7.4 Flat Belts Example 1, page 1 of 2

Transcription

1 7.4 Flat elts

2 7.4 Flat elts Example 1, page 1 of 2 1. Determine the minimum number of turns of rope that will allow the 5-lb force to support the 600-lb block, if the coefficient of static friction is lb 1 Tensions in the rope 5 lb } n turns 600 lb 600 lb 2 Impending motion (The motion can't be up, since the 5-lb force is too small to lift the 600-lb block.)

3 7.4 Flat elts Example 1, page 2 of 2 3 pply the equation for belt friction: 4 T 2 = T 1 e (1) Since = 0.15, Eq. 1 becomes 600 lb = (5 lb) e In this equation, T 2 is the tension in the direction of impending motion T 2 = 600 lb The other tension, T 1, is in the direction opposite the impending motion, so T 1 = 5 lb Solving gives = radians. If n is the number of turns of rope, then the number of radians is 2 n. Equating this to gives an equation for n: 2 n = = rad Solving for n gives n = 5.08 turns Rounding off to the next higher integer then gives n = 6 ns.

4 7.4 Flat elts Example 2, page 1 of 2 2. If the coefficient of static friction between the rope and the fixed circular drums and is 0.2, determine the largest value of the force P that can be applied without moving the 150-lb weight upwards. 60 P 1 Tensions in the rope on either side of drum = T Impending motion (The rope must move to the right if the 150-lb weight moves up) lb ecause slip impends between the rope and the drum, we can apply the equation for belt friction: T 2 = T 1 e (1) Rope is horizontal. where T 2 is the tension in the direction of impending motion. In our particular problem, T 2 = T and the tension opposing the impending motion is 150 lb T 1 = 150 lb Using these results and = 0.2 and = 2 in Eq. 1 gives T = (150)e0.2( /2) = lb

5 7.4 Flat elts Example 2, page 2 of 2 3 Tensions in the rope on either side of drum Impending motion 4 Flat-belt friction equation: T 2 = T 1 e P so, 60 5 Geometry P = (205.4 lb)e (2) T = lb = 60 = rad Using = 60 = rad in Eq. 2 gives P = (205.4)e / = 253 lb ns.

6 + 7.4 Flat elts Example 3, page 1 of 3 3. Determine the smallest force P applied to the handle of the band brake that will prevent the drum from rotating when the 15 lb ft moment is applied. The coefficient of static friction is 0.25, and the weight of lever arm C can be neglected. 15 lb-ft D 6 in. 80 o P C 2 Impending slip of band relative to drum (n observer on the drum would see the belt move down.) 1 Free-body diagram of drum 15 lb ft = 15 lb ft 12 in./ft 6 in. = 180 lb in. D D x T D y T 80 o 10 in. 15 in. 3 Equilibrium equation for drum M D = 0: T (6 in.) T (6 in.) 180 lb in. = 0 (1)

7 7.4 Flat elts Example 3, page 2 of 3 4 The belt-friction equation is in general T 2 = T 1 e (2) where T 2 is the tension in the direction of impending slip. In our particular problem, T 2 = T and the tension opposing the impending motion is Eq. (2) becomes T 1 = T T = T e (3) 5 Geometry = = 190 = (190/180) rad 80 = rad 80 Eq. 3 becomes, with and 0.25 T = T e 0.25(3.316) Solving Eqs. 1 and 4 simultaneously gives T = lb T = lb

8 + 7.4 Flat elts Example 3, page 3 of 3 6 Free-body diagram of lever arm C T = lb T sin 80 = ( lb) sin 80 P x T cos 80 y 10 in. 15 in. 7 Equilibrium equation for lever arm M = 0: [( lb) sin 80 ](10 in.) P(10 in in.) = 0 (5) Solving gives P = 9.15 lb ns.

