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# Cap 3, Novena Edc

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CHAPTER 3 80 mm PROBLEM 3.1 A foot valve for a pneumatic system is hinged at B. Knowing that a - 28°, determine the moment of the 1 6-N force about Point B by resolving the force into horizontal and vertical components. SOLUTION Note that and =a- 20° = 28° -20° = 8° Fx = (1 6 N)cos 8° = 1 5.8443 N F v =(16N)sin8° = 2.2268N ^£*. \©* \u2014 tL> Kl * U^r^-^T^P,, /|t^^^u^d ^r^^^^-, Also x = (0. 1 7 m) cos 20° = 0. 1 59748 m y = (0.17 m)sin 20° = 0.058143 m. >n^y c Noting that the direction of the moment of each force component about B is counterclockwise, MB =xFy +yFx = (0.1 59748 m)(2.2268N) +(0.058143 m)(l 5.8443 N) = 1.277 N-m or MB =1.277N-m^)4 PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No pail of this Manual may be displayed, reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 153 PROBLEM 3.2 A foot valve for a pneumatic system is hinged at B. Knowing that a = 28°, determine the moment of the 1 6-N force about Point B by resolving the force into components along ABC and in a direction perpendicular to ABC. SOLUTION First resolve the 4-lb force into components P and Q, where g = (16 N) sin 28° = 7.5115 N Then MB = rmQ °^7^ = (0.17m)(7.5115N) = 1.277N-m or MB = 1.277 N-m^^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies* Inc. All rights reserved. No part of (his Manual may be displayed, reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 154 200 in.., 25" -100 mm- 200 mm i '"> mm PROBLEM 3.3 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D. SOLUTION (a) O.2. PC 0>T-**> F v =(300N)cos25° = 27.1.89 N F y =(300 N) sin 25° = 126.785 N F = (27 1 .89 N)i + (1 26.785 N) j r = ZM = -(0.1m)i-(0.2m)j MD =rxF MD = HO. 1 m)i - (0.2 m)j] x [(271 .89 N)i + (1.26.785 N)j] = -(12.6785 N \u2022 m)k + (54.378 N \u2022 m)k = (41.700 N-m)k M =41.7N-nO^ (b) The smallest force Q at B must be perpendicular to DB at 45°^£L MD =Q(DB) 41 .700 N m = (2(0.28284 m) Q = 147.4 N ^L 45° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorspermittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 155 200 mn. 25° \u2022-lOOnim-* -200 mm *. 125 il \u2022 C H PROBLEM3.4 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D. SOLUTION (a) See Problem 3.3 for the figure and analysis leading to the determination ofMd M =41.7N-m^H Cl^n 0>\7.Siy\ c = at. (b) Since C is horizontal C = Ci r = DC = (0.2 m)i - (0. 125 m)j M D =rxCi = C(0.l25m)k 4l.7N-m = (0.l25m)(C) C = 333.60 N (c) The smallest force C must be perpendicular to DC; thus, it forms a with the vertical C = 334N < tan6^ 0.125 m 0.2 m a = 32.0° M D = C(£>C); DC = V( -2 m) 2 + (°- 125 m)' = 0.23585 m 41.70 Nm = C(0.23585m) C = 176.8 N^L 5HX)°< PROPRIETARY MATERIAL. \u20ac5 20)0 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed, reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 156 PROBLEM 3.5 An 8-1b force P is applied to a shift lever. Determine the moment of V about B when a is equal to. 259. SOLUTION First note P x = (8 lb) cos 25° = 7.2505 lb /^ =(8 lb) sin 25° = 3.3809 lb Noting that the direction of the moment of each force component about B is clockwise, have = -(8in.)(3.3809 1b) - (22 in.)(7.2505 lb) = -186.6 lb -in. i*22. \u2022**. or Mj =186.6 lb -in. J) ^ PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permittedby McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual, you are using it without permission. 157 PROBLEM 3.6 For the shift lever shown, determine the magnitude and the direction of the smallest force P that has a 21 0-lb \u2022 in. clockwise moment about B. 22 in. SOLUTION For P to be minimum it must be perpendicular to the line joining Points A. and B. Thus, a = e , 8 in ^T^s*. K 22 in. / = 19.98° i and MB =dP^n p ZZ i«. Where d = rAIB fl$ u =^m.y+(22m.y JB w = 23.409 in. Then _210Ib-in. ' min " 23.409 in. -8.97 lb Pmin =8.97 lb^ 19.98° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers andeducatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 158 PROBLEM 3.7 An 1 1 -lb force P is applied to a shift lever. The moment of P about B is clockwise and has a magnitude of250 lb \u2022 in. Determine the value ofa. 22 in. SOLUTION By definition where and also Then or or and MB =rmPsm.6 = a + (9Q°-tf>) _i 8 in. <p - tan" 19.9831' 22 in. r f/fl =V(8in.) 2 +(22in.)2 = 23.409 in. 250lb-in = (23.409in.)(lllb) xsin(tf + 90° -19.9831°) sin (« + 70.01 69°) = 0.97088 a + 70.0 169° = 76. 1391° a+ 70.0169° = 103.861° a = 6.12° 33.8° < PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educatorspermittedby McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual, you are using it without permission. 159 PROBLEM 3.8 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if a ~ .1 0°, (c) the smallest force P that creates the same moment about B. V\u2014Ar 4 in. n SOLUTION (a) We have MB =raBFN (4 in.)(200 lb) 800 lb -in. or MB =H00\b-m.)< A- m. (/;) By definition MB ~rA/BPsin 6» = 10° + (180°~70°) = 120° Then 800 lb \u2022 in. = (18 in.) x Psin 1 20° or P = 51.3 lb < (c) For P to be minimum, it must be perpendicular to the