7.1 Friction: asic pplications
7.1 Friction: asic pplications Example 1, page 1 of 2 1. The uniform ladder is 2-m long and makes an angle of = 60 with the floor. If the wall at is smooth and the coefficient of static friction at is = 0.3, determine if the ladder can remain in the position shown. Mass of ladder = 10 kg 1 Free-body diagram 2 No friction force is present because the wall is smooth. 3 N ecause the ladder is uniform, the weight acts through the center. 1 m 1 m (2 m)(sin 60 ) = 1.732 m 6 Equations of equilibrium: Fx = 0: N f = 0 Weight = mg = (10 kg)(9.81 m/s 2 ) = 98.1 N 60 f + ++ F y = 0: N 98.1 N = 0 M = 0: (98.1 N)(0.5 m) N (1.732 m) = 0 (1 m)(cos 60 ) = 0.5 m N 5 4 Impending motion The friction force must be drawn in a direction opposing the impending motion. 7 Results of solving the above equations of equilibrium: N = 98.1 N N = 28.3 N f = 28.3 N
7.1 Friction: asic pplications Example 1, page 2 of 2 8 Compute the maximum possible friction force that the surfaces can develop at point. f -max N = (0.3)(98.1 N) = 29.4 N To determine whether or not the ladder will stay in the original position, the friction force found from the equilibrium equations, f = 28.3 N, must be compared with the maximum force that the surfaces at can develop: Since f -max = 29.4 N f = 28.3 N < 29.4 N = f -max the surfaces are able to develop enough friction force and the ladder will stay in equilibrium.
+ + 7.1 Friction: asic pplications Example 2, page 1 of 2 2. The uniform ladder is 2-m long and the wall at is smooth. If the coefficient of static friction at is = 0.2, determine the smallest angle for which the ladder can remain in the position shown. 1 Free-body diagram 2 No friction force is present because the wall is smooth. Mass of ladder = 10 kg N 1 m 5 Equations of equilibrium: Weight = mg = (10 kg)(9.81 m/s 2 ) = 98.1 N 1 m f (2 m) sin + Fx = 0: N f = 0 (1) N 3 Impending motion F y = 0: N 98.1 N = 0 (2) 6 M = 0: (98.1 N)(1 m) cos N (2 m) sin = 0 (3) Three equations, but four unknowns: N, f, N, and n additional equation is needed. (1 m) cos 4 The friction force must be drawn in a direction opposing the motion.
+ 7.1 Friction: asic pplications Example 2, page 2 of 2 7 The fourth equation comes from the condition of impending slip at point, because if slip is just about to occur, then the friction force, f, is at its maximum value, which is N : f f -max N = (0.2)N (4) Three of the four equations are linear but the moment equation, Eq. 3, is nonlinear because cos and sin appear. M = 0: (98.1 N)(1m) cos - N (2 m) sin = 0 (Eq. 3 repeated) The easiest way to solve these equations is to use the general equation solver on a calculator. lternatively, manipulate the equations as follows. First note that Eq. 2 implies that N = 98.1 N Then using this value for N in Eq. 4 gives 8 Using f = 19.62 N in Eq. 1 gives N = f = 19.62 N Using N = 19.62 N in Eq. 3 gives (98.1 N) cos (19.62 N)(2) sin = 0 Dividing through by cos and rearranging gives Replacing the left-hand side of this equation by tan gives tan = 2.5 which implies sin = 98.1 = 2.5 cos (19.62)(2) 68.2 ns. f = N = (0.2)(98.1 N) = 19.62 N
+ + 7.1 Friction: asic pplications Example 3, page 1 of 2 3. The uniform ladder is 2-m long. The coefficient of static friction at is = 0.6 and at is = 0.4. Determine the smallest angle,, for which the ladder can remain in the position shown. 1 Free-body diagram 3 Friction force, f, is present and opposes the possible motion. 5 + Mass of ladder = 10 kg Equations of equilibrium: Fx = 0: N f = 0 (1) F y = 0: N + f 98.1 N = 0 (2) N f Weight = mg = (10 kg)(9.81 m/s 2 ) = 98.1 N 1 m (2 m) cos 2 1 m N Impending motion (2 m) sin f (1 m) cos 4 Friction force opposes the possible motion. M = 0: (98.1 N)(1 m) cos f (2 m) cos N (2 m) sin = 0 (3) 6 Three equilibrium equations but five unknowns: N, N, f, f, and two more equations are needed.