9 + 7.4 Flat elts Example 4, page 1 of 3 4. The uniform beam C weighs 40 lb. The coefficient of static friction between the cord and the fixed drum D is 0.3. Determine the smallest value of the weight W for which the beam will remain horizontal. 1 Free-body diagram of block D T (Tension in the cord) W C 4 ft 4 ft 2 If end of the beam is about to move down, then block is about to lose contact with the beam. Thus the normal force N is zero. W N = 0 3 Equilibrium equation for block F y = 0: T W = 0 (1)

10 + 7.4 Flat elts Example 4, page 2 of 3 4 Free-body diagram of beam 5 T is the tension in the cord. y N = 0 C y 6 The 40-lb weight of the beam acts through the center of the span. 40 lb 4 ft 4 ft C Cx x 8 Tensions in cord on either side of drum 7 Equilibrium equation for the beam T = 20 lb T M C = 0: (40 lb)(4 ft) T (4 ft + 4 ft) = 0 (2) Solving gives T = 20 lb 9 Impending motion ecause the cord is on the verge of slipping, we can apply the equation for belt friction: T 2 = T 1 e (3) where T 2 = 20 lb is the tension in the direction of impending slip and T 1 = T is the tension opposite the direction of impending motion

11 7.4 Flat elts Example 4, page 3 of 3 10 Geometry = T T 11 With and 0.3, Eq. 3 becomes T 2 = T 1 e (Eq. 3 repeated) 20 lb 0.3 Solving gives T = 7.79 lb Eq. 1 then gives W = T = 7.79 lb ns.

12 + 7.4 Flat elts Example 5, page 1 of 4 5. Determine the largest value W of the weight of block for which neither block will move. The coefficients of static friction are 0.2 between the blocks and the planes, and 0.25 between the cord and the drum. y 1 Free-body diagram of block T f x 2 Impending motion of block (Since we are to determine the "largest value of the weight" of block, block must be about to move down the plane. Thus block must be about to move up the plane.) 80 lb W 80 lb 70 N Equilibrium equations for block : Fx = 0: T f (80 lb) sin (1) F y = 0: N (80 lb) cos = 0 (2) Slip impends, so f = f -max = N = (0.2)N (3)

13 7.4 Flat elts Example 5, page 2 of 4 4 Geometry = 70 6 Tensions in cord on either side of drum 20 T T = lb Impending motion (lock moves down plane) 5 Solving Eqs. 1, 2, and 3, with T = lb = 70 gives 7 ecause the cord is about to slip over the drum, we can apply the equation for belt friction: f = 5.47 lb T 2 = T 1 e drum (4) N = lb where T 2 = T is the tension in the direction of impending slip, and T 1 = lb is the tension opposite to the direction of impending motion.

14 7.4 Flat elts Example 5, page 3 of 4 8 Geometry = = = rad 3 9 Substituting = 2 /3, T 2 = T, T 1 = lb, 0.25 in Eq. 4 gives and drum T = (80.65 lb) e Evaluating the right hand side yields T = lb

15 + 7.4 Flat elts Example 5, page 4 of 4 10 Free-body diagram of block 12 Equilibrium equations for block : y T = lb x + Fx' =0: N W cos = 0 (5) F y' = 0: lb + f W sin = 0 (6) f = f -max f = N N 50 W Impending motion 13 = (0.2)N (7) Solving Eqs. 5, 6, and 7, with = 50 gives 11 Geometry f = 27.4 lb N = lb W = 213 lb ns. =

16 7.4 Flat elts Example 6, page 1 of 4 6. motor attached to pulley drives the pulley clockwise with a 200 lb-in. torque. The flat belt then overcomes the resisting torque T at pulley and rotates the pulley clockwise. Determine the minimum tension that can exist in the belt without causing the belt to slip at pulley. lso determine the corresponding resisting torque T. The coefficient of static friction between the belt and the pulleys is 0.3. T 4 in. 200 lb in. 7 in. Driving pulley 22 in. Driven pulley

17 + 7.4 Flat elts Example 6, page 2 of 4 1 Free-body diagram of pulley 3 Equilibrium equation for pulley C T C M = 0: T D (4 in.) T C (4 in.) 200 lb in. = 0 (1) 4 in. x 200 lb in. ecause the belt is about to slip on the pulley, the belt friction equation applies: D y T D T 2 = T 1 e where T 2 = T D, the tension in the direction of impending motion, and T 1 = T C, so 2 Impending motion of belt relative to pulley (Since pulley is driven by a clockwise torque, the pulley would slip in a clockwise sense relative to the belt. Thus an observer at point D on the pulley would see the belt move in the direction shown.) T D = T C e (2)

