Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
6 Friction 1 Objectives Students must be able to • Utilize theory of dry friction – Describe theory of dry friction – Describe physical meanings of frictional effects – Describe and differentiate between static and kinetic coefficients of friction – Describe the angles of frictions – Add friction into the analyses of objects and structures in equilibrium 2 Objectives Students must be able to • Describe and analyze machines with frictions – Wedges – Threads, screws – Belts – Disks and clutches – Collar, pivot, thrust and journal bearings • Outline rolling resistance – Describe the physical meanings of rolling resistance – Differentiate between frictions and rolling resistance 3 Topic in textbook • Section A: Frictional Phenomena – Characteristics, theory, coefficient of friction, angle of friction • Section B: Applications – Wedges – Screws – Journal Bearings – Thrust Bearings; Disk friction – Flexible Belts – Rolling Resistance We will study this Part first. 4 Dry Friction Force of resistance acting on a body which prevents or retards slipping of the body relative to a surface with which it is in contact. Friction exists? roughnesses of the contacting surfaces. Magnitude: friction’s magnitude limitation will be discussed later Direction: tangent to the contacting surface and opposed to the relative motion or tendency for motion Line of Action (Point of application): contact surface 5 Equilibrium Friction Model [ F 0] F P N W [ M 0] x h P P h N W y W In equilibrium FBD is correct? a/2 a/2 x P modeling h F The DN at right side is supporting force more than its left side. x N If x > a/2 ? • Frictional force F • The application point (x) of N increases with force P Slipping and/or Tipping Effect P N The object is toppling (not in equilibrium) toppling Slipping x-limit F-limit x F 6 a/2 a/2 Motion P h F x N N • Slipping / Sliding – Relative sliding (translation motion) between two surfaces • Toppling / Tippling – Fall over (rotation) about the edge – Topple, tipping, rolling, tumble, trip 7 Experiment for determining Friction mg P m P FBD F F Static friction (no motion) Object at rest (no motion) Fk = kN Kinetic friction (motion) Fs:max = sN k , s N “impending motion”’ (on the verge of motion) P Fs S N S : coefficient of static friction Object with motion (steady state) Fk k N k : coefficient of kinetic friction : constant on 2 certain contacting surfaces s k (generally) 8 no change in k Angle of Friction however P, mg are Not depend on N Fk k N k tan 1 not depend on P,v,a Fk tan 1 k N k k = arctan(k) = angle of kinetic friction k k Fs S N (object in motion) ? tan 1 ( s ) Fs s tan N 1 S N mg s F 1 1 s max tan s tan N s = arctan(s) = angle of (max) static friction s s (object at rest) P N F R P Fs s 9 Dry Friction Dry Friction Characteristics • Frictional force acts tangentially to the contacting surfaces, opposing the relative or tendency for motion. • Fs is independent of the area of contact, provided that the normal pressure is not very low nor great enough for deformation of the surfaces. • In equilibrium: Impending slipping: Slipping: Very low velocity: = s = k k s 10 Dry Friction Impending Motion Static Friction Typical Values Contact Materials Metal / Wood / Leather / Leather / Aluminum / ice wood wood metal Aluminum μs 0.03 – 0.05 0.30 – 0.70 0.20 – 0.50 0.30 – 0.60 1.10 – 1.70 How to obtain values of s ? 