7.1 Friction: asic pplications Example 3, page 2 of 2 7 The two additional equations come from the condition of impending slip at points and, because if slip is just about to occur, then the friction forces, f and f, are at their maximum values, N and N f f -max N = 0.6N (4) f f -max N = 0.4N (5) Four of the five equations are linear but the moment equation Eq. 3 is nonlinear (sin and cos are present). To solve these equations, use the general equation solver on your calculator or manipulate the equation algebraically and use a trig identity such as tan = sin /cos. Results of solving the five equations (three equilibrium and two friction equations) given above: N = 79.1 N N = 47.5 N f = 47.5 N f = 19.0 N = 32.3 ns.
7.1 Friction: asic pplications Example 4, page 1 of 3 4. Four round pegs,, C, and D are attached to the bracket and loosely straddle the vertical pole. When a 100-N force is applied as shown, the bracket rotates slightly and friction forces develop between pegs, C, and the pole. If the coefficient of static friction between the pegs and the pole is determine the smallest value of for which the bracket will support the load. Neglect the effect of the rotation of the bracket on the distances shown. 1 Movement of bracket exaggerated for clarity 100 N D 50 mm Peg D loses contact with the pole 100 N racket rotates a small amount 100 mm C D 300 mm C Peg loses contact with the pole
++ + 7.1 Friction: asic pplications Example 4, page 2 of 3 2 Free-body diagram P = 100 N 4 The friction forces, f and f C, resist the motion by pushing the bracket up. 3 Impending motion of bracket C N N C f 100 mm 6 7 s the bracket inclines slightly, the pegs at and D lose contact with the pole. That is why no forces appear at and D on the free-body diagram. Equations of equilibrium: f C 300 mm 50 mm 5 The normal forces, N and N C, are directed from the pole to the pegs. Fx = 0: N N C = 0 (1) F y = 0: f + f C 100 N = 0 (2) M C = 0: (100 N)(300 mm) + f (50 mm) N (100 mm) = 0 (3)
7.1 Friction: asic pplications Example 4, page 3 of 3 8 There are only three equations of equilibrium but four unknowns (f, N, f C, and N C ), so at least one more equation is needed. The additional equation comes from the condition of impending slip at, but if the bracket is going to slip at, it will also slip at C. So we have two additional equations and one additional unknown, : f = f -max N (4) f C = f C-max N C (5) Solving Eqs. 1-5 gives the results below (Note that Eqs. 4 and 5 are nonlinear because multiplies N and N C ): f = 50 N N = 325 N f C = 50 N N C = 325 N = 0.154 ns.
+ 7.1 Friction: asic pplications Example 5, page 1 of 2 5. rm C acts as a brake on the wheel. The coefficient of static friction at is = 0.4. Determine the largest moment M that can act on the wheel without causing it to slip. Radius = 200 mm O M C 100 N 300 mm 400 mm 1 O x Free-body diagram of wheel. O O y M f 2 200 mm N Impending motion of point on outer surface of wheel 3 The friction force f opposes the motion. 200 mm 4 Equation of moment equilibrium for the wheel (Since we were not asked to compute the reactions Ox and O y, we do not need to write the force-equilibrium equations.): M O = 0: f (200 mm) M = 0 (1)
+ 7.1 Friction: asic pplications Example 5, page 2 of 2 5 Free-body diagram of arm C. 6 The sense of the friction force on the brake can be determined by Newton's Third Law (equal and opposite to the force on the wheel). N f y 100 N 300 mm 400 mm x 200 mm 7 Equation of moment equilibrium for the brake (Since we were not asked to compute the reactions x and y, we do not need to write the force-equilibrium equations.): M = 0: f (200 mm) N (400 mm) + (100 N)(300 mm + 400 mm) = 0 (2) 8 The third equation follows from the condition that slip impends at : Thus far we have two equations but three unknowns (M, f, and N ), so another equation is needed. f f -max N = 0.4N (3) Solving Eqs. 1-3 simultaneously yields f = 87.5 N N = 218.8 N M = 17 500 N mm = 17.5 N m ns.
