HW_7_Selected_Solutions

HW_7_Selected_Solutions - Statics 19 281 a Problem 437i The...

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Unformatted text preview: Statics 19 281 a Problem 437i The ball of a trailer hitch is subjected to a force F. If failure occurs when the moment of F about point A reaches 10,000in.-lb, determine the largest value F may have and specify the value of oz for which the moment about A is the largest. Note that the value of F you determine must not produce a moment r about A that exceeds 10,000 in.-lb for any possible value ofoz. Solution The maximum moment for a given value of force F occurs when the moment arm is at its maximum. Using the Pythagorean theorem in the figure to the right, the maximum value of the moment arm d is given by d = ,/(9in.)2 + (6in.)2 =10.8in. (1) We then use Eq. (4.1) on page 182 to determine the maximum force F, given that the moment of this force about point A may not exceed 10,000 in.-lb. Thus, MA = Fd = F(10.8in.) = l0,000in.-lb => F = 9261b. (2) The angle a at which this maximum force acts is then found to be dcosfi = 9m. :> ,8 = 33.6° (3) a: 180°—(180°~fl—90°)= 123.6°. (4) This solutions manual. to any prmt ot cletttoniu fonn. remains Ibo property ol McGt’tanlll. lnc [I ma) be used aml/ol possessed onl) h} pL‘l'llli\\||lll August of McGt‘awllill. and must be sunnmletml upon teqttCst of.\1L'Gra\\AHt|1. An) duplication or distribution. ellth in print or electronic lorm_ uulmut the permission of McGrttw-Htll. I\ |)l'0hlhllt‘ll 0. C05 :_._3_:_ _ 0 fl h ‘l0.8n%fi‘ ob 90°+ 33.60 = 513.40 {04: my? Y, +> EMA: lo 000 we F‘oL‘: lo 000;” It» £300.83: '0 000;” 45 an Wm— 1F=94wb 288 Solutions Manual Probiam 4.3M A flat rectangular plate is subjected to the forces shown, where all forces are parallel to the x or y axis. If F = 200N and P : 300N, determine the resultant moment-of all forces about the (a) z, axis. (b) a axis, which is parallel to the z axis. Solution Part (a) We use Eq. (4.1) on page 182 to find F(300 mm) — F(500 mm) + P(200 mm) — P(600 mm) (1) —l.6x105N-mm = ~160N-m. (2) In writing the above expression, positive moment is taken to be in the positive 1 direction, according to the right-hand rule. Part (b) We use Eq. (4.1) again to find — —F(3OO mm) + F(200 mm) — P(500 mm) (3) ——1.7x105N-mm=—170N-m. » (4) In writing the ab0ve expression, positive moment is taken to be in the positive a direction, according to the right-hand rule. ' ———————_—-%—__—__—— This solutions nitmuttl. In an} punt in electronic lonn. temttins the ptopeny of McGttnxrllill. Inc. It Inn) be med and/or possessed only by permisslon AllgUSI ol .\1c(jr.i\\-Hi||, and llllhl be surrendered upon request of McGtmvllill. Any duplication ot tll>lllh|lllltn (‘llht‘l' in punt or electronic loin]. \xnliout the pelmmion of McGttnvllill ts prohibited. Max : -@OON)(3OOWV9 +(é00NXQon-‘y9 ‘(éOQNYSbomM> +* O W '1 Mg: _]7o ooo Nmm J W My»: 328 Solutions Manual m Problem 4.4M The top view of a workpiece that fits loosely in a fixture for drilling is shown. The drill bit has two edges that apply in—plane cutting forces F to the workpiece. (a) If F = 600 N, determine the forces Q between the workpiece and fixture so that the resultant couple moment is zero when a = 30°. (b) Does your answer for Q from Part (a) change if 01 has different value? If yes. then repeat Part (a) with a = 60°. Solution Part (a) The workpiece is subjected to two couples. Equation (4.12) on page 209 is used, with positive moment being counterclockwise, to determine the resultant moment of the two couples as M = Q(150mm)»F(30mm). (1) As instructed in the problem statement, M = 0. Thus. we set Eq. (1) equal to zero, and using F = 600 N, we solve for Q to obtain Q = 120 N. (2) The answer obtained in Part (a) is valid for any value of a. Part (b) M This solutions manual. in An) print or slat-trunk lonn. remains the prnpcn) ol McGrnurlhll. Int. ll mu) be mud and/on possessed only by permisst AUgUSl 24. tut McCrmvHill. and must be Mll'lL‘llLlCIELl upon request 0] McGrmvllill. Any duplication \Il distribution. L‘lllk‘l