2. Force, Friction and Equilibrium

3. Equilibrium: Until motion begins, all forces on the mower are balanced. Friction in wheel bearings and on the ground oppose the lateral motion.

4. Objectives: After completing this module, 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.

5. 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. Static Friction: No relative motion. Kinetic Friction: Relative motion Kinetic Friction: Relative motion.

6. 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, n The force required to overcome static or kinetic friction is proportional to the normal force, n. f k =  k n f s =  s n n 12 N 6 N n 8 N 4 N n

7. 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.

8. Friction forces are independent of temperature, provided no chemical or structural variations occur. 4 N4 N4 N4 N 4 N Heat can sometimes cause surfaces to become deformed or sticky. In such cases, temperature can be a factor.

9. 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

10. Example 8: A 750kg car traveling at 90.0km/h (25.0m/s) locks its brakes and skids to rest after traveling 35.0m. Find (a)the acceleration and (b)the frictional force. (a)Use the kinematic equation without the time,

11. (b)Now we know what causes the acceleration. It is the forces that act on the car. To make applying Newton’s Second Law easier, the forces are drawn on the coordinate system with their tails at the origin. This is called the “free-body diagram.” Now apply the Second Law to each direction separately Free body diagram

12. The Static Friction Force In this module, when we use the following equation, we refer only to the maximum value of static friction and simply write: f s =  s 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

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

14. Friction and Acceleration When P is greater than the maximum f s the resultant force produces acceleration. Note that the kinetic friction force remains constant even as the velocity increases. P fkfk Changing Speed This case will be discussed later. f k =  k n a

15. 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 s = 0 n fsfsfsfs P W +

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

17. EXAMPLE 1(Cont.):  s = 0.5, W = 250 N. Find P to overcome f s (max). Now we know n = 250 N. 7. For this case 7. For this case: P – f s = 0 6. Next we find f s from: f s =  s n = 0.5 (250 N) P = f s = 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. n fsfs P 250 N +  s = 0.5

18. 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 n - W = 0 n = W Now: f k =  k n =  k W  F x = 0; P - f k = 0 P = f k =  k W P = (0.3)(250 N) P = 75.0 N fkfk n P mg +  k = 0.3

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

20. Review of Free-body Diagrams: For Friction Problems: Read problem; draw and label sketch.Read problem; draw and label sketch. Construct force diagram for each object, vectors at origin of x,y axes. Choose x or y axis along motion or impending motion.Construct force diagram for each object, vectors at origin of x,y axes. Choose x or y axis along motion or impending motion. Dot in rectangles and label x and y compo- nents opposite and adjacent to angles.Dot in rectangles and label x and y compo- nents opposite and adjacent to angles. Label all components; choose positive direction.Label all components; choose positive direction. For Friction Problems: Read problem; draw and label sketch.Read problem; draw and label sketch. Construct force diagram for each object, vectors at origin of x,y axes. Choose x or y axis along motion or impending motion.Construct force diagram for each object, vectors at origin of x,y axes. Choose x or y axis along motion or impending motion. Dot in rectangles and label x and y compo- nents opposite and adjacent to angles.Dot in rectangles and label x and y compo- nents opposite and adjacent to angles. Label all components; choose positive direction.Label all components; choose positive direction.

21. For Friction in Equilibrium: Read, draw and label problem.Read, draw and label problem. Draw free-body diagram for each body.Draw free-body diagram for each body. Choose x or y-axis along motion or impending motion and choose direction of motion as positive.Choose x or y-axis along motion or impending motion and choose direction of motion as positive. Identify the normal force and write one of following:Identify the normal force and write one of following: f s =  s n or f k =  k n f s =  s n or f k =  k n For equilibrium, we write for each axis:For equilibrium, we write for each axis:  F x = 0  F y = 0  F x = 0  F y = 0 Solve for unknown quantities.Solve for unknown quantities. Read, draw and label problem.Read, draw and label problem. Draw free-body diagram for each body.Draw free-body diagram for each body. Choose x or y-axis along motion or impending motion and choose direction of motion as positive.Choose x or y-axis along motion or impending motion and choose direction of motion as positive. Identify the normal force and write one of following:Identify the normal force and write one of following: f s =  s n or f k =  k n f s =  s n or f k =  k n For equilibrium, we write for each axis:For equilibrium, we write for each axis:  F x = 0  F y = 0  F x = 0  F y = 0 Solve for unknown quantities.Solve for unknown quantities.

22. You try it Example 2. A force of 80 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 acceleration will force P produce?

23. m Example. A force of P-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 = ? 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

24. Example (Cont.). P = ?; 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 constant speed, horizontal forces are balanced.

25. Example (Cont.). P = ?; 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 – 0.643P; n = 300 N – 0.643P Solve for n in terms of P

26. Example 2 (Cont.). P = ?; W = 300 N; u k = 0.2. 5. Apply  F x = 0 to con- stant horizontal motion.  F x = 0.766P – f k = 0 f k =  k n = (0.2)(300 N - 0.643P) 0.766P – f k = 0; 40 0 P 300 N n fkfk + 0.766P0.643P n = 300 N – 0.643P 0.766P – (60 N – 0.129P) = 0 f k = (0.2)(300 N - 0.643P) = 60 N – 0.129P

27. Example (Cont.). P = ?; W = 300 N; u k = 0.2. 40 0 P 300 N n fkfk + 0.766P0.643P 0.766P – (60 N – 0.129P )=0 6. Solve for unknown P. 0.766P – 60 N + 0.129P =0 0.766P + 0.129P = 60 N If P = 67 N, the block will be dragged at a constant speed. P = 67.0 N 0.766P + 0.129P = 60 N 0.895P = 60 N P = 67.0 N

28. Example 9: A student drags a 20.0kg laundry bag up a 30.0° ramp at a constant speed by exerting a force of 150.0N along the ramp. Find the forces that act on the laundry bag. The four forces on the laundry bag are the pull of the student, the weight of the bag (gravity), normal force and friction. The pull of the student is given, Gravity can be found using the mass/weight rule,

29. Using the free body diagram to carefully apply the Second Law to each direction separately, Putting in the numbers, For the y-direction, Notice that the normal force is not equal to the weight.

30. Example: The angle of an incline is slowly increased until a wooden block on it begins to slide. Given this angle is 52.0°. Find the coefficient of static friction. Using the free body diagram, apply the Second Law to each direction, Free Body Diagram Plugging in the numbers,

32. xy Example : 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

33. Example (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

34. 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:

35. 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. Equilibrium exists if speed is constant, but f k does not get larger as the speed is increased.Equilibrium exists if speed is constant, but f k does not get larger as the speed is increased. n fkfk P W

36. Summary: Important Points (Cont.) Choose an x or y-axis along the direction of motion or impending motion.Choose an x or y-axis along the direction of motion or impending motion. fkfk n P W +  k = 0.3 The  F will be zero along the x-axis and along the y-axis. In this figure, we have:

37. 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

38. Summary Static Friction: No relative motion. Kinetic Friction: Relative motion. f k =  k n f s ≤  s n Procedure for solution of equilibrium problems is the same for each case:

39. CONCLUSION: Friction and Equilibrium