If you a empt to slide one... 1 of 1 16-Sep-12 19:29 APPENDIX B If you attempt to slide one solid object across another, the sliding is resisted by interactions between the surfaces of the two objects. This resistance is referred to as friction force and is oriented parallel to the two surfaces in a direction opposite the direction of (pending) sliding. For example, if you push on a crate of weight W as it rests on the ground, as in Figure B.1a, the friction force exerted by the ground on the crate is in the direction opposite to the sliding direction. An equal and opposite friction force (not shown in Figure B.1a) is exerted by the crate on the ground (per Newton's third law). Figure B.1 (a) Person pushing a crate and the crate's free-body diagram; (b) friction force F friction versus push force P Copyright 2007 John Wiley & Sons, Inc. All rights reserved.
Coulomb Fric on Model 1 of 2 16-Sep-12 19:30 Coulomb Friction Model Friction is a complicated phenomenon that is not fully understood and is an area of continuing research. 1 Current thinking is that friction results from the microscopic irregularities (called asperities) present in all surfaces and from molecular attraction. On a macroscopic scale, friction between surfaces results in the behavior shown in Figure B.1b; an increase in the magnitude of force P is balanced by an increase in the friction force (F friction ) up to a maximum value (P cr ) and the crate remains stationary. An attempt to increase the magnitude of force P beyond P cr results in the crate sliding to the right. We can generalize the behavior in Figure B.1b to say that the friction force (F friction ) is always parallel to the contacting surfaces of the two objects and is directed so as to oppose their relative motion. Furthermore, the friction force is related to and limited by the normal contact force (F normal ) and the characteristics (e.g., smoothness) of the objects' surfaces, it is perpendicular to the normal force, and normal contact force must be present in order for friction force to be present (but not vice versa). We can model the behavior of contacting surfaces of many systems with the elementary friction law proposed by Coulumb in 1781. The Coulumb friction law states: 1. If the two solid objects remain stationary relative to one another (i.e., the two objects do not slide relative to one another), the friction force is such that (B1) 2. where µ s is called the coefficient of static friction. Sample values of µ s are given in Table B.1 and should only be used if values from experiments on the actual system are not available. For specific cases the values in Table B.1 may be incorrect by more than 100%. If the friction force is such that (B2) there is a state of impending sliding of the two objects relative to one another. Therefore, the product µ s F normal is the largest magnitude of friction force that can exist at the contacting surfaces without there being sliding. When this maximum friction force has been developed, the angle θ between F normal and the resultant R (= F normal + F friction ) is called the angle of friction, and θ = tan 1 µ s (Figure B.2). 3. If there is sliding between the two contacting surfaces, the friction force is given by (B3) where µ k is the coefficient of kinetic friction. In general, for given contacting surfaces, µ k < µ s (which is consistent with the behavior depicted in Figure B.1b). Sample values of µ k are given in Table B.1.
oulomb Fric on Model 2 of 2 16-Sep-12 19:30 Figure B.2 The resultant force R when friction is at its limit of µ s F normal Table B.1 Sample Coefficients of Friction * Materials µ s µ k Mild steel on mild steel 0.74 0.57 Aluminum on mild steel 0.61 0.47 Copper on mild steel 0.53 0.36 Cast iron on cast iron 1.10 0.15 Brake material on cast iron 0.40 0.30 Leather on cast iron 0.60 0.56 Rubber on metal 0.40 0.30 Rubber on wood 0.40 0.30 Rubber on pavement 0.90 0.80 Leather on oak 0.61 0.52 Glass on nickel 0.78 0.56 Copyright 2007 John Wiley & Sons, Inc. All rights reserved.
se of the Coulomb Fric on Model in Sta c Analysis of 1 16-Sep-12 19:30 Use of the Coulomb Friction Model in Static Analysis You might be wondering how to use the Coulomb friction law in static analysis. We illustrate its use by considering questions you might ask about the crate sitting on the ground in Figure B.1. For example, say that you are interested in Finding F friction for a Given Value of P If the Crate is in Equilibrium: You would apply the conditions of equilibrium to calculate the size of F friction, followed by using B1 as a check; if the calculated value of F friction is less than or equal to µ s F normal (i.e., F friction µ s F normal ), the crate is stationary relative to the ground. If, on the other hand, the calculated value of F friction is greater than µ s F normal (i.e., F friction > µ s F normal ), the crate is not stationary relative to the ground and will slide (and accelerate) to the right. This means that equilibrium is not possible with the given values of P and W (weight of the crate) and the character of the contacting surfaces (as indicated in the value of µ s ). Finding the Maximum Allowable Magnitude of P Such That the Crate will not Slide: You would apply the conditions of equilibrium to calculate the magnitude of P when F friction is equal to µ s F normal. Finding the Magnitude of P Such That the Crate Moves to the Right at a Constant Speed: You would apply the conditions of equilibrium to calculate the magnitude of P when F friction is equal to µ k F normal. The Coulumb friction law was used when we considered the bicycle in Chapter 2. In the free-body diagram of the bicycle in Figure 2.1 we included F friction where the ground contacts the rear tire. We were implicitly assuming in the analysis of the bicycle that our calculated value of F friction was less than µ s F normal so that the rear tire would not slide relative to ground. This is probably a pretty good assumption (unless we are considering bicycling across a frozen lake!); a check of this assumption should be added to increase the completeness of the analysis. Without dry friction between the tire and the ground, a bicycle would not function. We also factored in the presence of friction when we considered the bridge in Chapter 3. In calculating the required minimum weight of the anchorages that hold the main cables of the Golden Gate Bridge we made an explicit check of weight necessary to prevent sliding by applying B1. The anchorages are able to do their job because of dry friction. Finally, we considered how friction affects the performance of a gear train in Chapter 9. Dry friction between shafts and bearings increases the moment (often referred to as torque) necessary to turn gear trains. Therefore, in gear trains one would hope to reduce friction between shafts and bearings to a very small value (ideally to µ s = 0); this can be done with the use of lubricants and/or more sophisticated bearing systems. The discussion above about contacting surfaces is about dry friction. When a fluid film separates the surfaces, the contact is fully lubricated, and principles from fluid mechanics are required to describe the interaction of one object relative to the other. We don't develop this topic in this book. Copyright 2007 John Wiley & Sons, Inc. All rights reserved.