11 Sample 6/1 Determine the maximum angle which the adjustable incline may have before the block of mass m begins to slip. The coefficient of static friction between the block and the inclined surface is s. Fx 0 : Fy 0 : W=mg y H/2 x F Three force member F Fmax s N x “Impending Slip”: N 3 eq. , 3 unknowns mg sin max s (mg cos max ) (for slipping) Ans Possibility of toppling? [ M 0 ] x Fmax s N s (mg cosmax ) max tan 1 s H x tan 2 H mg cos ( x ) mg sin ( ) 0 2 H x tan 2 F F mg sin 0 N mg cos 0 x W=mg H tan 2 L 2 L/2 H /2 max,toppling tan 1 12 6/125 A uniform block of mass m is at rest on an incline z. Determine the maximum force P>0 that can be applied to the block in the direction shown before slipping begins. The coefficient of static friction between the block and the incline is s Consider only when z f s N y x F 0 : N mg cos 0 F 0 : mg sin f sin 0 f sin mg sin F 0 : P f cos 0 f cos P z x mg y tan s y f P mg sin 2 P N At this max P, object is about to move at which * direction? ( ?) P mg cos ( s tan ) 2 N mg cos s mg cos Pmax mg cos ( s tan ) mg sin tan tan Pmax ( s tan ) * 13 Dry Friction Example Friction 2 #1 Will this crate slide or topple over? 14 Fx 0 P F Fy 0 W N M 0 Ph xN Known : W , h, s 1st Eq. 4 Unknowns: P, F, N, x 2nd Eq. 3rd Eq. Two possibilities 1) about to slip F s N 2) about to tip 4th Eq. b x 2 4th Eq. use Pmin as answer It’s time consuming, Better to know it exactly P sW * N W * x* s h Check condition: s b x 2 * ? bW P 2 h N† W bW F† 2 h † Check condition: F † ? s N † The block of mass m is homogeneous, moving at a constant velocity. The coefficient of kinetic friction is k - Determine a) the greatest value that h may have so that the block will not tip over b) The location of point C on the bottom face of the block through with the resultant of the friction and normal force act if h = H/2. k 16 known values : m, H , b, k Moving With no acceleration - The block of mass m is homogeneous, moving at a constant velocity. The coefficient of kinetic friction is k - Determine =? k Solution 1 a) the greatest value that h may have so that the block will not tip over on the verge of tipping F k N tan 1 k (certain value) Moving The box is in Equilibrium P mg h=? Three force member in Statics Concurrent at one point OR F F N N on the verge of tipping (no acceleration) 1 b h ( ) tan 2 1 b ( ) k 2 parallel Ans 17 known values : m, H , b, k Moving with constant velocity =H/2 - The block of mass m is homogeneous. - The block is moving at a constant velocity. - Determine Solution 1 b) The location of point C on the bottom face of the block through with the resultant of the friction and normal force act if h = H/2. C =? N k F k N tan k Moving The box is in Equilibrium (no acceleration) Three force member P mg F x N h=H/2 Concurrent at one point