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# Suggested_Ch3_sol - 3.7 Compute the moment of the 100-lb...

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Unformatted text preview: ' 3.7 Compute the moment of the 100-lb force about A, (a) by using the deﬁnition of i the moment of a force, (b) by resolving the force into horizontal and vertical . components, (0) by resolving the force into components along AB and in the __ direction perpendicular to AB. .' i ' :___:.I ' ' I | . i__I..__.I : I._I_I I._ ._. ._.i_r. _.l _. .. i . i ._ l—_ . . I ~ I II-i—i————— : __ ﬁCﬂLUéjﬁfj ﬁeﬁmiiﬂaﬁ 45mm£ﬂ|+ ii .___ i ' . _ | | - I _ M r “a x F _a_| ! in. I ‘9:~¢+90‘1+60‘}I72.6¢ i I N :(/3)(/00)5fn /72.6°- LL— A \ a I ._ I ___.. ._ : _;.__ h 4) 3 _! _! A6:F5=+/2‘:I3;n. 9U I cut 47: ﬁzz.” ~. .5“me _:__.____l . I D I I ‘ - ' II—=—:—'—!r--i ' "I ' I "— I" -- '“——"—‘...._ In. _ . . :dquLFﬂhal-aiiiiL! I I—!--|-- ﬂﬁiwﬂxﬁnm :'_i 2—I —I =V3i‘53‘JXK-3ééiﬁwi _:__..._i . .._L l !--E— =[02)(50)—(S)(86.e)]f ! .-8I5.I3Ii'. ' : 600-,g33).{§ . _.. . :. :. ' I : - __|_ _ '___I_:_l _ "'— lml' jﬁ |_| | | l I I _ .. _ I i. ‘ ; I _ /6.’.n; i _ 3.11 Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 300 lb and that c = 18 in., determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied (a) at point A, (b) at point C. —.' | | | l i" I: -'!h|'_j: _. ___l_ ._._ _ ..! i.—5_..___ .. _ . .. .__E_ _. I ._£{q§1. .MmiﬂﬂMﬁﬂLd—ilJLE-Lﬁﬁ: enampmm+mg+s . _!_ _.._ i ._i_.__..__=....._.___._ - LL79; r, .- \_ ' .. .._.____!__!__i . g fﬂmbﬁﬁt fa ‘— 54- (22.:vm')_‘b‘(/2m)ﬁ _ _ _ _ __ I_.:__.E.__. M6 ; :z Xf: {—22.53—12ﬂxﬂ8051llv 3) .. ' .! __ ' z " 'i—L—W : -900? +£Jéoje «.— —-(32HO/b»in):@:~(270/an)f. I .ll. nfL;;;:TiTm. %:2w%#2 4- : .. I - - I ‘ ___: ._._ ._ . ! . . _ . . _. (.1511. _.ﬂvlim w_'l”_&rli§=g_ﬁ__.13r.usf%.4r_¢tgh¢_gane%¢ﬁj__mf IF... ;_ . _.. _ w...- I i I: I i ;._ '. .. ...| - ' ' _!__!.___.__.;.__H._J___L__LVi I |\' ' - . - . . .. .___I_;.__ .5 _i . _.:_ f ,(/80/b)1+{2"°”’)fz’ ._ _____ ____ g . _i_ _ I __ 445; 170/544”) 4 {_ . _'_ _—___ ___ | | ' : '_' "Ii—___ '— _ ____:": Li‘ V——I“'“f‘i‘ .__ __nr_h__l._ =J.11__.ip::_iwf I 'fCL‘ﬂEF— Whr‘ t _ .,;____:_ _;_i I_:i r;f' LL _L f_"ﬂf'*—‘*——' “"tldﬁ:—i'—5” _;Jr‘““m:g ——1 "—-—r* ‘ I __L' _L;L_ _ _* “‘F'FP—‘ :' l . | —— — "1— ___i ——I =— _:w=_:::::r _ __im—f__;:LK::ﬂ' _ft‘ _ ___ ___. _._r _I__j___: ____ i _ ' ' __ !_ _ : I -’ I‘" ‘I I __ “ f_ﬂ -_ _ri __ _. __ _i____ I __:.___i_ I | _I _i _.E__ ..