18 + 7.4 Flat elts Example 6, page 3 of 4 4 Geometry Radius = 7 in. 4 in. 6 Free-body diagram of pulley 4 in. C 7 in. 4 in. = 3 in. T C = lb 22 in. From triangle C, T cos ( 2 ) = 3 in. 22 in. Solving gives = = rad. x y 7 in. So Eq. 2, with = 0.3, becomes T D = T C e 0.3(2.868) (3) T D = lb 5 Solving Eqs. 1 and 3 simultaneously gives T C = lb T D = lb ns. 7 Equilibrium equation for pulley M = 0: T + (36.65 lb)7 in. (86.65 lb)7 in. = 0 Solving gives T = 350 lb in. ns.

19 7.4 Flat elts Example 6, page 4 of 4 8 Finally, we must check that pulley does not slip. ut this follows from the observation that pulley has a larger angle of wrap than pulley. Thus it must be able to carry a maximum possible tension larger than the 86.5 lb maximum tension that pulley carries. 9 More precisely, because ' > lb tension ' we know that e ' > e. Multiplying through by lb gives lb tension (36.65 lb)e ' > (36.65 lb)e T max-, the maximum possible tension that pulley could support without slipping lb, by Eq. 2 Thus T max- > lb and pulley won't slip.

20 + 7.4 Flat elts Example 7, page 1 of 2 7. If the coefficient of static friction between the fixed drums D and E and the ropes is 0.35, determine the largest weight W that can be supported. 1 Free-body diagram of block C D T E T D E C W 100 lb 50 lb C 2 Equilibrium equation for block C F y = 0: T E + T D W = 0 (1) W

21 7.4 Flat elts Example 7, page 2 of 2 3 Rope tensions acting on drum D = 6 Rope tensions acting on drum E (not drum D) = E D 50 lb T E 100 lb T D Impending motion 4 Impending motion (Since we are to determine the largest value of weight of block C that can be supported, block C must be about to move down, so the rope connected to C must also be about to move down.) 7 Flat-belt friction equation: or, T 2 = T 1 e 5 ecause slip is about to occur, the belt-friction equation applies: T 2 = T 1 e Here, T 2 = T D, T 1 = 100 lb, 0.35, and so 8 T E = (50 lb)e 0.35 = lb (3) Using the results of Eqs. 2 and 3 in Eq. 1 gives T D = (100 lb)e = lb (2) W = 450 lb ns.

22 7.4 Flat elts Example 8, page 1 of 3 8. Pulley is rotating under the action of a 6 N-m torque. This motion is transmitted through a flat belt to drive pulley, which is in turn acted upon by a resisting torque T (the "load" on pulley ). The coefficient of static friction between the belt and the pulleys is Determine a) the maximum possible value of T, b) the maximum force in the belt, and c) the corresponding force required in the spring C. 80 mm T 6 N m C 80 mm Driven pulley Driving pulley

23 + 7.4 Flat elts Example 8, page 2 of 3 1 Free-body diagram of pulley 3 Equilibrium equations for pulley T 2 6 N-m D + + Fx = 0: x T 1 T 2 = 0 (1) F y = 0: y = 0 (2) 80 mm M = 0: T 2 (0.08 m) T 1 (0.08 m) 6 N m = 0 (3) = x 4 Flat-belt friction equation: y T 2 = T 1 e 2 T 1 Impending motion (Since the driving torque acting on pulley is clockwise, if slip is about to occur, then pulley will slip clockwise relative to the belt. n observer located on point D on pulley would see the belt move in the direction shown.) = T 1 e (4) Solving Eqs. 1-4 simultaneously gives x = N y = 0 T 1 = N T 2 = N ns.

24 + 7.4 Flat elts Example 8, page 3 of 3 5 Free-body diagram of member C 7 Free-body diagram of pulley x = N F spring T 2 = N 6 + Equilibrium equations for member C Fx = 0: F spring N = 0 (4) 80 mm x T Solving gives F spring = N ns. y T 1 = N 8 Equilibrium equations for pulley M = 0: T + (24.11 N)(0.08 m) 99 (0.08 m) = 0 Solving gives T = 6 N m ns. That is, the maximum resisting torque equals the driving torque.

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