OR parallel H x tan 2 H k 2 Ans 18 Inequality: hard to deal Dry Friction’s Problem F s N F 0 M O static friction 0 + or F k N Equilibrium eq. kinetic friction fricitional eq. Friction • Kinetic motion is known. F 0 M O 0 Equilibrium eq. + Moving at constant vel. F k N Kinetic friction 20 F 0 M O 0 F s N + Static friction not assuming In static equilibrium No. of unknown must = No. of Equilibrium eq. “impending motion” friction can support equilibrium Static equilibrium? Assume: static equilibrium. F s N (usually) : can't be used F 0 M O 0 Equilibrium eq. Only equilibrium eq. can be used to determine unknown values. Solve for F (and also N) using only equilibrium eq. Assumption Checking Friction is determined by equilibrium eq’s. Static friction If F s N Static friction However, F must <= N F s N F Fmax s N O.K. F Fmax Assumption is not true! 21 Motion occurs. F = kN F 0 M O 0 Equilibrium eq. + F s N Static friction Find min to hold equilibrium total unknown < equilibrium + frictional eq. Find min P to initiate the motion unknown : Impending motions occur at both point in the same time. F 0 M O 0 Equilibrium eq. F = s N static friction unknown : F N F N A A B B P FA N A Fc N c Bx By Both can be used total unknown together to = determine equilibrium + unknown frictional eq. values. impending motions do not occur in both point at the same time. several possibilities to slip Assumption needs to be make and that assumption should be checked later. 22 500N and 100N Equilibrium? Sample 6/3 Determine the friction force acting on the block shown if P = 500N and P = 100N. The block is initially at rest. s 0.2 k 0.17 Equilibrium state is unknown Assume: Body in equilibrium (not assuming the impending motion) 2 Eq , 2 unknown Fx 0 : P cos F mg sin Fy 0 : N P sin mg cos 0 F ? s N impending motion P=500N Assumption Checking: P=100N Fy 0 : Not valid! still valid! OK ! | F | s N 0.2(1093) 219 N F 242N Object not in Equilibrium Fx 0 : N 1093N F 134.3N N 956N Contradict! 0 F s N Friction must be enough to maintain Correct? equilibrium | F | s N 0.20(956) 191.2 N N P sin mg cos 0 N 956N F k N 0.17(956) 162.5N 23 Dry Friction Example Hibbeler Ex 8-1 #1 The uniform crate has a mass of 20 kg. If a force P = 80 N is applied to the crate, determine whether it remains in equilibrium. The coefficient of static friction = 0.3. Equilibrium State is not known 24 Example Hibbeler Ex 8-1 #2 Assume: Equilibrium Fx 0 80cos30 F 0 F 69.3 N F s NC (70.8 N) Fy 0 80 sin30 196.2 NC 0 NC 236 N MO 0 80 sin30(0.4) 80cos30(0.2) NC ( x ) 0 x 0.0908 m |x| < 0.4 m (physical boundary of toppling) x 0.4 m The crate will not topple. F s NC (70.8 N) The crate, is close to but doesn't not slip. 25 Dry Friction Example Bedford Ex 9.5 #1 Suppose that = 10° and the coefficient of friction between the surface of the wedge and the log are s = 0.22 and k = 0.20. Neglect the weight of the wedge. – If the wedge is driven into the log at a constant rate by vertical force F, what are the magnitude of the normal forces exerted on the log by the wedge? – Will the wedge remain in place in the log when the force is removed? 