__ __I __ __ |___ _ ﬁ_e_ _ ___. _ _ _ .. _ . _ _ P bl 3 53 __i_ 3.53 Four lg-inediameter pegs are attached to a board as shown. Two strings are I I m em ' l passed around the pegs and pulled with the forces indicated. (a) Determine the resultant couple acting on the board. (b) If only one string is used, around which . ' pegs should it pass and in what directions should it be pulled to create the same couple with the minimum tension in the string? (c) What is the value of that _ minimum tension? -'L—-—;—-—-?-"i"“ i 3:4" i.'.:i:}___|” ' !__: ! ._ _. _I__._ I - (a) Realm “cage. 43> Mgféo/bX/OS in)“ I ‘ + (#0¢)[13.57inr). =7 530 lb-in, + Sew/b, in. _ M 2! I=-7Q--/b¥.in. . . . . . . f mm. _, _ _ _ I . ....__§_....__ . WE?“ ﬂhi? one Shr-ﬂgﬁ Hamel Q“! _ , . _;_._..____i . maﬁa: b: Ming, mime a .U _._ = . - ---- -!——-- Th. gain. - I .2. : . . . -- -- ' "—'!' : ' E Direct/931 ‘oﬁ 19¢“. :- "i1" 5'55”? ‘ . . . _. ' ‘ ' —"— - (C) 77’"? £5ch ﬁgmenrfﬁ-é- .Cenf‘ﬂ " ' 1" ';‘(/1M‘l’+(‘f-M)‘ =' IS‘fn- I _ 1 I . _. . nemﬁﬂc II Pcffa-Cndi'wiar Album»: d ngn;‘fhg i . _. __.. . . . _. __.._.._l ! ‘ ; [ff/r. I‘ﬂi) =.-: in, I _ .. _ _ - . ._ ._ we must 1/34“ II I _I I _I I - - ---—!- ; - - 'M’=‘F4J. //.70/b-rn.:F(/f6~3/") F: 70-”1 _ 7' : _ L: :i ._____ I- I 3.66 A force and couple act as shown on a square plate of side a = 25 in. Knowing I _ .!._ that P = 60 lb, Q = 40 lb, and a= 50°, replace the given force and couple by a single F _ L force applied at a point located (a) on line AB, (b) on line AC. In each case _ i determine the distance from A to the point of application of the force. . .‘ Tﬁegfw'n Ewe and couple are replica-A bu} rm _._ __. eq oil/altar Force-Couple System al’ H S I l 1 I I kin! :._.___. J3 51::{60 Man 55“ I -: 353376? If: __l_.l ‘ n . +3 Mg = ﬁg. — an 445.963 meal—(40 mam) --— I . ‘r: =/M.o7 firm“) .I—i_:—_|—;_:!"_ i. . -. ' 45.6163 — if 60 lb /. i H "" E‘p‘lolaﬁinglmgmén-ft ﬁfkou’f ANN? Mdué v‘ .Lq:___ ﬂrpowm (a): [47.07[5'7n.=(45963/\$)X1 1:32am. _'_'_' _._ i . . . f: 50/5 150° a‘ﬁpaiwt 3,11%. fie-from A, I ' -— its} - {—7)} Fonts» mid/#4071 MA. Sham/I93 , 3:3.871'11. f :50 IE 50° 62f Faint 3.87m. bE/DVJ - .—=—3— Pmblam 335 3.76 Four packages are transported at constant speed fromA to B by the conveyor. — _ At the instant shown, determine the resultant of the loading and the location of its _ line of action. We ‘Firsf dclerrnim H4: e ua‘mlrnf‘ form" could: 3: EH” :1? HI, 'H’Ie'n Fe etiuivalenf smile (I - _ I _ Iﬂﬂ» lb I I I +l R: Hoo+2301~150+ 500:/300 . '_ I l l | l .l _i l I __; __ | l l l I l I —-- 1Q M’K=(#oo)(z)+(zso) 6)+(l\$‘0ﬂ0)+(§00)(15) ._l - - A = 800+IST00+II€ooIl1503:7l3I00/5-fﬁ .. _. l I l l | l _ II ._ ﬁasco/Ml: ﬂag/“0059) '_.!. uric. E L 5:1300 ' lb d ! . ! . __ I I . I I . .. __ _.