ther Examples of Fric on in Sta c Analysis 1 of 5 16-Sep-12 19:31 Other Examples of Friction in Static Analysis We now present the analysis of two simple systems that operate because of dry friction. Wedges. A wedge is a simple machine used to make adjustments in the position of one object relative to another. Wedges can also be used to apply a large force. For example, the wedge in Figure B.3 can be used to raise the block. To calculate the maximum force P required to raise the block we apply equilibrium conditions when the wedge is just about to move to the right (and therefore begins to raise the block). When the wedge is just about to move, we can describe the relationship between friction and normal force using B2. More specifically, for the contacting surfaces along A and B (Figure B.4) we write (B4a) (B4b) Figure B.3 A wedge used to raise a block. Support is such that surface at wall is frictionless. Figure B.4 (a) Free-body diagram of block; (b) free-body diagram of wedge The application of equilibrium equation 7.5B applied to the block (Figure B.4a) then allows us to write (B5) which can be solved for F A,normal :
Other Examples of Fric on in Sta c Analysis 2 of 5 16-Sep-12 19:31 (B6) Application of equilibrium equations 7.5A and 7.5B to wedge (assuming that its mass is very small) (Figure B.4b) then allows us to write (B7a) (B7b) Equations B7a and B7b in combination with B6 can be solved for P: (B8) Plots of B8 are shown in Figure B.5a for various values of static friction and illustrate that the force P required to lift a 1-N block increases with the static friction coefficient and with the wedge angle. Notice that for a small coefficient of static friction and wedge angle P is less than one (e.g., µ s = 0.1, α = 10, P = 0.38), whereas for a larger coefficient of static friction and wedge angle P is much greater than one (e.g., µ s = 0.7, a = 20, P = 2.10). This means that the proper selection of coefficient of static friction and wedge angle will allow one to lift a large weight (W) with a (relatively) small force P. Figure B.5 (a) The maximum force P necessary to lift a 1-N block for various coefficients of static friction; (b) conditions under which a wedge-block system is self-locking
ther Examples of Fric on in Sta c Analysis 3 of 5 16-Sep-12 19:31 Under a certain condition, the wedge is self-locking; this means that the wedge will remain in place (holding up the block) even when there is no applied force P. This might be advantageous if you need to keep an object (like the block) in a raised position for an extended period of time. We can determine the condition under which a wedge block system will be self-locking by imposing equilibrium on the block's and wedge's free-body diagrams in Figure B.6 (notice how these free-body diagrams are different from the ones in Figure B.4). Figure B.6 Free-body diagrams of block and wedge with no force P applied For the block we write: (B9a) For the wedge we write (B9b) (B9c) From B9a, B9b and B9c we find that for the wedge to be self-locking when (B10) Figure B.5b shows a plot of wedge angle (α) versus static coefficient of friction (as determined from B10). Convince yourself that the correct regions of this plot have been marked as the self-locking region and the non-self-locking region. Belts. Belts are used to connect mechanical components to one another, often with the intention of transferring power from one component to another. For example, a rubber belt connects a pulley on the crankshaft of an automobile engine to a pulley on the water pump (Figure B.7a). A belt operates because of friction between it and the pulleys that it connects. To calculate the condition under which a belt will not slip on the pulleys, we consider equilibrium of a portion of the belt, as shown in Figure B.7b: (B11a)
Other Examples of Fric on in Sta c Analysis 4 of 5 16-Sep-12 19:31 (B11b) If θ is small, we can write (B12) Figure B.7 (a) A belt connects the crack shaft and the water pump on an automobile engine; (b) free-body diagram of belt segment; (c) free-body diagram of pulley and belt Substituting B12 into B11a and B11b, and recognizing that in the limit each term can be written as a differential, we have (B13a) (B13b)
Other Examples of Fric on in Sta c Analysis 5 of 5 16-Sep-12 19:31 Combining B13a and B13b we write (B14) which we then integrate between the limits of T 1 to T 2 and 0 to β (wrap angle of belt, in radians, Figure B.7c): (B15) (B16a) Equation B16a can also be presented as (B16b) Figure B.8 plots this expression and shows that the greater the belt wrap (larger values of β), the greater the ratio of T 2 /T 1 before belt slipping occurs. Equations B16a and B16b (as well as Figure B.8) also work for cables wrapped around drums. Figure B.8 The ratio T 2 /T 1 versus belt wrap angle (β) Copyright 2007 John Wiley & Sons, Inc. All rights reserved.