26 = 10° s = 0.22 and k = 0.20. Neglect the weight of the wedge. Symmetry about the center of the wedge Moving down at constant rate Fy 0 2N sin( ) 2k N cos( ) F 0 2 2 2 N 2 s* N symmetric/mirror concept F 2sin( ) 2k ( ) 2 2 F N 1.75F # 10 10 2sin( ) 2(0.20)( ) 2 2 27 = 10° s = 0.22 and k = 0.20. Neglect the weight of the wedge. Will the wedge remain in place in the log when the force is removed? Impending Motion * Think of minimum friction coefficient s (On the verge of slipping) that still “self-lock” the wedge If s s* the wedge will remain in place If s s* the wedge will move out. Symmetry about the center of the wedge s* N s* N symmetric/mirror concept Fy 0 2N sin( / 2) 2s*N cos( / 2) 0 s* tan( / 2) 0.087 As s s* , wedge will remain in place. # 28 Dry Friction Example Bedford 9.20 #1 The coefficient of static friction between the two boxes and between the lower box and the inclined surface is s. What is the largest angle for which the lower box will not slip. 29 Impending Motion at same time ? The coefficient of static friction between the two boxes and between the lower box and the inclined surface is s. What is the largest angle for which the lower box will not slip. FBD of upper block Fy 0 N2 W cos 0 N2 W cos FBD of lower block Fy 0 N2 W cos N1 0 N1 2W cos Fx 0 s N2 s N1 W sin 0 s 3W cos W sin s tan / 3 or tan1(3 s ) Unknown: N1 N2 T # Eqs : 2obj 2eq (don't think of tipping - no moment eq 30 Bedford 9.30 (ex) Impending Motion at same time The cylinder has weight W. The coefficient of static friction between the cylinder and the floor and between the cylinder and the wall is s. What is the largest couple M that can be applied to the stationary cylinder without causing it to rotate. Fy 0 s Nw Nf W 0 Fx 0 Nw s Nf 0 MO 0 M s (Nw Ns )r 0 sW Nw , 2 1 s Impending rotation W Nf 1 s2 (1 s ) M sWr 1 s2 # 31 Bedford 9.33 (Ex) The disk of weight W and radius R is held in equilibrium on the circular surface by a couple M. The coefficient of static friction between the disk and the surface is s. Find that the largest value M can have without causing the disk to slip. Fx 0 Fs W sin 0 Fy 0 N W cos 0 MO 0 M FsR 0 s tan , N W cos cos 1 tan2 1, tan =s (Impending slip) Fs s N M s NR sWR cos M sWR 1 s2 32 Dry Friction Example Bedford 9.166 #1 Each of the uniform 1-m bars has a mass of 4 kg. The coefficient of static friction between the bar and the surface at B is 0.2. If the system is in equilibrium, what is the magnitude of the friction force exerted on the bar at B. 33 Dry Friction Example Bedford 9.166 #2 Fx 0 Ox Ax 0 Fy 0 Oy Ay 4g 0 MO 0 Ay (1cos 45) Ax (1sin 45) 4g (0.5 cos 45) 0 34 Dry Friction Example Bedford 9.166 #3 Fx 0 Ax F cos 30 N sin 30 0 Fy 0 Ay 4g F sin 30 N cos 30 0 MB 0 Ay (1cos 45) Ax (1sin 45) 4g (0.5 cos 45) 0 Ay 0, Ax 2g, Oy 4g, Ox 2g N 43.8 N, F 2.63 N # 35 Dry Friction Example Bedford 9.166 #4 symmetry? 