Engineering Mechanics: Friction (PDF)

Engineering Mechanics Friction in Action What is friction? Friction is a retarding force that opposes motion. Friction types: ❑ Static friction Dry or Coulombic ❑ Kinetic friction friction ❑ Fluid friction Sources of dry fri...

Engineering Mechanics Friction in Action What is friction? Friction is a retarding force that opposes motion. Friction types: ❑ Static friction Dry or Coulombic ❑ Kinetic friction friction ❑ Fluid friction Sources of dry friction ❑ Asperities between contacting surfaces ❑ Interactions at the atomic level Tribology studies sources of friction, lubrication, wear and tear etc. Coefficient of Friction At impending motion: F = µN This is the maximum force for a given N. μ is coefficient of friction. ❑ for impending relative motion μ = μs ❑ for actual relative motion μ = μk ❑ µk < µs The Laws of Dry Friction. Coefficients of Friction Maximum static-friction force: Kinetic-friction force: Maximum static-friction force and kinetic- friction force are: - proportional to normal force - dependent on type and condition of contact surfaces - independent of contact area  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units The Laws of Dry Friction. Coefficients of Friction Four situations can occur when a rigid body is in contact with a horizontal surface: No friction, No motion, Motion impending, Motion, (Px = 0) (Px < F) (Px = Fm) (Px > Fm) © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 5  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units 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 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units Angles of Friction Consider block of weight W resting on board with variable inclination angle θ. No friction No motion Motion impending Motion © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 7  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units Problems involving Dry Friction All applied forces known All applied forces known Coefficient of static friction is known Coefficient of static friction Motion is impending is known Motion is impending Determine value of coefficient Determine whether body will of static friction. Determine magnitude or remain at rest or slide direction of one of the applied forces © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 8  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units Sample Problem 8.1 SOLUTION: Draw the free body diagram for the block. Remember that the friction force is opposite the direction of impending motion. Determine values of friction force and normal reaction force from plane required to maintain equilibrium. Calculate maximum friction force and A 100-N force acts as shown on a 300-N compare with friction force required block placed on an inclined plane. The for equilibrium. If it is greater, block coefficients of friction between the block will not slide. and plane are µs = 0.25 and µk = 0.20. If maximum friction force is less than Determine whether the block is in friction force required for equilibrium and find the value of the equilibrium, block will slide. friction force. Calculate kinetic-friction force. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units Sample Problem 8.1 SOLUTION: Determine values of friction force and normal reaction force from plane required to maintain equilibrium. What does the sign tell you about the assumed direction of Calculate maximum friction force and compare impending motion? with friction force required for equilibrium. What does this solution imply about the block? The block will slide down the plane. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 10  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units Sample Problem 8.1 If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 11 How to invoke laws of friction? Key: Always keep track of the number of unknowns and the number of equilibrium equation. Read : Beer and Johnston 8th Ed. pgs 441-442 Problem type 1: ❑ Like a usual equilibrium problem ❑ Solve using what we learnt earlier ❑ Only a couple of changes: Verify if the surface is capable of handling the load OR Find the minimum friction coefficient required. OR Given slipping occurs find the friction coefficient. Friction law need to be used at the very end.  Problem 1 The friction tongs shown are used to lift a 750-lb casting. Knowing that h = 35in. determine the smallest allowable value of the coefficient of static friction between the casting and blocks D and D’.  SI Version The friction tongs shown are used to lift a 350 kg casting. Knowing that h = 864 mm, determine the smallest allowable value of the coefficient of static friction between the casting and blocks D and D’. 