2. Objectives: After completing these notes, you should be able to: Define and calculate the coefficients of kinetic and static friction, and give the relationship of friction to the normal force.Define and calculate the coefficients of kinetic and static friction, and give the relationship of friction to the normal force. Apply the concepts of static and kinetic friction to problems involving constant motion or impending motion.Apply the concepts of static and kinetic friction to problems involving constant motion or impending motion. Define and calculate the coefficients of kinetic and static friction, and give the relationship of friction to the normal force.Define and calculate the coefficients of kinetic and static friction, and give the relationship of friction to the normal force. Apply the concepts of static and kinetic friction to problems involving constant motion or impending motion.Apply the concepts of static and kinetic friction to problems involving constant motion or impending motion.

3. Friction Forces When two surfaces are in contact, friction forces oppose relative motion or impending motion. P Friction forces are parallel to the surfaces in contact and oppose motion or impending motion.

4. 2 N2 N2 N2 N Friction and the Normal Force 4 N The force required to overcome static or kinetic friction is proportional to the normal force, F n The force required to overcome static or kinetic friction is proportional to the normal force, F n. F f =  k F n F f ≤  s F n n 12 N 6 N n 8 N 4 N n

5. Friction forces are independent of area. 4 N4 N4 N4 N 4 N4 N4 N4 N If the total mass pulled is constant, the same force (4 N) is required to overcome friction even with twice the area of contact. For this to be true, it is essential that ALL other variables be rigidly controlled.

6. Friction forces are independent of speed. 2 N The force of kinetic friction is the same at 5 m/s as it is for 20 m/s. Again, we must assume that there are no chemical or mechanical changes due to speed. 5 m/s 20 m/s

7. Friction Forces -coefficient of the surfaces in contact,  Friction is dependent on: Static friction opposes the intended motion of two surfaces in contact but at rest relative to one another. Kinetic friction opposes motion of two surfaces in contact that are moving relative to one another. FaFa PHYSICS FsFs -normal force, F n FaFa PHYSICS FkFk velocity Kinetic friction is less than static friction. FsFs book pulled wheel driven book dragged

8. surfaces in contact ss kk leather-soled shoes on wood0.30.2 rubber-soled shoes on wood0.90.7 climbing boots on rock1.00.8 shoes on ice0.10.05 auto tires on dry concrete1.00.8 auto tires on wet concrete0.70.5 auto tires on icy concrete0.30.02 waxed skis on dry snow0.080.04 waxed skis on wet snow0.140.1 wood on wood0.40.2 glass on glass0.90.4 steel on steel - dry0.60.4 steel on steel - greased0.10.05 synovial joints in humans0.010.003 Coefficients of Friction

9. The Static Friction Force In this unit, when we use the following equation, we refer only to the maximum value of static friction and simply write: F f =  s F n When an attempt is made to move an object on a surface, static friction slowly increases to a MAXIMUM value When an attempt is made to move an object on a surface, static friction slowly increases to a MAXIMUM value. n fsfs P W F f ≤  s F n

10. Constant or Impending Motion For motion that is impending and for motion at constant speed, the resultant force is zero and  F = 0. (Equilibrium) F FfFf F – F f = 0 Rest F FfFf F – F f = 0 Constant Speed Here the weight and normal forces are balanced and do not affect motion.

11. Friction and Acceleration When F is greater than the maximum F F the resultant force produces acceleration. Note that the kinetic friction force remains constant even as the velocity increases or decreases. Kinetic is a hard equality. FF Changing Speed This case we will have to use Newton 2 nd law F f =  k F n a

12. EXAMPLE 1: If  k = 0.3 and  s = 0.5, what horizontal pull P is required to just start a 250-N block moving? 1. Draw sketch and free- body diagram as shown. 2. List givens and label what is to be found:  k = 0.3;  s = 0.5; W = 250 N Find: P = ? to just start 3. Recognize for impending motion: P – F fs = 0 FnFn FfFfFfFf P mg +

13. EXAMPLE 1(Cont.):  s = 0.5, mg = 250 N. Find P to overcome F f (max). Static friction applies. 4. To find P we need to know F fs, which is: 5. To find F n : FnFn FfFf P 250 N + For this case: P – F fs = 0 F f =  s F n F n = ?  F y = 0 F n – mg = 0 mg = 250 N F n = 250 N (Continued)

14. EXAMPLE 1(Cont.):  s = 0.5, W = 250 N. Find P to overcome F f (max). Now we know F n = 250 N. 7. For this case 7. For this case: P – F f = 0 6. Next we find F f from: F f =  s F n = 0.5 (250 N) P = F f = 0.5 (250 N) P = 125 N This force (125 N) is needed to just start motion. Next we consider P needed for constant speed. FnFnFnFn FfFf P 250 N +  s = 0.5