36 Dry Friction Example Hibbeler Ex 8-3 #1 The rod with weight W is about to slip on rough surfaces at A and B. Find coefficient of static friction. Direction of N? 38 Dry Friction Example Hibbeler Ex 8-3 #2 Impending slip FA s N A FB s NB 3 equilibrium equations, 3 unknowns (N A , NB , s ) Fx 0 Fy 0 M A 0 s N A s NB cos 30 NB sin30 0 (1) N A W NB cos 30 s NB sin30 0 (2) NB l W cos 30(l / 2) 0 (3) 39 Dry Friction Example Hibbeler Ex 8-3 #3 From (3) NB 0.4330W From (1) & (2) 0.2165 s2 s 0.2165 s 0.228 # 40 6/9 For a 20 jaw opening, what is the minimum coefficient of static friction between the jaws and the tube which will enable tongs to grip the tube with out slipping Solution 1 y M M F1 N1 x F N2 o F2 A x 0: F1 (r) F2 (r) 0 F1 F2 0: N1 (l ) N2 (l ) 0 N1 N 2 0: N1 sin N2 sin F1 cos F2 cos 0 2( N1 ) sin 2( F1 ) cos F1 (tan ) N1 tan s ( s ) min tan tan(20 / 2) 0.176 (...= ... F2 (tan ) N 2 ) F1 (tan ) N1 ( F1 )max s N1 41 6/9 For a 20 jaw opening, what is the minimum coefficient of static friction between the jaws and the tube which will enable tongs to grip the tube with out slipping Solution 2 y The object is in equilibrium N1 F1 R1 N2 Two force systems x R1 F1 N1 R2 F2 N 2 R2 F2 2 20o 10o F2 F 1 tan N2 N1 tan s F s N F s N ( s ) min tan tan10o 42 Sample 6/5 Find the maximum value which P may have before any slipping takes place. ? Possible ways to slip 1) Middle object is going to move lonely. 2) Middle + buttom object is going to move together. y T N1 x F1 F1 y N2 x y x F2 F3 F F1 F2 P m2 g sin 0 f 2 2 N 2 0: Fx 0 : F2 F3 m3 g sin 0 f 3 3 N 3 N1 m1g cos Fy 0 : N2 N1 m2 g cos 0 N2 N1 m2 g cos (m1 m2 ) g cos N3 N2 f1 1 N1 Fy 0 : N1 m1g cos 0 N1 m2 g F2 T F1 m1g sin 0 x m1g P Fx 0 : m3 g Fy 0 : N 3 N 2 m3 g cos 0 N 3 N 2 m3 g cos (m1 m2 m3 ) g cos 44 Friction F3 can support the equilibrium of 40-kg object Check the Assumption Assume: 1) Middle object is going to move lonely. Impending motion y T N1 F2 ( F2 )max 2 N 2 2 (m1 m2 ) g cos ) x F1 m1g y x T F1 m1g sin 0 F 0: F1 F2 P m2 g sin 0 0: F2 F3 m3 g sin 0 F x N2 N1 P y x F2 F3 2 friction eq F3 F2 m3 g sin 468N ( F3 )max 3 N 3 3 (m1 m2 m3 ) g cos 459 N N3 N2 3 equilibrium eq P [ 1m 1 2 (m1 m2 )]g cos m2 g sin 103.074N m2 g F2 5 unknowns Fx 0 : x F1 F1 ( F1 )max 1N1 1 (m1g cos ) m3 g F3 ( F3 ) max Impossible, the assumption is wrong 45 F2 can support the 50-kg and 40-kg as one body Assume: 2) Middle + bottom object is going to move together. F1 ( F1 )max 1N1 1 (m1g cos ) F3 ( F3 )max 3 N 3 3 (m1 m2 m3 ) g cos Impending motion y T N1 x F1 F1 y F 0: F1 F2 P m2 g sin 0 x 0: F2 F3 m3 g sin 0 F x N2 F2 3 (m1m2 m3 ) g cos m3 g sin 263N N1 P m2 g y x F2 N2 ( F2 )max 2 (m1m2 ) g cos 272 N F2 ( F2 )max N3 F3 T F1 m1g sin 0 x m1g F2 Fx 0 : m3 g OK P 1m1g cos F2 m2 g sin 93.8N Ans 46 Assume: ? 3) 40-kg object moves lonely --– possible? F2 ( F2 )max 2 N 2 2 (m1 m2 ) g cos F3 ( F3 )max 3 N 3 3 (m1 m2 m3 ) g cos y T N1 x F1 T F1 m1g sin 0 F 0: F1 F2 P m2 g sin 0 0: F2 F3 m3 g sin 0 x m1g F1 y x F x N2 N1 P Equation without unknown (little chance to be true) m2 g F2 Fx 0 : y x F2 N3 F3 N2 m3 g 47 C 0.3 y + 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. x f B NC 0 B 0.6 unknown: 8 fc Nc f B N B N B fC W1 0 0.6 fC 0.8N C M 0.3W1 0 A 0.4 NC W1 50 fC fB NB NB fB fA N A N B W2 0 Motion Possibility? f A fB 0 1. Slip at C, B 0.3FB x( N B W2 ) 0 