___ . . __I__I..___ . i I. - _ _.__. : .| ! I_ I___I..__ 4/) _ (£16330 lb _______l__ - E :1300lb ; H K ' __:_ . :H SOD/Mt A I hm 6% '" ' l l— 'If'i Ii '. ':I—|—I"5"—' ME: £3: Ilsoolbatf: I300 15M 4:13.67 HI 5' l f” “5” °F ’3' q I . .i i .I ._ _ . .. I _ _. ; |___._ - = . - .. i r I I ' —'- 3.77 Determine the distance from point A to the line of action of the resultant of the - ._. __. _. three forces shown when (a) a =1 m, (b) a = 1.5 m, (c) a = 2.5 m. ._ I in miter—._- _'l'.lﬂ.'l'l£.‘39.l_ I) irimir'tlas. tangy-35;-_~f.¢.2l£;jf_%r_eflEwl atléﬂ ' __;LH_E_»_w._.bJ pm! eauziugjﬁwi Er'rauafILaEpI/mszngt. :__. , . We hm: +1 K=~MN~MN+ Hemp/ow ‘ ' if C II +3 Mf :(ZQNXBan)—(%Nle)—(8£Ma = Z-ﬁNw-(B-WM ; 'i‘ "— Mtge; 4*, 2 ' "H _I ._. . . I.-. ._. H- “3 Iiiﬁﬂowﬁeﬁ ME . . ... __. . _ . ----' R I ' Fl _—__- :1 d =‘E“ -A _ . . 171.; (a’a='m=mn‘=8(’)'z=6 d=%‘%§”=aém< :_ ' "' ' ' ' " "'"'—; 1(5) d=/.5m; Mg: 8(I.s')-2=Io d:/04’<’”°"‘=m 4 ' ._. . . . _.. ._-: A I . _ '1(C'-a:2.5‘m: MK:8 2.5-2:76 d =/3*’"’"”:/,8m 3 ,. t ) W 4 . , . . . .... ___ . I_____. — — , ' 3.82 A bracket is subjected to the system of forces and couples shown. Find the ' ___ Ffﬂhlem 34-3; ___;_. resultant of the system and the point of intersection of its line of action with (a) line __|______ _ : ‘ ; _AB,(b)lineBC,(c)lineCD. " l .. I ' ' ' ' I __ __ ._.HE_-£'_F_.r1ﬁ_i_ﬂ~3p£a.{ta_ +E7E [email protected]_c<:§;§_.am:i .c.9_u.z_p_j_e:_: . . E:le 9.-!'r.£ni__£:1fi_-E_Ef.tioﬁpzf.<...f5.y:5'i‘f M. -___.ﬁw‘li -3. .__. _ . . T _... __._._. ;.__ . '——' R.— 3a_z +5.0: ~85; + 2553’ :(80/b)_€ -(éo/b)é i - ___: | . _ .! ' . i. I R1 I \ 1—1j_1_i I L ’ib lat-3&4; "a j £43, :63 x 30; +Izix253‘ 40042 — Work ' : ~ Lacie +30% —2oo if. woof = {#80 lb. Mi? 5 :in. 3 [11. I I I I ' i I {3335}; W? must haw—g . . .___!__l__ i i __ Jan 1;: = gﬂ‘xaoj i4“: -- i 4/801? = “80\$,ng ' T'im— ' - .. [(80/[9-‘71-_ 6;". I I ' ' 5 _ .5.-. i _. .____ . . 5 Hill-i"- . EI£+¥0 J1- _EI.S_:_Q1'_A. . ' - I ! | - . . ____=. ____| We mus? have [email protected]=zgx(—eoa) 4604é:—60x4e -(e. M3 #7150 _ ‘ — 1-6—0—_ 8m. . I = _;_i. I. T_:_I___ |_.;__,_ . .'_;__..; ;. .___ - Point offm‘crsechb z’brghtm‘ 5 l_._ Wamustham I : :66 x E 4an law '} ? =[(/2in)_é + 331* WOW-5 (“M31 ~#Bo{é:—7zof-sogg 650\$; 5—240 = — 5 in. i V I l ——— ‘ Point magnate», Wm opts-3mm“; a «q — ...
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Suggested_Ch3_sol - 3.7 Compute the moment of the 100-lb...

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