Exercises 1 of 6 16-Sep-12 19:31 EXERCISES B1. What is the magnitude of the horizontal force P in EB1 that must be exerted on the 100-kg block to cause the block to move if the coefficient of static friction is 0.25? EB1 B2. What is the maximum angle θ for which the block of mass m in EB2 will not slide down the incline if the coefficient of static friction is 0.30 and P = 0? EB2 B3. If the static and kinetic coefficients of friction are 0.35 and 0.20, respectively, and P = 0 determine the friction force acting on the block in EB2 if a. b. θ = 10 θ = 25 B4. If the static and kinetic coefficients of friction are 0.60 and 0.45, respectively, and the angle of the incline is 15, determine the friction force acting on the block in EB2 if a. b. P = 10 N P = 60 N B5. If the static coefficient of friction between the block and the incline in EB2 is 0.60 and the angle of the incline is 15,
xercises 2 of 6 16-Sep-12 19:31 determine the range of values of the force P for which the block will not slide up or down the incline. B6. The coefficient of static friction between block A and its incline is 0.25 in EB6. What must the minimum coefficient of static friction between block B and its incline be if the blocks are in equilibrium? The mass of block B is twice that of block A. If the coefficient of friction is less than this minimum, in which direction will the blocks slide? EB6 B7. Determine an expression for the force P as a function of wedge angle α and static coefficient µ s required to raise the block in EB7 if there is friction acting along Surfaces 1, 2, and 3. Plot your answer and compare with equation B8. EB7 B8. Determine an expression for the force P as a function of wedge angle α and static coefficient µ s required to remove the wedge in EB8 if there is friction acting along surfaces 1 and 2. Plot your answer. EB8 B9. Determine an expression for the force P as a function of wedge and angle α and static coefficient µ s required to remove the wedge in EB9 if there is friction acting along surfaces 1, 2, and 3. Plot your answer and compare with the answer found in Exercise B8.
xercises of 6 16-Sep-12 19:31 EB9 B10. Determine the magnitude of the minimum horizontal force P that must be applied to the wedge B to raise the block A in EB10. The coefficient of friction between all surfaces is 0.15. Is the wedge self-locking? EB10 B11. Determine the magnitude of the minimum downward force P in EB11 that must be applied to the wedge B to move the block A. The coefficient of friction between all surfaces is 0.15. Is the wedge self-locking? EB11 B12. What is the minimum vertical force P in EB12, applied to wedge E, necessary to push end C of the bar CD to the right? The coefficient of static friction at all surfaces is 0.15. Assume that the pins at A and D are frictionless. If the pins at A and D are not frictionless, would the required force increase, decrease, or remain the same? (Do not do any calculations support your answer with qualitative reasoning.)
Exercises 4 of 6 16-Sep-12 19:31 EB12 B13. The rope connecting the 6-kg block A with block B passes over a fixed cylinder in EB13. Determine the largest and smallest masses of block B for which static equilibrium is possible if the coefficient of static friction is 0.30. EB13 B14. A cable is completely wrapped around the horizontal shaft in EB14. One end is attached to a 60-kg crate, and a tensile force T pulls the other. The coefficient of static friction between the shaft and the cable is 0.30. Determine a. b. the minimum tension T for which the crate will not descend the maximum tension T for which the crate will not rise EB14
xercises 5 of 6 16-Sep-12 19:31 B15. A ship is secured by wrapping a rope around a capstan in EB15. If a dockworker can apply a 170-N force to counteract a 7-kN force by the ship, determine the number of complete turns of the rope about the capstan required to keep the rope from slipping if the coefficient of static friction is 0.30. EB15 B16. Determine the maximum moment M A in EB16 that a motor may apply to pulley A without exceeding the maximum allowable belt tension of 200 N. Also determine the corresponding moment M B exerted on pulley B by its drive shaft if the system is in equilibrium. The coefficient of static friction between the belt and the pulleys is 0.30. EB16 B17. It is known that the maximum moment M A in EB17 that a motor can apply to pulley A without causing the belt to slip over either pulley is 20 N m. The coefficient of static friction is 0.40. For this moment, determine a. b. c. the tension in the belt on either side of pulley A the moment M B on pulley B for equilibrium whether slip is impending at pulley A or pulley B
xercises 6 of 6 16-Sep-12 19:31 EB17 Copyright 2007 John Wiley & Sons, Inc. All rights reserved.
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