M x W2 20 NA M fA NA x 2. Slip at C, A f C C N C 3. Slip at A, B ? f B B N B 4. other… f A AN A 48 C 0.3 y + 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. x B 0.6 Assume f B NC 0 2. Slip at C, A N B fC W1 0 0.6 fC 0.8N C M 0.3W1 0 A 0.4 NC W1 50 f A fB 0 fC 0.3FB x( N B W2 ) 0 M fB f C C N C NB NB fB fA x N A N B W2 0 W2 20 NA 8 unknowns, 6+2 equation. Don’t solve 8 eq paralleling! Solve only 6 eq. excluding f C C N C f B B N B Assumption f A AN A Checking f A AN A 49 C 0.3 y + x B 0.6 8-56 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. Assume f B NC 0 N B fC W1 0 0.6 fC 0.8N C M 0.3W1 0 Block is tipping A 0.4 NA W1 W2 = 62.5 N 1 A C N A N B W2 0 x 0.3 A = 0.12 m f A fB 0 M (0.8 0.6 C ) W1 A CW2 = 42.5 N 0.3FB x( N B W2 ) 0 NB 1 A C f B NC f A A f C A C 2. Slip at C, A W1 W2 = 25 N 1 A C W1 W2 = 7.5 N 1 A C f C C N C f B B N B Assumption Checking f A AN A A (W1 W2 ) 1 A C 0.3W1 = 9.5 N-m f C C N C 25 ? (0.6)(42.5) = 37.5 f A AN A Correct? 50 C 0.3 y + 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. x B 0.6 f B NC 0 unknown: fc Nc f B N B N B fC W1 0 0.6 fC 0.8N C M 0.3W1 0 A 0.4 NC W1 50 f A fB 0 fC f C C N C NB NB fB fA x W2 20 NA f B B N B Possibility? 1. Slip at C, B 0.3FB x( N B W2 ) 0 M fB N A N B W2 0 M fA NA x 2. Slip at C, A 3. Slip at A, B ? 4 C2 6 4. C-Slip, other…x-limit f A AN A 5. B-Slip, x-limit x 0.1 6. A-Slip, x-limit 51 C 0.3 y + 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. x B 0.6 Assume f B NC 0 4. C-Slip, x-limit N B fC W1 0 0.6 fC 0.8N C M 0.3W1 0 A 0.4 NC W1 50 f A fB 0 fC 0.3FB x( N B W2 ) 0 M fB f C C N C NB NB fB fA x N A N B W2 0 W2 20 NA f B B N B f A AN A x 0.1 8 unknowns, 6+2 equation. Don’t solve 8 eq paralleling! Solve only 6 eq. excluding f C C N C Assumption Checking x 0.1 52 C 0.3 y + 8-61 The uniform 50-N slender rod rests on the top center of the 20-N block. Determine the largest couple moment M which can be applied to the rod without causing motion of the rod. x B 0.6 Assume f B NC 0 4. C-Slip, x-limit N B fC W1 0 0.6 fC 0.8N C M 0.3W1 0 N 3(W1 W2 ) A 3 C A 0.4 N A N B W2 0 x 0.1 W1 W2 3 C = 21.21 N f A f B NC W W2 N B W1 C 1 3 C = 43.64 N fC C (W1 W2 ) 3 C = 6.36 N f A fB 0 M 0.3FB x( N B W2 ) 0 f C C N C f B B N B f A AN A x 0.1 = 63.64 N (0.8 0.6C )(W1 W2 ) 3 C 0.3W1 = 5.78 N-m f C C N C Assumption Checking Ans 21.21 <=0.6*43.64=26.18 21.21 <=0.4*63.64=25.46 x 0.1 53 54 s 0.05 6/28 Determine the minimum angle at which slipping does not occur at either contact point A or C. The coefficient of static friction at both A and C is 0.05. (Note: you should assume the mass of each bar – why?) known : m s 2mg Bx Bx N1 F1 N2 By unknown : mg F2 By F1 F2 N1 N 2 Bx By 2mg Equilibrium : 3 2 6 Friction : 2 mg N1 F1 N2 F2 55 s 0.05 F F 2mg mg N1 N2 F1 2 force F2 2mg Bx N1 F1 6/28 Determine the minimum angle at which slipping does not occur at either contact point A or C. The coefficient of static friction at both A and C is 0.05 F1 F2 x 0: F1 F2 0 y 0: N1 N2 3mg 0 N1 N2 3mg M B 0 : N1(l cos ) N 2 (l cos ) N1 N 2 l F1 (l sin ) F2 (l sin ) mg( cos ) 0 2 M B 0 : N1 (l cos ) F1 (l sin ) 0 F1 By 1 ( N1 ) tan N1 5 mg 4 mg 2 N2 7 mg 4 1 5 ( mg ) tan 4 ( ) (tan ) ( F1 F2 ) ... A or C is going to slip first? s ( N1 ) ( F1 ) max s ( N 2 ) ( F2 ) max F2 F1 1 5 ( mg ) tan 4 s ( N1 ) ( F1 ) max min tan 1 ( 1 s 1 ( N1 ) tan min ) tan 1 (1 / 0.05) 63.435 56 P applies at pin B, AB or BC? 8-61 The end C of the two-bar linkage rests on the top center of the 50-kg cylinder. Determine the largest vertical force P which can be applied at B without causing motion. Neglect the mass of the bars. P applies at pin B C 0.6 P From solving equilibrium eqs. NC P P fc tan fC C N C ? fC ( E 0.3 2 force P tan 1 ) P 0.577 P tan C NC 0.6P NC fc f C always f C ,max However value P are, BC will not slip at C. 57 8-61 The end C of the two-bar linkage rests on the top center of the 50-kg cylinder. Determine the largest vertical force P which can be applied at B without causing motion. Neglect the mass of the bars. Max P without causing motion NC P fC P tan Slipping or tipping? C 0.6 W mg 50(9.81) f E f C P cot N E NC W P W fE x( P mg ) 0.2 f C 0 E 0.3 Slipping? f E E N E P cot E ( P mg ) P E mg 531.227 cot E x NE Tipping (x=0.05)? 0.05( P mg ) 0.2 P cot 0 P 5mg 374.6 20cot 5 Ans minimum value 58 6/23 The 10-kg cylinder is initially placed at V-Block. If s 0.5 , find (1) friction force F acting at each side when P=0 (2) The P to start sliding the cylinder up y F1 F2 ? Assume: body in equilibrium mg (incomplete) [ M y ,origin 0] F1 F1 30 y y [ Fx 0 ] x P z 2N cos 45 2F N1 N2 No tendency to move in z-axis direction !! Fz 0 N1 N2 =N F y 0 2 N cos45 mg cos30 N = 601 2 F 10(9.81) sin 30 0 F 24.5N 45 F1 2F mg cos 30 45 F2 Assumption Checking Fmax s N 0.5(60.1) 30.0 N OK (2) Find P min to move F x 0: impending motion P 2 F 10(9.81)sin 30 0 60 P 2(0.5)(60.1) 10(9.81)sin 30 109.1N A C B D Determine the minimum coefficient of static friction at each point of contact so that the pile does not collapse. weight W, radius r Think of Equilibrium to find out the relation of F and N FA FB N A NB F N 2N cos30 2F sin30 - W 0 FC FA F ( s , A ) min N W 2 sin 30 1 cos 30 N sin30 F cos30 F 0 F sin 30 W sin 30 N 1 cos 30 2 1 cos 30 NC W N cos30 F sin 30 0 1 sin 30 ( s ,C ) min 3 1 cos 30 NC 3W 2 FC 61 6/108 Describe what should happen Posibility? 1) A B and C don’t move 2) A move, B and C don’t move 3) B move (so do A), C don’t move Many possibilities? 4) C move (so do A and B) ? tan 1 C C 0.35 B 0.20 A 0.30 15o Yes: C not moving No, C is moving. C may or may not moving depends on friction at A and B. 15 ? tan 1 (0.35) 19.29 C not moving 15 tan 1 (0.2) 11.30 B may or may not moving depends on friction at A. 15 tan 1 (0.3) 16.69 A may or may not moving depends on friction at A. 62 6/108 Describe what should happen Assume 1) A and B don’t move, C don’t move f B s N B = 75.806 N B mB g cos = 379.03 mC g y mA g N2 mB g x fC N1 = 58.403 f A mA g sin N1 = 125.507 NB fB fA N1 mB g sin f B NC Assumption Checking NA f C ? A N A mB g 125.507 mA g A (mA g cos N1 ) Assumption is wrong ! fB fA NA NB So, A and B move and C don’t move f A k , A N A f B k , B N B 63 Recommend Problem 6/16 6/125 6/23 6/41 6/40 6/37 6/108 64