7.1 Friction: asic pplications Example 6, page 1 of 4 6. The uniform block is initially at rest when a 10-lb force is applied. The coefficient of static friction between the block and the plane is = 0.6. Determine if the block will move. 1 ft 10 lb 20 lb (weight) 2 ft
7.1 Friction: asic pplications Example 6, page 2 of 4 1 Free-body diagram 1 ft 4 Free-body diagram showing resultant forces 10 lb 10 lb 20 lb 20 lb 2 ft 2 N d 3 f d The distributed friction force f d opposes possible slip to the right; the distributed normal force N d opposes possible tipping of the block. s the block is pushed to the right by the 10-lb force, the floor opposes the possible motion by providing a distributed reaction force. The component of this reaction force parallel to the floor is the distributed friction force f d, and the component normal to the floor is the distributed normal force N d. 5 The resultant of the distributed force f d is f. 6 f 0.5 ft x The resultant of the distributed force N d is N. ecause N d opposes possible tipping of the block, it is not uniform but is greater near the right-hand side of the base of the block to balance the tendency to tip. Thus the resultant N does not act at the middle of the base but instead acts at some unknown distance, x, from the middle. N
+ ++ 7.1 Friction: asic pplications Example 6, page 3 of 4 7 Equations of equilibrium: Fx = 0: 10 lb f = 0 F y = 0: N 20 lb = 0 M = 0: (20 lb)(0.5 ft) (10 lb)(2 ft) + N(0.5 ft + x) = 0 8 Solving these equations gives f = 10 lb N = 20 lb x = 1 ft 9 These are the values required if the system is to stay in equilibrium, that is, not move. To determine if the system can produce the 10-lb friction force f required to keep the system in equilibrium, we have to compare f with the maximum possible value of the friction force: f max N = (0.6)(20 lb) = 12 lb ecause f = 10 lb is less than the 12 lb maximum possible force, the surfaces can develop enough force to balance forces in the x direction (thus the block will not slide to the right).
7.1 Friction: asic pplications Example 6, page 4 of 4 10 lb 20 lb 10 We next consider whether or not the block will tip. Recall that solving the equilibrium equations gave the result x = 1 ft. That is, to maintain equilibrium, the normal force N must act at the location shown, 0.5 ft to the right of the block. ut this is impossible because N is the normal force from the ground acting up on the block; the farthest N can act is at the right hand corner,. Thus the block will tip because N cannot act far enough to the right to prevent it. f 0.5 ft 0.5 ft x = 1 ft N (impossible location because outside the base of the block)
7.1 Friction: asic pplications Example 7, page 1 of 6 7. The uniform block is initially at rest when the force P is applied. The coefficient of static friction between the block and the plane is = 0.6. Determine the minimum value of P that will cause the block to move. 1 ft P 20 lb (weight) 2 ft
7.1 Friction: asic pplications Example 7, page 2 of 6 1 Free-body diagram 1 ft 3 Free-body diagram showing resultant forces P P 20 lb 20 lb 2 ft f d 4 The resultant of the distributed force f d is f. f 0.5 ft x N 2 N d s the block is pushed to the right by the force P, the floor opposes the possible motion by providing a distributed reaction force. The component of this reaction force parallel to the floor is the distributed friction force, f d, and the component normal to the floor is the distributed normal force N d. 5 The resultant of the distributed force N d is N. ecause N d opposes possible tipping of the block, it is not uniform but is greater near the right-hand side of the base of the block to balance the tendency to tip. Thus the resultant N does not act at the middle of the base but instead acts at some unknown distance, x, from the middle.