15. EXAMPLE 1(Cont.): If  k = 0.3 and  s = 0.5, what horizontal pull P is required to move with constant speed? (Overcoming kinetic friction)  F y = m a y = 0 F n - mg = 0 F n = mg Now: F f =  k F n =  k mg  F x = 0; P - f k = 0 P = F f =  k mg P = (0.3)(250 N) P = 75.0 N FfFf FnFnFnFn P mg +  k = 0.3

16. The Normal Force and Weight The normal force is NOT always equal to the weight. The following are examples: 30 0 P m FnFnFnFn mg Here the normal force is less than weight due to upward component of P.  P FnFnFnFn mg Here the normal force is equal to only the compo- nent of weight perpendi- cular to the plane.

17. m Example 2. A force of 100 N drags a 300-N block by a rope at an angle of 40 0 above the horizontal surface. If u k = 0.2, what force P will produce constant speed? 1. Draw and label a sketch of the problem. 40 0 P =100 fkfk n W = 300 N 2. Draw free-body diagram. The force P is to be replaced by its com- ponents P x and P y. 40 0 P W n fkfk + W PxPxPxPx P cos 40 0 PyPyPyPy PyPyPyPy P sin 40 0

18. Example 2 (Cont.). P = 100; W = 300 N; u k = 0.2. 3. Find components of P: 40 0 P mg n fkfk + P cos 40 0 P sin 40 0 P x = P cos 40 0 = 0.766P P y = P sin 40 0 = 0.643P P x = 0.766P; P y = 0.643P Note: Vertical forces are balanced, and for horizontal forces are unbalanced.

19. Example 2 (Cont.). P = 100; W = 300 N; u k = 0.2 4. Apply Equilibrium con- ditions to vertical axis. 40 0 P 300 N n fkfk + 0.766P 0.643P  F y = 0 P x = 0.766P P y = P x = 0.766P P y = 0.643P n + 0.643P – 300 N= 0 [P y and n are up (+)] n = 300 N – 64.3P; n = 223 N Solve for n

20. Example 2 (Cont.). P = 100; W = 300 N; u k = 0.2. 5. Apply  F x = 0 to con- stant horizontal motion.  F x = 0.766P – f k = ma f k =  k n = (0.2)(223) 76.6 – f k = ma; 40 0 P 300 N n fkfk + 0.766P0.643P n = 223N 76.6-44.7= ma f k = (0.2)(223) = 44.7N

21. Example 2 (Cont.).P = 100N; W = 300 N; u k = 0.2. 40 0 P 300 N n fkfk + 0.766P0.643P 76.6-44.7= 31.9N = ma 6. Solve for mass. 300N= mg; m=30kg 31.9N=30 a If P = 100 N, the block will be accelerated. a = 1.06 m/s 2 a = 31.9/30 a = 1.06 m/s 2

22. xy Example 3: What push P up the incline is needed to move a 230-N block up the incline at constant speed if  k = 0.3? 60 0 Step 1: Draw free-body including forces, angles and components. P 230 N fkfk n 60 0 W cos 60 0 W sin 60 0 Step 2:  F y = 0 n – W cos 60 0 = 0 n = (230 N) cos 60 0 n = 115 N W =230 N P

23. Example 3 (Cont.): Find P to give move up the incline (W = 230 N). 60 0 Step 3. Apply  F x = 0 x y P W fkfk n 60 0 W cos 60 0 W sin 60 0 n = 115 N W = 230 N P - f k - W sin 60 0 = 0 f k =  k n = 0.2(115 N) f k = 23 N, P = ? P - 23 N - (230 N)sin 60 0 = 0 P - 23 N - 199 N= 0 P = 222 N

24. Summary: Important Points to Consider When Solving Friction Problems. The maximum force of static friction is the force required to just start motion.The maximum force of static friction is the force required to just start motion. n fsfs P W Equilibrium exists at that instant:

25. Summary: Important Points (Cont.) The force of kinetic friction is that force required to maintain constant motion.The force of kinetic friction is that force required to maintain constant motion. f k does not get larger as the speed is increased.f k does not get larger as the speed is increased. n fkfk P W

26. Summary: Important Points (Cont.) Remember the normal force n is not always equal to the weight of an object.Remember the normal force n is not always equal to the weight of an object. It is necessary to draw the free-body diagram and sum forces to solve for the correct n value. 30 0 P m n W  P n W

27. Summary Static Friction: No relative motion. Kinetic Friction: Relative motion. f k =  k n f s ≤  s n

28. CONCLUSION: Chapter 4B Friction and Equilibrium