PLANE FRICTION. Climber on ''Spud Boy'' Clark Canyon, California

Transcription

1 LAE FRICTIO Climber on ''Spud Boy'' Clark Canyon, California

2 Friction What are the benefits and pitfalls of high or low friction? FRICTIO Can t live with it and sure can t live without it Disadvantages Low friction means slipping Uses energy (loss in engines, etc.) Creates wear on surfaces (abrasion) (i.e. worn out tires) Advantages Friction means traction or stability (tires) Allows for life as we know it (sit/stand) Makes a predictable model of how to use it (i.e.: striking a match) 2

3 Forces of Friction Consider a block of weight 'W' resting on another block. A force '' is applied such that motion is impending The application of the equations of equilibrium dictate: A force '' opposite and equal to 'W' exists. This is called the normal * force and is always perpendicular to the surface supporting weight 'W' A friction force, Ff, opposes ''. This force always opposes the direction of impending motion. It is also parallel to the surface supporting weight 'W' (Impending Motion) W Ff * In this context, 'normal' is defined as a geometric perpendicular 3

4 Coefficient of Static Friction The first law of friction states the force of friction is proportional to the normal force. That is: F f = otice this is a linear relationship. The larger the normal force, the greater the force of friction by an equal factor (Impending Motion) W Example ush a 200 lbf box across the floor ush a 20 lbf box across the floor Ff The force to overcome developed friction Ff will be 10 times greater 4

5 Coefficient of Static Friction Vectors 'Ff' and '' form resultant 'R'. The resultant vector 'R' is at angle φ with the normal, the friction angle The tangent of the friction angle is known as the coefficient of static friction and is denoted by the symbol µs A low value of µs denotes a smooth surface. It is also affected by whether or not the surfaces are wet, and if so, wetted with what (Impending Motion) W Ff R φ s = F f 5

6 Coefficients of Friction Values of µs can be found in the appendix to this online text. They are also found in a number of handbooks and on various Internet websites. However, it is important to understand the following Values of friction coefficients are approximate at best. They are intended only for guidance Exercise care when using friction coefficients. Multiple independent references should be used For any specific application the ideal method of determining the coefficient of friction is through testing 6

7 Coefficients of Friction Coefficients of friction are sensitive to atmospheric dust and humidity, oxide films, surface finish, velocity of sliding, temperature, vibration, and extent of contamination. In many cases the degree of contamination is perhaps the most important single variable A short table is included to illustrate how the coefficient of friction is affected by surface films. When a metal surface is perfectly clean in a vacuum, the friction is much higher than the normal accepted value and seizure can easily occur Effect of films on coefficient of static friction Material Clean Dry Thick Oxide Film Sulfide Film Steel-Steel Copper-Copper

8 Solving a Friction roblem Friction problems can be solved in one of three ways Analytically Trigonometrically Graphically The analytical method is a simple sum of forces analysis. It is the preferred method in most cases The trigonometric method requires the construction of a force triangle. All unknowns are solved using basic trigonometric identities. This method can be easier to apply for certain types of friction problems than the analytical method 8

9 9 Solving a Friction roblem Graphical solutions also require the construction of a force triangle. However, it MUST be drawn to scale Such solutions will generally have a single vector with a known magnitude and direction. The other two vectors will have known directions but unknown magnitudes Start by drawing the vector with known magnitude and direction to scale Then draw the lines of action for the vectors with unknown magnitiude. Make sure they are drawn in a proper direction and tip-to-tail Where the two unknowns intersect defines the magnitude of the two unknowns. These magnitudes can be measured with a ruler. Although not as accurate as the other two methods, it is well within typical engineering tolerances

10 Solving a Friction roblem Analytically M2 M1 What must be the mass of M1 for impending motion to occur? µs = 0.3 M2 = 10 kg (assume pulley is frictionless) W W = M 2 g = 10 kg m/ s 2 = 98.1 = M 1 g Ff F x = 0 = F f F y = 0 = = F f and = 98.1 F f = = = 29.4 FBD of M2 = 29.4 M 2 = g = 29.4 = 3.0 kg m/ s 10

11 φ Solving a Friction roblem W=98.1 R Ff Trigonometrically To solve this same problem trigonometrically, we need to resolve and Ff into vector R and determine angle φ. Draw a force triangle using W, and R and identify angle φ. As illustrated on the previous page... W = M 2 g = 10 kg m/ s 2 = 98.1 = M 1 g φ Sketch the force triangle. It need not be drawn to scale. Solve the force triangle using trigonometric identities R α = tan = 16.7 o = 98.1 tan 16.7 = 29.4 R = = M 2 = g = = 3.0 kg m/ s 11

12 Solving a Friction roblem Graphically A graphic solution starts similarly to a trigonometric solution. A major exception is the force triangle is drawn to scale. As such, select a scale in force For this example, select a scale of 1 = 10 and draw the force triangle. This means angle φ must also be measured correctly Draw the vertical weight vector 9.81 inches long φ R 10 Vector '' has a known direction but unknown length. Draw vector '', placing its tail on the tip of 'W', some arbitrary length to the right. Vector R has a known direction but unknown length. Draw this vector, placing the tip of 'R' on the tail of 'W', until it intersects '' With a ruler, measure vectors '' and 'R'. They should be about 2-7/8 and 10-1/4 in respectively. Apply your scale to arrive at: Apply your scale of 1 in = 10 to arrive at: = 29 and R = 102 M 2 = g = 29 = 2.96 kg m / s 12