+ ++ 7.1 Friction: asic pplications Example 7, page 3 of 6 6 Equations of equilibrium: Fx = 0: P f = 0 (1) F y = 0: N 20 lb = 0 (2) 9 Case 1: Sliding 1 ft 7 M = 0: (20 lb)(0.5 ft) P(2 ft) + N(0.5 ft + x) = 0 (3) Three equations but four unknowns (P, f, N and x), so one more equation is needed. P 8 The fourth equation comes from considering possible impending motion. There are two cases to consider: sliding and tipping. 20 lb (weight)
7.1 Friction: asic pplications Example 7, page 4 of 6 10 Case 2: Tipping 11 P 12 We have to analyze each case separately. Let's (arbitrarily) choose Case 1 first. If sliding impends, then f = f max N = 0.6N (4) Solving Eqs. 1-4 simultaneously gives P = 12 lb 20 lb (weight) N = 20 lb f = 12 lb x = 1.2 ft
7.1 Friction: asic pplications Example 7, page 5 of 6 13 Free body diagram for Case 1 (Sliding impends) P = 12 lb 20 lb 14 ut this diagram shows that the only way the equilibrium equations for Case 1 can be satisfied is if the normal force N lies to the right of the block (x = 1.2 ft). Since this is impossible, the Case 1 assumption that sliding impends must be incorrect. f = 12 lb 0.5 ft N = 20 lb x = 1.2 ft
7.1 Friction: asic pplications Example 7, page 6 of 6 15 P Free body diagram for Case 2 (Tipping impends) 0.5 ft 0.5 ft 16 Since the block is just about to tip, it loses contact with the floor except at the corner, where the normal force N is concentrated. Since N acts at the corner, we know x = 0.5 ft (5) 20 lb Solving the equilibrium equations, Eqs. 1, 2, and 3, simultaneously with Eq. 5 gives f = 5 lb N = 20 lb P = 5 lb ns. f x N Since there were only two possibilities, sliding and tipping, and we eliminated sliding, we know that the above result P = 5 lb is correct. However, we can also check our work by verifying that the friction force f is less than the maximum possible value: f = 5 lb < f max N = (0.6)(20 lb) = 12 lb. (OK)
7.1 Friction: asic pplications Example 8, page 1 of 3 8. The cylinder is initially at rest when a horizontal force P is applied. The coefficients of static friction at and are = 0.3 and = 0.6. Determine the minimum value of P that will cause the cylinder to move. P Radius = 0.2 m 0.3 m 20 kg Weight = mg = (20 kg)(9.81 m/s 2 ) = 196.2 N 0.3 m P 1 Free-body diagram f N 2 0.2 m Possible motion of point on cylinder. Force P tends to rotate the cylinder clockwise. 5 f The friction force from the floor opposes the motion of point on the cylinder. N 3 The friction force from the wall opposes the motion of point on the cylinder. 4 Possible motion of point.
+ + 7.1 Friction: asic pplications Example 8, page 2 of 3 6 + Equilibrium equations Fx = 0: P + f N = 0 (1) F y = 0: 196.2 N + f + N = 0 (2) M = 0: f (0.2 m) + N (0.2 m) P(0.3 m) = 0 (3) There are three equations and five unknowns (P, f, N, f, N ), so two more equations are needed. The two additional equations come from considering possible impending motion. There are two cases to consider: 7 We have to analyze each case separately. Let's (arbitrarily) choose Case 1 first. Thus if the cylinder is about to slip about its center, then slip impends simultaneously at points and, so f = f -max N = 0.3N (4) f = f -max N = 0.6N (5) Solving Eqs. 1-5 simultaneously gives P = 554 N Case 1 The cylinder spins about its center. Case 2 The cylinder rolls up the wall without slipping. f = 34.6 N f = 312 N N = 115 N N = 519 N negative normal force, N, is impossible (The floor can't pull down on the cylinder), so the assumption of slip at both and must be wrong.
7.1 Friction: asic pplications Example 8, page 3 of 3 8 Next consider Case 2 the cylinder is about to roll up the wall. Thus the cylinder is about to lose contact with the floor at point, and so the friction and normal forces there are zero: f = 0 (6) N = 0 (7) Solving the equilibrium equations, Eqs. 1, 2, and 3, simultaneously with Eqs. 6 and 7 gives f = 0 N = 0 f = 196 N N = 392 N P = 392 N ns. Since there were only two possibilities, spinning about the cylinder center or rolling up the wall, and we eliminated spinning, the above result P = 392 N must be correct. However, we can also check our work by comparing the friction force, f, with the maximum possible value: f = 196 N < f -max N = (0.6)(392 N) = 235 N (OK)
7.1 Friction: asic pplications Example 9, page 1 of 4 9. The small block rests on top of the large block. The coefficients of static friction are shown in the figure. Determine the smallest value of applied force P that will keep block from sliding down the inclined plane. Frictionless pulley Cord = 0.3 = 0.2 60 kg 10 kg P 30
+ 7.1 Friction: asic pplications Example 9, page 2 of 4 1 Free-body diagram of block Weight = mg = (10 kg)(9.81 m/s 2 ) = 98.1 N 4 Tension in cord, T Impending motion of block relative to block (If block moves down the plane, block must move up the plane.) Normal force from block N 5 f y 3 The numerical value of will be calculated later. P 30 The friction force from block opposes the impending motion of block up the incline. x 2 It's convenient to use an inclined xy coordinate system. 6 Equations of equilibrium for block. We assume that the blocks will not tip because they are much longer than they are high; thus no moment equation is needed (Since no dimensions are given, we could not write a moment equation even if we wanted to). + Fx =0: P T + f + (98.1 N) sin = 0 (1) F y = 0: N (98.1 N) cos = 0 (2)
7.1 Friction: asic pplications Example 9, page 3 of 4 7 30 Geometry = 90 60 = 30 y 60 30 x 10 Friction force from block opposes motion of block. T 8 Free-body diagram of block f N y 13 = 30 Weight = mg = (60 kg)(9.81 m/s 2 ) = 588.6 N [Weight of block alone (Note that the weight of block is not included because block is not part of this free-body. The effect of the weight of block is transmitted through the normal force, N.)] 12 f Friction force from inclined plane opposes N motion of block. Normal force from inclined plane 30 9 Impending motion of block relative to block. x 11 Impending motion of block relative to inclined plane.
