CHAPTER 24 SIMPLE MACHINES EXERCISE 106, Page 240 1. A simple machine raises a load of 825 N through a distance of 0.3 m. The effort is 250 N and moves through a distance of 3.3 m. Determine: (a) the, (b) the, (c) the efficiency of the machine at this load. Force ratio = load 825 N = 3.3 effort 250 N Movement ratio = dis tan ce moved by theeffort 3.3m = 11 dis tan ce moved by theload 0.3m Efficiency = 100% = 3.3 100 = 30% 11 2. The efficiency of a simple machine is 50%. If a load of 1.2 kn is raised by an effort of 300 N, determine the. Force ratio = load 1200 N = 4 effort 300 N Efficiency = 4 4 from which, = = 8 efficiency 50 0.5 100 3. An effort of 10 N applied to a simple machine moves a load of 40 N through a distance of 100 mm, the efficiency at this load being 80%. Calculate: (a) the, (b) the distance moved by the effort. Force ratio = load 40 N = 4 effort 10 N 292
(a) Efficiency = (b) Movement ratio = distancemovedbytheeffort dis tan ce moved by theload 4 4 from which, = = 5 efficiency 80 0.8 100 from which, the distance moved by the effort = distance moved by the load = 5 100 = 500 mm 4. The effort required to raise a load using a simple machine, for various values of load is as shown: Load F l (N) 2050 4120 7410 8240 10300 Effort F e (N) 252 340 465 505 580 If the for the machine is 30, determine (a) the law of the machine, (b) the limiting, (c) the limiting efficiency. The load/effort graph is shown below. 293
(a) The law of the machine is F e = a F l + b where gradient of curve, a = AB 570 170 400 4 = 0.04 BC 10000 10000 100 and intercept, b = 170. Hence, the law of the machine is: F e = 0.04 F l + 170 (b) Limiting = 1 1 = 25 a 0.04 (c) Limiting efficiency = 1 1 100% a 0.04 30 = 83.3% 5. For the data given in question 4, determine the values of and efficiency for each value of the load. Hence plot graphs of effort, and efficiency to a base of load. From the graphs, determine the effort required to raise a load of 6 kn and the efficiency at this load. Load F l (N) 2050 4120 7410 8240 10300 Effort F e (N) 252 340 465 505 580 Efficiency = Force ratio = load effort 8.13 12.12 15.94 16.32 17.76 27.1% 40.4% 53.1% 54.4% 59.2% Graphs of load/effort, load/ and load/efficiency are shown below. From the graph, when the load is 6 kn, i.e. 6000 N effort = 410 N and efficiency = 48% 294
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EXERCISE 107, Page 242 1. A pulley system consists of four pulleys in an upper block and three pulleys in a lower block. Make a sketch of this arrangement showing how a of 7 may be obtained. If the is 4.2, what is the efficiency of the pulley. 296
Efficiency = 4.2 100% = 60% 7 2. A three-pulley lifting system is used to raise a load of 4.5 kn. Determine the effort required to raise this load when losses are neglected. If the actual effort required is 1.6 kn, determine the efficiency of the pulley system at this load. Load = 4.5 kn and = n = 3 When losses are neglected, efficiency = 100% = from which, = load effort = i.e. and 4.5kN effort effort = 4.5kN 3 = 3 = 1.5 kn If the actual effort required is 1.6 kn, efficiency = 100% = load 4.5 effort 100% 1.6 100% 3 = 93.75% 297
EXERCISE 108, Page 243 1. Sketch a simple screw-jack. The single-start screw of such a jack has a lead of 6 mm and the effective length of the operating bar from the centre of the screw is 300 mm. Calculate the load which can be raised by an effort of 150 N if the efficiency at this load is 20%. A simple screw-jack is shown below, where lead, L = 6 mm and radius, r = 300 mm Movement ratio = 2 r 2 (300) 100 L 6 Efficiency = i.e. 20 100 100 from which, = 20(100 ) 20 100 Force ratio = load effort from which, load = effort = 20 150 N = 3000 = 9425 N = 9.425 kn 2. A load of 1.7 kn is lifted by a screw-jack having a single-start screw of lead 5 mm. The effort is applied at the end of an arm of effective length 320 mm from the centre of the screw. Calculate the effort required if the efficiency at this load is 25%. 298
Movement ratio = 2 r 2 (320) 128 L 5 Efficiency = i.e. 25 100 128 from which, = 25(128 ) 32 100 Force ratio = load effort from which, effort = load 1.7 kn 1700 32 32 = 16.91 N 299
EXERCISE 109, Page 245 1. The driver gear of a gear system has 28 teeth and meshes with a follower gear having 168 teeth. Determine the and the speed of the follower when the driver gear rotates at 60 revolutions per second. Movement ratio = teeth on follower 168 = 6 teethondriver 28 Also, = speed of driver speed of follower i.e. 6 = 60 rev / s speed of follower from which, the speed of the follower = 60 6 = 10 rev/s 2. A compound gear train has a 30-tooth driver gear A, meshing with a 90-tooth follower gear B. Mounted on the same shaft as B and attached to it is a gear C with 60 teeth, meshing with a gear D on the output shaft having 120 teeth. Calculate the movement and s if the overall efficiency of the gears is 72%. The speed of D = T T A C speed of A T B T D speed of A TB TD Movement ratio = = speed of D T T A C 90 120 = 6 30 60 The efficiency of any simple machine = 100% from which, = efficiency = 72 100 6 = 4.32 300
3. A compound gear train is as shown on page 223. The is 6 and the numbers of teeth on gears A, C and D are 25, 100 and 60, respectively. Determine the number of teeth on gear B and the when the efficiency is 60%. speed of A TB TD Movement ratio = speed of D T T A C i.e. B 6 = T 60 25 100 from which, number of teeth on B, T B = 6 25 100 60 = 250 Efficiency = i.e. 60 100 6 from which, = 60 6 100 = 3.6 301
EXERCISE 110, Page 246 1. In a second-order lever system, the is 2.5. If the load is at a distance of 0.5 m from the fulcrum, find the distance that the effort acts from the fulcrum if losses are negligible. Force ratio = dis tan ceof effort from fulcrum dis tan ceof load from fulcrum i.e. 2.5 = x 0.5 Hence, the distance that the effort acts from the fulcrum, x = 2.5 0.5 = 1.25 m 2. A lever AB is 2 m long and the fulcrum is at a point 0.5 m from B. Find the effort to be applied at A to raise a load of 0.75 kn at B when losses are negligible. Clockwise moment = anticlockwise moment F 1.5 0.5 0.75 e Hence, effort at A, F e = ( 05. )( 075. ) 15. = 0.25 kn or 250 N 3. The load on a third-order lever system is at a distance of 750 mm from the fulcrum and the effort required to just move the load is 1 kn when applied at a distance of 250 mm from the fulcrum. Determine the value of the load and the if losses are negligible. 302
Clockwise moment = anticlockwise moment F 750 1 250 l i.e. F l = 250 1 kn = 333.3 N 750 3 Force ratio = dis tan ceof effort from fulcrum 250 = 1 dis tan ceof load from fulcrum 750 3 EXERCISE 111, Page 246 Answers found from within the text of the chapter, pages 238 to 246. EXERCISE 112, Page 247 1. (b) 2. (f) 3. (c) 4. (d) 5. (b) 6. (a) 7. (b) 8. (d) 9. (c) 10. (d) 11. (d) 12. (b) 303