216 mm 162 mm 288 mm 350 kg 144 mm 14 Problem 2 A woman pedals her bicycle up a 5-percent grade and a slippery road at a steady speed. The woman and bicycle have a combined mass of 82 kg. with mass center at G. If the rear wheel is on the verge of slipping, determine the coefficient of friction µs between the rear tire and the road. If the coefficient of friction is doubled, what would be the friction force acting on the rear wheel? (why may we neglect friction under the front wheel) Problem 3 The light bar is used to support the 50-kg block in its vertical guides. If the coefficient of static friction is 0.3 at the upper end of the bar and 0.4 at the lower end of the bar, find the friction force acting on each end for x = 75mm. Also find the maximum value of x for which the bar will not slip.  Stone Crusher Problem Two large cylinders, each of radius r = 500 mm rotate in opposite directions and form the main elements of a crusher for stone aggregates. The distance d is set equal to the maximum desired size of the crushed aggregate. If d = 20 mm, µs = 0.3, determine the size of the largest stones which will be pulled through the crusher by friction alone. Assume the stone to be symmetrically placed on both wheels, and that the weight of the stone is negligible compared to the contact forces. r r s 20  Stone Crusher Problem - Solution N F F W F N N FBD of Aggregate FBD of Crusher Cylinder From FBD of aggregate it can be seen that the aggregate will pass through the crusher cylinders due to its self weight and the friction forces from the two cylinders. Since the aggregate passes through the crusher due to friction alone, it implies that the self weight of the aggregate is neglected in this case. Thus the aggregate reduces to a two-force body subjected to reaction forces from the cylinders at two points only and along the same line. 21  Stone Crusher Problem - Solution θ N R R R F FBD of Aggregate r = 500 mm d = 20 mm FBD of Crusher Cylinder µs = 0.3 The aggregate and cylinder will be at the point of slipping when the maximum size aggregate passes. 22 Stone Crusher Problem - Solution S/2 θ r = 500 mm r  +  d/2 d = 20 mm µs = 0.3 Centerline  of  Crusher   Cylinders From  the  arrangement  of  the  crusher  and  the  maximum  aggregate 23  Problem Type 2 Number of equations is less than number of unknowns. Motion of the body is impending. You are asked to obtain: ❑ Force/Torque required to start the impending motion. ❑ Some distance, angle etc. for the impending motion. Need to use the law of friction at impending surface. Careful about the sign of forces. Problem 4 The coefficients of friction are µs = 0.4 and µk = 0.3 between all surfaces of contact. Determine the force P for which motion of the 30-kg block is impending if cable AB (a) is attached as shown, (b) is removed. Assume that the cord is inextensible and the pulleys are well oiled.  Problem 40 The moveable bracket shown may be placed at any height on the 30 mm diameter pipe. If the coefficient of friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load can be supported. Neglect the weight of the bracket. 27  Problem 40 - Solution Given- Diameter of the pipe=30mm; y Height of the bracket=60mm. x µ =0.25 FBD of Bracket As the bracket is loaded as shown, the reaction forces on the bracket by the pipe is at only 2 points as shown. 28  Problem 40 - Solution Given- Diameter of the pipe=30mm; Height of the bracket=60mm. µ =0.25 FBD of Bracket 29 Problem Type 3 Similar to Problem Type 2, but with one significant difference There can be multiple modes of slipping. Which particular contact at which the impending slippage occurs have to be decided by ❑ Trial and error ❑ Inspection ❑ Physical intuition. Ultimately it must be checked that everything is consistent, i.e., the force on surfaces other than slipping surfaces should be less than µN. Caution: Be careful about the direction of forces Problem 4a A light metal panel is welded to two short sleeves of 0.025 m inside diameter that can slide on a fixed horizontal rod. The coefficient of friction between the sleeves and the rod are μs = 0.4 and μk = 0.3. A cord attached to corner C is used to move the panel along the rod. Knowing that the cord lies in the same vertical plane as the panel, determine the range of values of θ for which the panel will be in impending motion to the right. Assume that sleeves make contact with the rod at the exterior points A and B. A B C  Metal Panel Problem – Two cases NA NB NA NB FA FB FA FB θ P P θ FBD – Case (i) FBD – Case (ii) Small angle Large angle There will two possible situations of loads acting on the panel as shown in the FBDs above: 32  Metal Panel Problem - Solution Case (i) From equilibrium conditions of the panel we get d y h x b b = 400 mm h = 612.5 mm d = 25 mm If θ > 21.80 the panel will move to the right under a net force of Therefore for case (i) the panel will move to the right if 33  Metal Panel Problem - Solution Case (ii) From equilibrium conditions of the panel we get d h Can also take Moment w.r.t B b b = 400 mm h = 625 mm d = 25 mm 34 Problem 4b For the previous problem 4a, assuming that the cord is attached at point E at a distance x = 0.1 m from the corner C. (b) Determine the largest value of x for which the panel can be moved to the right. E x Problem 5 A block of mass mo is placed between the vertical wall and the small ideal roller at the upper end A of the uniform slender bar of mass m. The lower end B of the bar rests on the horizontal surface. If the coefficient of static friction is μs at B and also between the block and the wall, determine a general expression for the minimum value θmin of θ for which the block will remain in equilibrium. Evaluate the expression for μs = 0.5 and m/mo = 10. For these conditions, check for possible slipping at B. Problem 6 The 12-lb slender rod AB is pinned at A and rests on the 36-lb cylinder C. Knowing that the diameter of the cylinder is 12.5 in. and that the coefficient of static friction is 0.35 between all surfaces of contact, determine the largest magnitude of the force P for which equilibrium is maintained Problem 9 Determine the force P required to move the two identical rollers up the incline. Each role weighs 30lb, and the coefficient of friction at all contacting surfaces is 0.2. Problem 5 Two slender rods of negligible weight are pin- connected at A and attached to the 18-lb block B and the 80-lb block C as shown. The coefficient of static friction is 0.55 between all surfaces of contact. Determine the range of values of P for which equilibrium is maintained Problem 3 What is the force F to hold two cylinders, each having a mass of 50 kg? Take the coefficient of friction equal to 0.2 for all surfaces of contact. Problem Type 4 Whether there is an impending slipping or tipping. If slipping occurs: ❑ F = µN at the point of slipping If tipping occurs ❑ Reactions at all points other than the point of tipping is equal to zero. Note that this case is just a special case of what we have seen earlier. Problem 7 A homogenous block of weight W rests on a horizontal plane and is subjected to the horizontal force P as shown. If the coefficient of friction is μ, determine the greatest value which h may have so that the block will slide without tipping. Problem 8 The device shown prevents clockwise rotation in the horizontal plane of the central wheel by means of frictional locking of the two small rollers. For given values of R and r and for a common coefficient of friction μ   at all contact surfaces, determine the range of values of d for which the device will operate as described. Neglect weight of the two rollers. QUESTION? SUGGESTIONS? Problem 1 What is the height h of the step so that the force P will role the cylinder of weight 25kg over the step without impending slippage at the point of contact A. Take the coefficient of friction to be equal to 0.3. Problem 1b A car is stopped with its front wheels resting against a curb when its driver starts the engine and tries to drive over the curb. Knowing that the radius of the wheels is 280 mm, that the coefficient of static friction between the tires and the pavement is 0.85, and that the weight of the car is equally distributed over its front and rear wheels, determine the largest curb height h that the car can negotiate, assuming (a) front wheel drive, and (b) rear wheel drive. Problem 2 A 5 deg wedge is used to lift the 1000-lb cylinder as shown. If the coefficient of friction is ¼ for all surfaces, determine the force P required to move the wedge. Problem 3 θ = 35 deg A slender rod of length L is lodged between peg C and the vertical wall and supports a load P at end A. Knowing that and that the coefficient of static friction is 0.20 at both B and C, determine the range of values of the ratio L/a for which equilibrium is maintained. 