7.1 Friction: asic pplications Example 9, page 4 of 4 14 Equilibrium equations for block + + Fx =0: (588.6 N) sin 30 f f T = 0 (3) F y = 0: (588.6 N) cos 30 + N N = 0 (4) Four equations in six unknowns (T, P, f, N, f, N ). Two more equations come from the condition of impending sliding between the blocks and between block and the plane: T Free body diagram of block repeated f f 588.6 N N = 30 y f f -max N = 0.2N (5) f f -max N = 0.3N (6) Solving Eqs. 1-6 simultaneously gives f = 119 N N = 595 N f = 25 N N = 85 N T = 150 N N 30 x P = 75 N ns.
+ 7.1 Friction: asic pplications Example 10, page 1 of 8 10. The three blocks are stationary when the force P is applied. The coefficients of static friction for each pair of surfaces are given in the figure. Determine the smallest value of P for which motion will occur. The blocks are sufficiently long that tipping will not occur. P 10 kg 10 kg 10 kg C = 0.8 C = 0.3 C = 0.15 1 Free-body diagram of block Weight = mg = (10 kg)(9.81 m/s 2 ) = 98.1 N 4 Equilibrium equations for block : P + Fx = 0: P f = 0 (1) f F y = 0: N 98.1 N = 0 N 3 2 Impending motion of block relative to Friction opposes the motion The last equation gives N = 98.1 N (2)
+ 7.1 Friction: asic pplications Example 10, page 2 of 8 5 Free-body diagram of block 8 Impending motion of block relative to block (n observer on would see moving in this direction.) N = 98.1 N f 9 Weight = 98.1 N f C N C Friction force opposes relative motion 6 Impending motion of block relative to C 7 Friction force opposes relative motion 10 + Equilibrium equations for block : Fx = 0: f f C = 0 (3) F y = 0: 98.1 N 98.1 N + N C = 0 The last equation gives N C = 196.2 N (4)
+ 7.1 Friction: asic pplications Example 10, page 3 of 8 11 Free-body diagram of block C 14 Impending motion of block C relative to block (n observer on would see C moving in this direction.) Weight = 98.1 N N C = 196.2 N N C C f C f C 15 Friction force opposes relative motion 12 Impending motion of block C relative to floor 13 Friction force opposes relative motion 16 + Equilibrium equations for block C: Fx = 0: f C f C = 0 (5) F y = 0: 98.1 N 196.2 N + N C = 0 The last equation gives N C = 294.3 N (6)
7.1 Friction: asic pplications Example 10, page 4 of 8 17 We now have six equilibrium equations but seven 18 The seventh equation comes from the condition of unknowns (P, f, N, f C, N C, f C, N C ), so another impending slip. We have to consider three cases: equation is needed. Case 1 C Stationary Impending motion Case 2 C Impending motion: blocks and move together Stationary Case 3 C Impending motion: blocks, and C move together
7.1 Friction: asic pplications Example 10, page 5 of 8 19 nalyze each case separately. 22 For block, Eq. 3 is 20 Case 1 f f C = 0 So for equilibrium, C Stationary f C = f = 78.5 N 21 Slip impends so f = f -max by Eq. 2 N = (0.8)(98.1 N) = 78.5 N (7) We have to check to see if the surfaces of contact between blocks and C develop enough friction force to keep block stationary. Let's compare this with the maximum possible friction force: f C-max CN C = (0.3)(196.2 N) = 58.9 N (8) So the surfaces can develop only 58.9 N while 78.5 N are needed for equilibrium. Thus block will move, contrary to our assumption for Case 1.
