1. 6.2 Friction

2. Chapter 6 Objectives  Calculate the weight of an object using the strength of gravity (g) and mass.  Describe the difference between mass and weight.  Describe at least three processes that cause friction.  Calculate the force of friction on an object when given the coefficient of friction and normal force.  Calculate the acceleration of an object including the effect of friction.  Draw a free-body diagram and solve one-dimensional equilibrium force problems.  Calculate the force or deformation of a spring when given the spring constant and either of the other two variables.

3. Chapter 6 Vocabulary  ball bearings  coefficient of friction  coefficient of static friction  compressed  deformation  dimensions  lubricant  normal force  prototype  restoring force  rolling friction  sliding friction  engineering  engineering cycle  extended  free-body diagram  g forces  Hooke’s law  spring  spring constant  static friction  subscript  viscous friction  weightless

4. Inv 6.2 Friction Investigation Key Question: What happens to the force of sliding friction as you add mass to a sled?

5. 6.2 Friction  Friction results from relative motion between objects.  Friction is a resistive force.  Describing friction as resistive means that it always works against the motion that produces it.

6. 6.2 Types of Friction  Static friction  Sliding friction  Rolling friction

7. 6.2 Types of Friction  Air friction  Viscous friction

9. 6.2 A model for friction  No single model or formula can accurately describe the many processes that create friction.  Some of the factors that affect friction include the type of material, the degree of roughness, and the presence of dirt or oil.  Even friction between two identical surfaces changes as the surfaces are polished by sliding across each other.

10. 6.2 A model for friction  The coefficient of friction is a ratio of the strength of sliding friction between two surfaces compared to the force holding the surfaces together, called the normal force.  The coefficient of friction is most often a number between zero and one.

11. 6.2 Dry sliding friction  The symbol for coefficient of friction is the Greek letter μ.  A coefficient of one means the force of friction is equal to the normal force.  A coefficient of zero means there is no friction no matter how much force is applied to squeeze the surfaces together. F f =  F n Normal force (N) Coefficient of friction Friction force (N)

12. 1.You are asked for the force of friction F f. 2.You are given weight F w, applied force F, and coefficient of sliding friction μ. 3.The normal force is the sum of forces pushing down on the floor, so use F f = μF n. 4.First, find the normal force: F n = 100 N + 10 N = 110 N  Use F f = μF n and substitute values: F f = (0.25)(110 N) = 27.5 N Calculate force of friction A 10-N force pushes down on a box that weighs 100 N. As the box is pushed horizontally, the coefficient of sliding friction is 0.25. Determine the force of friction resisting the motion.

13. 6.2 Calculating the force of friction  The normal force is the force perpendicular to two surfaces which are moving relative to each other.  In many problems, the normal force is the reaction in an action-reaction pair.

14. 6.2 Static friction  It takes a certain minimum amount of force to make an object start sliding.  The maximum net force that can be applied before an object starts sliding is called the force of static friction.

15. 6.2 Static Friction  The coefficient of static friction ( μ s ) relates the maximum force of static friction to the normal force.  It takes more force to break two surfaces loose than it does to keep them sliding once they are already moving. F f =  s F n Normal force (N) Coefficient of sliding friction Friction force (N)

16. 6.2 Table of friction coefficients

17. 1.You are asked for the force to overcome static friction F f 2.You are given the weight F w. Both surfaces are steel. 3.Use F f ≤ μ s F n 4.Substitute values: F f ≤ (0.74) (50 N) = ≤ 37 N Calculate the force of static friction A steel pot with a weight of 50 sits on a steel countertop. How much force does it take to start to slide the pot?

18. 6.2 Friction and motion  When calculating the acceleration of an object, the F that appears in Newton’s second law represents the net force.  Since the net force includes all of the forces acting on an object, it also includes the force of friction.  The real world is never friction-free, so any useful physics must incorporate friction into practical models of motion.

19. 1.You are asked for the acceleration a. 2.You are given the applied force F, the mass m, and the coefficient of rolling friction μ. 3.Use: a = F ÷ m, F f = μF n, F w = mg and g = 9.8 N/kg. Calculating the acceleration of a car including friction The engine applies a forward force of 1,000 newtons to a 500-kilogram car. Find the acceleration of the car if the coefficient of rolling friction is 0.07.

20.  The normal force equals the weight of the car:  F n = mg = (500 kg)(9.8 N/kg) = 4,900 N.  The friction force is: F f = (0.07)(4,900 N) = 343 N.  The acceleration is the net force divided by the mass:  a = (1,000 N – 343 N) ÷ 500 kg = 657 N ÷ 500 kg  a = 1.31 m/s 2 Calculating the acceleration of a car including friction

21. 6.2 Reducing the force of friction  Friction cannot be completely eliminated but it can be reduced.  A fluid used to reduce friction is called a lubricant.  In systems where there are axles, pulleys, and rotating objects, ball bearings are used to reduce friction.  Another method of reducing friction is to separate two surfaces with a cushion of air.

22. 6.2 Using friction  There are many applications where friction is both useful and necessary.  Friction between brake pads and the rim slows down a bicycle.  All-weather tires have treads, patterns of deep grooves to channel water away from the road-tire contact point.  Friction keeps nails and screws in place.  Cleats greatly increase the friction between the sports shoe and the ground.