3.46 < L/a < 13.63 Problem 4 The moveable bracket shown may be placed at any height on the 3-cm diameter pipe. If the coefficient of friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load can be supported. Neglect the weight of the bracket. Problem 5 Two slender rods of negligible weight are pin- connected at A and attached to the 18-lb block B and the 80-lb block C as shown. The coefficient of static friction is 0.55 between all surfaces of contact. Determine the range of values of P for which equilibrium is maintained  Problem 2 What is the minimum coefficient of friction required just to maintain the bracket and its 250 kg load? The center of gravity is 1.8m from the centerline. Problem 3 What is the force F to hold two cylinders, each having a mass of 50 kg? Take the coefficient of friction equal to 0.2 for all surfaces of contact. COMPLEX FRICTION  Square-Threaded Screws Square-threaded screws are frequently used in jacks, presses, etc. Analysis is similar to block on inclined plane. Recall that friction force does not depend on area of contact. Thread of base has been “unwrapped” and shown as straight line. Slope is 2πr horizontally by L vertically. Moment of force Q is equal to moment of force P. Impending motion Self-locking, solve Non-locking, solve upwards. Solve for Q. for Q to lower load. for Q to hold load. 62  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units Journal Bearings. Axle Friction Journal bearings provide lateral support to rotating shafts. Thrust bearings provide axial support Frictional resistance of fully lubricated bearings depends on clearances, speed and lubricant viscosity. Partially lubricated axles and bearings can be assumed to be in direct contact along a straight line. Forces acting on bearing are weight W of wheels and shaft, couple M to maintain motion, and reaction R of the bearing. Reaction is vertical and equal in magnitude to W. Reaction line of action does not pass through shaft center O; R is located to the right of O, resulting in a moment that is balanced by M. Physically, contact point is displaced as axle “climbs” in bearing. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 63  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units Thrust Bearings. Disk Friction Consider rotating hollow shaft: For full circle of radius R, © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 64 Belt Friction  Vector Mechanics for Engineers: Statics Tenth Edition n SI Units Belt Friction Relate T1 and T2 when belt is about to slide to right. Draw free-body diagram for element of belt Combine to eliminate ΔN, divide through by Δθ, In the limit as Δθ goes to zero, Separate variables and integrate from © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 66  Belt  Friction   Application capstan Diesel  engine motor  belt Band  Brake 67  Problem 46 A pulley requires 200 Nm torque to get it rotating in the direction as shown. The angle of wrap is π radians, and µs = 0.25. What is the minimum horizontal force F required to create enough tension in the belt so that it can rotate the pulley? T2 250 mm T1 68  Problem 46 - Solution T2 T T1 For torque T = 200 Nm in given direction, T2 > T1 Equating moments acting on the pulley we get: (T2-T1)*0.25m = 200Nm (i) Considering friction in the pulley the forces T1 and T2 can be correlated as: Combining T2 from eqn. (ii) in eqn. (i) we get: 69  Problem 46 - Solution T2 T T1 Putting µs = 0.25 and β = π radians in the above equation we get 1.193 T1 = 800N Equating forces acting on the pulley we get: F = T1+T2 70  Problem 47 A flat belt connects pulley A to pulley B. The coefficients of friction are µs = 0.25 and µk = 0.20 between both pulleys and the belt. Knowing that the maximum allowable tension in the belt is 600 N, determine the largest torque which can be exerted by the belt on pulley A. 71  Problem 47 - Solution T1 T2 = 600 N T1 = 355.4 N T2 = 600 N Since angle of contact is smaller, slippage will occur on pulley B first. Determine belt tensions based on pulley B. Equating moments acting on pulley A we get 72  Problem 48 Knowing that the coefficient of static friction is 0.25 between the rope and the horizontal pipe and 0.20 between the rope and the vertical pipe, determine the range of values of P for which equilibrium is maintained. N 73  Problem 48 - Solution T=100N µ1=0.25 (between rope & horizontal pipe) µ2=0.20 (between rope &vertical pipe) N Consider the sections of the rope as shown. Due to the friction, the tension in each part is different. P will be max if there is impending slip in the direction of P (i.e. P>T2>T1>T). P will be min if there is impending slip in the direction of T (i.e. PT µ1=0.25 (between rope & horizontal pipe) P = Pmin : PT µ1=0.25 (between rope & horizontal pipe) P = Pmin : PT1). T will be min if there is impending slip in the direction opposite to T (i.e. TT1) m=500kg, T = Tmin: (TT1) T = Tmin: (T