7.1 Friction: asic pplications Example 10, page 6 of 8 23 Case 2 Impending motion together 25 For block C, Eq. 5 is f C f C = 0 C Stationary So for equilibrium, f C = f C 24 Slip impends so = 58.9 N f C = f C-max Compare this with the maximum possible friction force by Eq. 8 f C-max CN C by Eq. 6 = 58.9 N = (0.15)(294.3 N) = 44.1 N (9) We have to check to see if the surfaces of contact between block C and the ground develop enough friction force to keep block C stationary. So the surfaces can develop only 44.1 N while 58.9 N are needed for equilibrium. Thus block C will move, contrary to our assumption for Case 2.
7.1 Friction: asic pplications Example 10, page 7 of 8 26 Case 3 28 Eq. 5 gives f C : C Impending motion together Thus f C f C = 0 f C = f C by Eq. 10 27 Slip impends so f C = f C-max = 44.1 N (10) and so = 44.1 N (11) by Eq. 8 by Eq. 9 44.1 N = f C < f C-max = 58.9 N (OK) We don't have to check that the surfaces of contact between blocks and and between and C develop enough friction to keep and in equilibrium, since there were only three cases of possible motion, and we showed that the first two cases were impossible. Nonetheless, we can verify that our work is correct by showing that the friction forces acting between and and between and C are less than their maximum possible values. Eq. 3 gives f : f f C = 0 Thus f = f C = 44.1 N (12) and so by Eq. 7 44.1 N = f < f -max = 78.5 N (OK)
7.1 Friction: asic pplications Example 10, page 8 of 8 29 30 31 Thus the surfaces of contact between blocks and and between and C can develop enough friction to keep blocks,, and C moving together as a unit. Finally, we can calculate P from Eq. 1: or, P f = 0 P = f = 44.1 N C ns. Why didn't we consider a case like this? Case 4 by Eq. 12 32 nswer: No matter what the impending motion is, there are only seven unknown forces (f, N, f C, N C, f C, N C, and P). Since these seven unknowns must satisfy the six equations of equilibrium, the unknowns can be chosen to satisfy only one additional equation a friction equation. In the unlikely event that the masses and 's just happen to have values such that the seven forces simultaneously satisfy the six equilibrium equations and two friction equations, then one of the eight equations must be redundant. pplying this reasoning to Case 4, we see that if forces exist that satisfy Case 4's equations, then these forces must be identical to the forces satisfying the equations for Case 1 (slip between and ) and Case 2 (slip between and C). Since solving Case 4 would give the same answer as solving Case 1 (or Case 2), we don't have to consider Case 4. similar argument can be made for other possible motions. Impending motion of relative to Impending motion of relative to C Stationary
7.1 Friction: asic pplications Example 11, page 1 of 9 11. The two cylinders shown are initially at rest when horizontal forces of magnitude P/2 are applied to the ends of the axle in the lower cylinder. The coefficients of static friction for each pair of surfaces are given in the figure. Determine the largest value of P that can be applied without moving the cylinders up the inclined plane. P/2 = 0.4 C = 0.5 Radius of each cylinder = 300 mm Mass of each cylinder = 50 kg C P/2 = 0.6 25
7.1 Friction: asic pplications Example 11, page 2 of 9 6 The friction force from the upper cylinder opposes the relative motion of point C on the lower cylinder. 1 Free-body diagram of lower cylinder N C f C C O y P 5 3 Impending motion of point C on lower cylinder relative to upper cylinder (n observer on the upper cylinder would see this motion as the lower cylinder moves). 4 Weight = mg = (50 kg)(9.81 m/s 2 ) = 490.5 N The numerical values of and will be calculated later. f 8 The friction force from the plane opposes the motion of point on the cylinder. N 7 Radius = 300 mm 25 Impending motion of point on cylinder. The x component of the applied force, P cos, is pushing the cylinder up the plane. 2 x It is convenient to use an inclined xy coordinate-system.
