However, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied.

Transcription

1 FRICTION 1

2 Introduction In preceding chapters, it was assumed that surfaces in contact were either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent relative motion between surfaces). Actually, no perfectly frictionless surface exists. For two surfaces in contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other. However, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied. The distinction between frictionless and rough is, therefore, a matter of degree. There are two types of friction: dry or Coulomb friction and fluid friction. Fluid friction applies to lubricated mechanisms. The present discussion is limited to dry friction between nonlubricated surfaces. 2

3 Example of friction 3

4 Dry Friction Friction is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body. Experiments show that frictional forces act tangent (parallel) to the contacting surface in a direction opposing the relative motion or tendency for motion. 4

5 For the body shown in the figure to be in equilibrium, the following must be true: F = P, N = W, and Wx = Ph. 5

6 Impending motion Motion F S = m s N coefficient of static friction F k = m k N coefficient of kinetic friction 6

7 The maximum friction force is attained just before the block begins to move (a situation that is called impending motion ). The value of the force is found using F s = m s N, where m s is called the coefficient of static friction. The value of m s depends on the materials in contact. 7

8 The Law of Dry Friction, Coefficients of Friction Block of weight W placed on horizontal surface. Forces acting on block are its weight and reaction of surface N. Small horizontal force P applied to block. For block to remain stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force. As P increases, the static-friction force F increases as well until it reaches a maximum value F m. F ms N m Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force F k. F mk N k 8

9 Maximum static-friction force: F ms N m Kinetic-friction force: F k m N m k k 0.75m s Maximum static-friction force and kineticfriction force are: - proportional to normal force - dependent on type and condition of contact surfaces - independent of contact area 9

10 Four situations can occur when a rigid body is in contact with a horizontal surface: No friction, (P x = 0) No motion, (P x < F m ) Motion impending, (P x = F m ) Motion, (P x > F m ) 10

11 Angles of Friction It is sometimes convenient to replace normal force N and friction force F by their resultant R: No friction No motion Motion impending Motion tan tan s s F N m m s ms N N tan tan k k F k N m k mk N N 11

12 Another examples will show how the angle of friction can be used to advantage in the analysis of certain types of problems. No friction No motion Motion impending Motion 12

13 Problems Involving Dry Friction All applied forces known Coefficient of static friction is known Determine whether body will remain at rest or slide All applied forces known Coefficient of static friction is known Motion is impending Motion is impending Determine value of coefficient of static friction. Determine magnitude or direction of one of the applied forces 13

14 Example Pushing the uniform crate that has a weight W and sits on the rough surface. Slip, tip over, or static? 14

15 Relation with moment Verge of slipping Tip over 15

16 Relation with moment 200 N μs = N Fmax = (0.3) (200 N) 0.2 m 80 N F + = 60 [N] Fx = 0 80N F = 0 F = 80 [N] 80 N 0.4 m O F x N = 200 N + Fy = 0 N = 200 N -200 N + N = 0 N = 200 [N] + M O 0 ; 80(0.4) 200(x) 0 x = 0.16 m 16

17 Relation with moment 200 N μs = N Fmax = (0.3) (200 N) 0.2 m = 60 [N] 80 N 80 N + Fx = m F 80N F = 0 F = 80 [N] O F N = 200 N + Fy = 0 x N = 200 N -200 N + N = 0 N = 200 [N] + M O 0 ; 80(0.6) 200(x) 0 x = 0.24 m tip over 17

18 Sample problem 1350N 450 N A 450 N force acts as shown on a 1350N block placed on a inclined plane. The coefficients of friction between the block and the plane are μs = 0.25 and μk = Determine whether the block is in equilibrium, and find the value of the friction force. 18

19 Solution Force required for equilibrium. 1350N We first determine the value of the friction force required to maintain equilibrium. Assuming that F is directed down and to the left, we draw the free-body diagram of the block and write 450 N 1350N The force F required to maintan equilibrium is an 360N force directed up and to the right ; the tendency of the block is thus to move down the plane. 450N 19

20 Maximum Friction force. 1350N 450 N The magnitude of the maximum friction force can be developed is 1350N Since the value of the force required to maintain equilbrium (360N) is larger than the maximum value which cn be obtained (270 N), equilibrium will not be maintained and the block will slide down the plane. 450N 20

21 Actual Value of Friction Force. 450 N 1350N The magnitude of the actual friction force is obtained as follows : 1350 N The sence of this force is opposite to the sense of motion ; the force is thus directed up and to the right : Factual = 216 N 450 N It should be noted that the forcee is opposite to the sense of motion ; the force F = 216 N N = 1080 N 21

22 Sample problem A support block is acted upon by two forces as shown. Knowing that the coefficient of friction between the block and the incline are μs = 0.35 and μk = 0.25, determine (a) the force P for which motion of the block up the incline is impending, (b) the friction force when the block is moving up (c) the smallest force P required to prevent the block from sliding down 22

23 Solution Free Body Diagram. For each part of the problem we draw a free-body diagram of the block and a force triangle including the 800N vertical force, the horizontal force P, and the force R exerted on the block by the incline. The direction of R must be determine in each separate case. We note that since P is perpendicular to the 800N force, the force triangle is a right triangle, which can easily be solved for P. In most other problems, however, the force triangle will be an oblique triangle and should be solved by applying the law of sines. 23

24 a. Force P for impending Motion of the Block Up the Incline P = (800 N) tan 44.29º = 780 N 24

25 b. Friction Force F when the Block Moves Up the incline R = (800 N) / cos 39.04º = N F = R sin Φk = ( N) sin 14.04º F = 250 N 25

26 c. Force P for impending Motion of the Block Down the Incline P = (800 N) tan 5.71º = 80 N 26

27 Sample problem The moveable bracket shown may be placed at any height on the 75mm diameter pipe. If the coefficient of friction between the pipe and bracket is 0.25, 150mm determine the minimum distance x at which the load can be supported. Neglect the weight of the bracket. 75mm 27

28 SOLUTION: When W is placed at minimum x, the bracket is about to slip and friction forces in upper and lower collars are at maximum value. 150 mm 75 mm 150 mm 75 mm 37.5 mm F F A B m N s m N s A B 0.25N 0.25N A B Apply conditions for static equilibrium to find minimum x. x F 0 : N N 0 B A F y 0 : FA FB W N A 0.25N B W 0 0.5N A W N A NB 2W M 0 : N 6 in. F 3 in. W x 1.5 in. 0 B 6 6 A A N N A N A W x W W W x NA (150 mm) FA (75mm) W (x mm) = NA 75 ( 0.25NA ) Wx W = (2W) (2W) Wx W = 0 x = 300 mm B N x A 12 in. 28

29 PROBLEMS 29

30 Problem Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ= 30º and P = 200 N 30

31 Solution Free Body Diagram 31

32 32

33 Problem Determine whether the 9 kg block shown is in equilibrium, and find the magnitude and direction of the friction force when P = 60 N and θ = 15º 33

34 Solution Free Body Diagram 34

35 35

36 Problem The coefficients of friction are μs = 0.40 and μk = 0.30 between all surface of contact. Determine the forces P for which motion of the 27kg block is impending if cable AB a) is attached as shown b) is removed 18kg 27kg 18kg 27kg 36

37 18kg 27kg 37

38 38

39 18kg 27kg 39

40 40

41 THE END 41