13 Forces of Friction The previous discussion and example were contingent upon a force '' of sufficient magnitude to cause impending motion. What happens if '' is less than that value and motion is not impending? In this case, the equations of equilibrium simply dicatate that W = and F =. However, F will be less than Ff as previously determined. As such, the definition and application of µs does not apply W F 13

14 Example 2 Motion ot Impending Assume a wooden crate filled with parts has a weight of 200 lbf and rests on a concrete floor. A force of = 62 lbf is applied Is motion impending? What is the apparent coefficient of static friction? How does it compare to tabulated values? Why? W F 14

15 Example 2 Motion ot Impending By inspection, we see that = 200 lbf and F = 62 lbf. From the appendix, we know that µs = Since: s = F f and must equal Ff for motion to be impending, then: = F f = s = lb f = 124 lb f Obviously, motion is not impending We can also calculate an apparent coefficient of friction 200 lbf s = F = 62 lb f 200 lb f = 0.31 Ff 62 lbf This apparent discrepency is because the coefficient of static friction is valid OLY if motion is impending. Under circumstances defined in this example, µs cannot be applied or calculated. 15

16 Example 3 Static Friction The Stillson wrench is used to assemble and dissassemble pipe Given the dimensions shown, what are the minimum values for coefficient of friction at each of the two jaws? B D A C E F 16

17 Example 3 Static Friction Assume members AB and DE are rigidly attached A friction force and its corresponding normal force develops at point 'A' Recognize the jaw assembly as a two-force member with forces acting through points 'A' and 'D'. Then equilibrium requires the forces at 'A' to be balanced by the forces at 'D' B F D φ R A E Since this is a two-force member, the geometry provides direction. The slope of the resultant is 4.5:1. Since only friction and normal forces act at point 'A', this slope is the tangent of and, in this case, is the friction angle. The minimum coefficient of friction is: A = tan = =

18 Example 3 Static Friction To determine the coefficient of friction of the lower jaw, we will need to develop the FBD of the handle-jaw assembly c We know the direction of vector D from the previous FBD is 4.5:1 (or o ) It is important to note that c and Fc are not equal to and F from the previous FBD. This part of the assembly is not a two-force member; the applied force '' must be considered C D 1 φ Fc 2 20 This requires we write the equations of equilibirum for this FBD before we can do the friction analysis F 18

19 Example 3 Static Friction From previous geometry: = tan = 12.52o M c = cos o D 1 sin o D 2 22 F y = cos o D F x = F c sin o D c = F c c From M c : = D Substituting into F x : F c = D D = D From F y : = D c C D 1 φ Fc = D D = F 19

20 Kinetic Friction We looked at a situation where force '' caused impending motion. This is the only time when the coefficient of static friction can be applied Then we looked at a scenario that did not result in impending motion. It was obvious the coefficient of friction cannot be used to aid in the solution of such problems But what happens when the value of '' exceeds the value required for impending motion? 20

21 Kinetic Friction To answer this, consider pushing a heavy wooden box across a concrete floor. Experience tells us it can take quite some effort to get the box moving, but once moving, we no longer need to apply as much force to keep it moving This illustrates the concept of sliding or kinetic friction. When the box is in motion, the coefficient of static friction does not apply. We must now introduce the concept of kinetic friction A coefficient of kinetic friction is: Fk W Motion k = F k Common values are tabulated in the appendix Let's investigate further the transition from static to sliding friction 21

22 Kinetic Friction Consider a weight resting on a flat surface with a normal force '' between the weight and the surface. Let's apply force Fapp gradually, starting at zero. A friction force, 'f', will develop and increase linearly while the block remains static. This is indicated by the sloping red line When 'f' reaches fs, motion will be impending. At this point, frictional force fs = µs As soon as 'f' exceeds fs, the weight will go in motion. The amount of force, Fapp required to maintain motion will decrease. et Force (Causing Motion) Friction Force fnet = Fapp - f Fapp = Applied force ff,k = µ = ormal force Static Friction Dynamic Friction Friction Force = -Applied Force et Force Applied Force (Fapp) fs = µs fk = µk 22

23 Kinetic Friction The decrease in Fapp implies two very important concepts First, since the normal force remains constant, the coefficient of kinetic friction must decrease Second, if force Fapp is not reduced, the mass will accelerate. Acceleration of a body is covered in a dynamics course If the body does go in motion and if Fapp is reduced to maintain a constant velocity, then the system can be analyzed using static methods et Force (Causing Motion) Friction Force fnet = Fapp - f Fapp = Applied force ff,k = µ = ormal force Static Friction Dynamic Friction Friction Force = -Applied Force et Force Applied Force (Fapp) fs = µs fk = µk 23

24 Example 4 - Kinetic Friction A pallet of oak was delivered to a carpenter shop. The floor of the shop is also of oak. The pallet has a weight of 1000 lbf and needs to be pushed from one end of the shop to the other. How much force is required to start the pallet moving? How much force is required to keep it moving? Motion 1000 lb s For motion to begin: s = 0.62 (From appendix) F y = = 1000 lb f F x = F s... = F s F s = = 620lb f 24 Fs... = 620 lb f

25 Example 4 - Kinetic Friction A pallet of oak was delivered to a carpenter shop. The floor of the shop is also of oak. The pallet has a weight of 1000 lbf and needs to be pushed from one end of the shop to the other. How much force is required to start the pallet moving? How much force is required to keep it moving? Motion 1000 lb k To sustain motion: k = 0.48 (From appendix) F y = = 1000 lb f F x = F k... = F k F k = = 480 lb f 25 Fk... = 480 lb f