7.1 Friction: asic pplications Example 11, page 3 of 9 + 9 Equilibrium equations for cylinder: Free-body diagram of lower cylinder repeated Fx = 0: (490.5 N) sin P cos + f + N C = 0 (1) y ++ F y = 0: 90.5 N) cos P sin + f C + N = 0 (2) M O = 0: f (300 mm) f C (300 mm) = 0 (3) N C C 490.5 N f C 10 Geometry O P = 90 65 = 25 y f 25 x 65 Radius = 300 mm = 25 N 25 x
7.1 Friction: asic pplications Example 11, page 4 of 9 11 Free-body diagram of upper cylinder Radius = 300 mm y Weight = 490.5 N f O = 25 P C f C 13 N C The friction force from the lower cylinder opposes the relative motion of point C on the upper cylinder.. x 15 The friction force from the plane opposes the motion up the plane. N 14 Impending motion of point as normal force N C pushes the upper cylinder up the plane. 12 Impending motion of point C on upper cylinder relative to lower cylinder (n observer on the lower cylinder would see this motion as the upper cylinder moves).
+ 7.1 Friction: asic pplications Example 11, page 5 of 9 16 Equilibrium equations Fx = 0: (490.5 N) sin 25 + f N C = 0 (4) Free-body diagram of upper cylinder repeated y ++ F y = 0: (490.5 N) cos 25 f C + N = 0 (5) M O' = 0: f (300 mm) f C (300 mm) = 0 (6) 490.5 N 25 17 Thus far we have six equation but seven unknowns (P, f, N, f, N, f C, N C ), so another equation is needed. The seventh equation comes from the condition of impending slip. We have to consider only two cases: f O P C f C N C x 1. slip occurs at point (and simultaneously rolling occurs about points and C). N Radius = 300 mm 2. slip occurs at point C (and simultaneously rolling occurs about points and ). Slip at point will be discussed later.
7.1 Friction: asic pplications Example 11, page 6 of 9 18 Case 1 efore motion O' fter motion C O 20 Slip For impending slip at, O' C O Displacement of point O (Point O moves up the plane) f = f -max N = 0.4N (7) 19 Rolling without slipping (The radial line O on the lower cylinder rotates through the same angle,, as the radial line O'C on the upper cylinder.)
7.1 Friction: asic pplications Example 11, page 7 of 9 21 Solving Eqs. 1-7 simultaneously gives f = 296 N f = 296 N f C = 296 N N = 618 N N = 741 N N C = 504 N P = 1111 N 22 We must check that the surfaces at and C can provide enough friction force to prevent slip and allow rolling: f 296 f -max N = (0.6)(618 N) = 371 N (OK) f C-max CN C = (0.5)(504 N) = 252 N (Not enough! We need f C = 296 N for equilibrium.) So the assumption of impending slip at is wrong.
7.1 Friction: asic pplications Example 11, page 8 of 9 23 Case 2 (Slip at C, rolling at and ) efore motion O' fter motion C O O' C 25 Rolling without slipping O Displacement of point O 26 Slip For impending slip at C, f C = f C-max CN C = 0.5N C (8) 24 Rolling without slipping
7.1 Friction: asic pplications Example 11, page 9 of 9 27 Solving the six equilibrium equations, Eqs. 1-6, plus Eq. 8 yields 29 f = 207 N N = 624 N f = 207 N N = 652 N What about slip occurring at point only? Well if the lower cylinder moves, then the upper cylinder must also move. ut the only way that the upper cylinder can move is if either 1) it slips at point, or 2) it slips at point C. Thus the case of slip impending at point alone is impossible and does not have to be considered. 28 f C = 207 N N C = 415 N P = 915 N ns. The above answers must be correct since we eliminated the only other possible case where slip impends. ut we can check our results by verifying that the friction forces at and are less than their maximum possible values. f = 207 N f -max N = (0.6)(624 N) = 374 N (OK) f = 207 N f -max N = (0.4)(652 N) = 261 N (OK) What about simultaneous slip at and? nswer: we have already found values of the seven unknowns in the problem that satisfy the six equilibrium equations and the equation for slip at. In the unlikely case that the seven values happen to satisfy an eighth equation (slip at ), then that equation must be redundant, and the solution for the eight equations is the same as we have already found for the seven equations. n analogous statement can be made for the case of simultaneous slip at and C.