Free Body Diagram in X direction (Box is in motion) ΣF = ma F 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 acceleration a = ΣF/ m gives a = 60N / 10 kg = 6 m/s2

Also known as breaking force: = the force that will begin movement Newton’s 3rd Law: Reaction Force Diagram Newton’s 2rd Law: ΣFnet = manet Free Body Diagram in X direction (with motion) F FR ΣF = ma F F = Push by fingers = 90 N FReaction = Push Back = -90 N No exception, F always = -FR fk F = Push = 90 N fk = Kinetic Friction = k FN = -30N FN = weight = mg box = 2kg * 10 m/s2= 20N brick = 10kg * 10 m/s2 = 100N Net Force = ΣF = ma ΣF = Fpush + fk = ma ΣF = 90N - 30N = 60N Also, a = ΣF/ m gives a = 60N / 10 kg = 6 m/s2 Static Friction Force s = .4 fs = s FN Box fs = .4 * 20N = 8N Brick fs = .4 * 100N = 40N Also known as breaking force: = the force that will begin movement = maximum Static Friction Force = s FN Kinetic Friction Force k = .3 fk = k FN Box fk = .3 * 20N = 6N Brick fk = .3 * 100N = 30N Which is easier to push? A stalled car or a rolling car? Why?

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 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=vo2+2axx. Find x Known Variables k = 0.0500 vo = 4.00 m/s vf = 0.00 m/s Unknown Variables ax = x =

Static and Kinetic Frictional Force 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=vo2+2axx. Find x Known Variables k = 0.0500 vo = 4.00 m/s vf = 0.00 m/s Unknown Variables ax = 0.49 m/s2 x =16.3m

Static and Kinetic Frictional Force 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=vo2+2axx. Find x 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.

Static and Kinetic Frictional Forces Static 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 Forces Static 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 Forces Static 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 Forces Static 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 Forces Static 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?

Problem Box: fs = s FN = .4 (10kg) ( 9.80m/s2) fs = 39 N (Breaking Force)

Problem Box: 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 of Newton’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 @ 270 Thrust - 103,000 @ 30 Lift - ??? @ 120 Drag - ??? @ 210 (drag is a friction)

List our Forces: Weight – 86,500 @ 270 Thrust - 103,000 @ 30 Lift - ??? @ 120 Drag - ??? @ 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 @ 240 Thrust - 103,000 @ 0 Lift - ??? @ 90 Drag - ??? @ 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.

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

W: Fwx = -mgsin10.0o (gravity pulls backwards at10.0o ) s = 0.350 Free Body Diagram for X comp Friction Truck Engine 10o Gravity x component givens W: Fwx = -mgsin10.0o (gravity pulls backwards at10.0o ) s = 0.350 f s = s FN =.35 mgcos10.0o Ftruck = max = Fx y component givens W: -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.

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)