ll‘l print or elecimnic Inrm. \\l[|10t|l the)t‘|’lllls\llilll‘lX1L’(illl\\rl’llll.lV prohibited. L1H, a) B600“ 61>? d: 30" 77: M: 0 EM :0 +5 0: @(fSOmM>— 600A}(30ww> Q fume; (produce a. Coaple {JO “4‘1 Fierce)”, 1 mt «:4 3 50 Okkinerj b) The, Tbs-3a couple: fife, :‘HJWNMOIMVL r‘ Jar Vale): a”! flat WM “5 :14 Ffi‘rr‘f &) \jcsfid, Statics 10 333 Problem 4,51 l A structure built in at point 0 supports 70 and 85N couples and a tip mo— ment. Determine the resultant couple moment, using both scalar and vector approaches. Solution From the geometry provided in the problem statement, we write vector expressions for the forces applied at points B and D as f3:(70N)%(—6i+3j—2l€). (l) 13D 2 (85N)%(—81‘— 12j+912). (2) For the scalar approach, we evaluate the moments ofeach of the x, y, and 1 components of the 70N and 85 N forces, taking positive moments to be in the positive coordinate directions according to the right—hand rule. Thus, using Eq. (4.12) on page 209, we obtain 2 3 A 6 A A M = 70N(§)2m1+70N(;)2mJ +0k 9 A A 8 A A +85N(fi)2ml+0j+85N(fi)2mk—2001N-m, (3) which provides M = (—501‘+ 120j + 8012)N-m. (4) For the vector approach, the following position vectors are needed 7AB=~2km, 7CD22jm. (5) We use Eq. (4.1 1) on page 209 to write M=7ABXfiB+7CDXfiD+ME (6) 70N-m ’ J k 85N-m ’ J k A _ = 7 0 0 —2 + 17 0 2 0 ~2001N-m (7) —6 3 —2 —8 — l 2 9 70 N-m A A _ A 85 N - m A A A A = [1(6)—j(~l2)+k(0)]+ [1(18)—j(0)+k(16)]—2001N-m. (8) which provides M=(—50i+120j+8012)N-m. (9) E This solunpns manual. In any print \n' electronic torin, temuins the pmpeny of MCGI'HW-Hl”. Inc It may be used and/«tr posscvetl only by pennixsinn AUngSl m NICGHHVHIH (Ind must he \Lll'lendt‘led upon request 0‘ McGi’tnvlhll Any duplication or lell'lhllllUIL either in print 0| cleennnu lm’in unlimit the permission ul .\lt(]rtm Hill. is prohibited w- MW ans-Ammo", “nu-m , ( Col/1+} SCa‘qr CAN-’2’]. COWC 8W4’m'f'éicj M0m6m+ @ E nan-mum WW ‘QOONMf ——-_ *9 /o+raJ 2573:6501 + 1.103 4., XOQNMJ @VB Vet-74 or‘ a Pram Scalw MeflmJ: Eb: (“605+3OJ‘ «24a 12>N A é=éwv%quaaw ES+GUJ‘SI/y 690$;'Jl)‘014 Veuhr; Fa Howij €21 L/{U E; :(';k>w‘ rcu: M EM=J J «.3;\ “a A a M _ A'5K@/+(rcc>>< FL>+ M; :f 3: C‘jk O 0—} + O 9» O +4200?) ‘60 30 “‘9'0 ‘L’O ~60 338 Solutions Alanna] Problem 4.561 Forces F and T exerted by air on a rotating airplane propeller are shown. Forces F lie in the .ry plane and are normal to the axis of each propeller blade, and thrust forces T act in the z direction. Show that the forces F can be represented as a couple or system of couples. Solution The figures shown at the right are Views looking down the axis toward the pro— peller; forces T are not shown and these are irrelevant to the question being asked. We resolve the forces F into horizontal and vertical components, as shown in the second figure at the right. We then ob— serve that the vertical force F cos 30° in the positive y direction and the vertical force F cos 30° in the negative y direction constitute a couple. The sum of the two horizontal forces F sin 30° in the negative x direction and the force F in the positive x direction constitute another couple. View of the propeller looking down the z axis ———%—————— 'l'lrrs solutions manual. in any print or electronic fomr. remains the pmpert) ol Mt‘Gr’tmrHrll. Int [I may be used and/or possessed oul) h) perrnrssum August 24. 2009 m McGrtm llrll and must be surrendered upon request of \cht'au-Hill \n} duplicutron m disrrrhulrou. either in prrnt or electrorrrt torrrr. without the per’rrrissum of 54thth ellill. is prohibited, O F R COS 30 : F?— F - J —' ’— LéIf—j: Fans 30° 30° F Slnfio": — F La FX“ F§§w30° Cauple~% Peirce: are, eqyaf F: #OWFOSH‘E ——-——e Z—F31’0 F5053$~F¢o$3433=c9 o v Foosgo :0 2 FA” \J stso" F"— F$;W30°‘F51\m3083 O < FslmSOO F— was) -F(0.5) : O / ((Q'Jéjie £19m Vaduz, le, roman COUPXQ {(0544 Horhow+5§ parCC‘S‘) F O , FCOS 35) \x \x' 300 S‘MBOO : 0.5 ...
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HW_7_Selected_Solutions - Statics 19 281 a Problem 437i The...

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