1. Friction And Forces in Two Dimensions

2. Review A force is a push or pull. A force is a push or pull. Newton’s 3 Laws of Motion: Newton’s 3 Laws of Motion: Objects resist changes in velocity unless acted upon by a net force. Objects resist changes in velocity unless acted upon by a net force. The magnitude of an object’s acceleration depends on the net force and inversely on the object’s mass. The magnitude of an object’s acceleration depends on the net force and inversely on the object’s mass. a = F net / m a = F net / m For every force there is an equal and opposite counter-force. For every force there is an equal and opposite counter-force.

3. Friction A force that opposes motion. A force that opposes motion. It was once thought that friction was caused by surface roughness. It was once thought that friction was caused by surface roughness. In this model, microscopic nooks and bumps on the surface of an object lock and slip against similar features on another surface. In this model, microscopic nooks and bumps on the surface of an object lock and slip against similar features on another surface. It is now known that this model is incorrect. It is now known that this model is incorrect. Friction is caused by brief electromagnetic interactions between charged particles in two surfaces. Friction is caused by brief electromagnetic interactions between charged particles in two surfaces.

4. Friction Two important types of friction: Two important types of friction: Static friction Static friction Prevents motion from beginning Prevents motion from beginning Kinetic friction Kinetic friction Prevents motion from continuing Prevents motion from continuing Static friction is usu. stronger than kinetic friction. Static friction is usu. stronger than kinetic friction. This is why, when you’re pushing a heavy object across the floor, it’s harder to start it moving than it is to keep it moving. This is why, when you’re pushing a heavy object across the floor, it’s harder to start it moving than it is to keep it moving.

5. Friction F f =  F N F f =  F N F f = Friction (N) F f = Friction (N)  = Coefficient of friction (no units)  = Coefficient of friction (no units) F N = Normal force = mg cos  F N = Normal force = mg cos  Coefficient of friction Coefficient of friction Unique for every pair of substances and set of conditions. Unique for every pair of substances and set of conditions. Different CoF for static and kinetic friction. Different CoF for static and kinetic friction.

6. Coefficients of Friction Material 1Material 2 StaticSliding Aluminum 1.05-1.351.4 AluminumMild Steel0.610.47 Cast Iron 1.10.15 CopperCast Iron1.050.29 CopperMild Steel0.530.36 CopperSteel (304 stainless)0.230.21 Glass 0.9 - 1.00.4 GlassNickel0.780.56 LeatherOak (Parallel grain)0.610.52 Nickel 0.7-1.10.53 OakOak (parallel grain)0.620.48 OakOak (cross grain)0.540.32 RubberAsphalt (Dry) 0.5-0.8 RubberAsphalt (Wet) 0.25-0.75 RubberConcrete (Dry) 0.6-0.85 RubberConcrete (Wet) 0.45-0.75

7. Coefficient of Friction CoF must be determined experimentally for any pair of surfaces. CoF must be determined experimentally for any pair of surfaces.  = F f / F N  = F f / F N

8. Sample Problem The coefficient of static friction between two blocks of wood is 0.500. What applied force is necessary to start a 250.- kg wooden box sliding along a horizontal wooden surface? The coefficient of static friction between two blocks of wood is 0.500. What applied force is necessary to start a 250.- kg wooden box sliding along a horizontal wooden surface? F a must be equal to or greater than F f,static. F a must be equal to or greater than F f,static. F f =  F N F f =  F N F N = F w = mg F N = F w = mg F N = (250. kg)(9.81 m/s 2 ) F N = (250. kg)(9.81 m/s 2 ) F N = 2450 N F N = 2450 N F f = (0.500)(2450 N) F f = (0.500)(2450 N) F f = 1230 N F f = 1230 N

9. Sample Problem 2 A box with a mass of 40.0 kg requires an applied force of 250. N to keep it moving at a constant speed along a horizontal surface. What is the CoF between the box and the floor? A box with a mass of 40.0 kg requires an applied force of 250. N to keep it moving at a constant speed along a horizontal surface. What is the CoF between the box and the floor? F f =  F N F f =  F N F f =  mg F f =  mg (250. N) =  (40.0 kg)(9.81 m/s 2 ) (250. N) =  (40.0 kg)(9.81 m/s 2 ) (250. N) =  (392 N) (250. N) =  (392 N)  = 0.638  = 0.638

10. In lab today We will calculate the coefficient of friction between a shoe and the surface of a ramp. We will calculate the coefficient of friction between a shoe and the surface of a ramp. You will volunteer the use of your shoes! You will volunteer the use of your shoes! Place the shoes on a ramp. Place the shoes on a ramp. Incline the ramp until the shoe begins to slide. Incline the ramp until the shoe begins to slide. Measure the angle at which the shoe begins to slide. Measure the angle at which the shoe begins to slide. Calculate the coefficient of friction using some simplifying math. Calculate the coefficient of friction using some simplifying math. Repeat the procedure but first wet the shoe’s sole and the board. Repeat the procedure but first wet the shoe’s sole and the board.