Beer, Johnston, Eisenberg Vector Mechanics for Engineers – Statics 8 ed Ch2.9-11_A_07

Vector Mechanics for Engineers: Statics w/CD-ROM

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Unformatted text preview: W Wm__L___WWLWWW J '0 V 1."' .; ~ ‘ ML 77 45J An irregulafly shaped machine component is held in the position : by three chmps. Kmmgfi’fmt F A = 940 N, determine the magni- : ‘ forces FE md FC WM by the other two clamps. fittou 21:35:07; FB 3‘.“ 50° ~F¢ s‘m70" Sm ‘ ° FE: Fa. i SM 50° '26: :c:°wou~, FE Cos 50° 413:; Cos #0“ $0¥S+i4u¥~2 V\w\_V_F(3>V "gram {F3 éfiuw‘kian ‘3‘- n #0" S} A 600 C05 500 - E; 005709 0:6)“40U ‘f‘ Pg ‘ ' .4, WWam‘m/WMWWWMWW WMWWWWWWWWW m ‘ «mi $0 \u} . s mm. 2”ch ific 2F ,c “Aw U gffiFgc'riffi k1 F3; (sasnfl 2:: ' f * ‘ g » (3.48:)TWO traffic signals are temporarily suspended from a cable showif Khowipg that the signal at B weighs 300 N, determine the weight 0 the signal at C. ‘ ' X: [go—30299.9 = (444° al..,,.a.,flmmm‘mm‘m,‘ nmflnmmmflmmmm,.nm‘mm‘m‘.‘ WMHW“..WWMW,Wy,MWW (Wow A A A V e . ‘ a . < WWWW , ‘M (TIT; “’0': S\ :c\\-\‘\\ dig‘véer‘ct—CL any.) 9/ 7 $6?an a an komkcr 0‘: SHCY‘ $95.14 div-x JV; Co" £4 n 22.55 I Two forces of magnitude TA = 6 ldps and Tc = 9 kips are applied i1 as shofim to a welded connection. Knowing that the connection is in equi- ‘ fibfium, determine the magnitudes of the forces TB and TD. TC TB = TA 4’ Tb Les LIOO : (”K-W33 + (N4 K}f$>C°5 40° PROBLEM 2.53 In a circus act, an aerialist performs a handstand-on a wheel while being pulled across a high wire ABC of length 8 m by. another performer as shown. Knowing that the tension in rope DE is 35‘N when the aerialist is \ held in equilibrium at (I: 2.5 m, determine (a) the weight of the V 5.. aerialist, (b) the tension in the wire. 1‘- +£W530m in “BC, ‘15 or 3"“%\L vTABL k ‘ d VOAML bécaose ‘\’ a whee. 0A 43kt at»: onl‘ chorms ‘H‘e ain‘t i3» do¢5 Ad; 5N9‘g 4% wire ln‘io u} 2 -P7egas‘ Vac can ‘H‘umk 0‘9 “Hag wk14.\ 0.5} On ?u\\\\)30 fik I fi-qsm“ wkeL\ +\q. Vaduz. is good «hi 9 0‘ E 5 t » oVPosr\L. L. .3-._ 7, “£36, are, Simfiaw *rtamfiks,30 UL ¢OLK Qm\<‘.u\oc\'€. I— Q:2.Sm 2 Z b:;.q5-z.5: 5.45m affix—AZ; L} mnA bI—tcl = ($4.3 \‘l‘nrx S409 C11 «AA. A... 20mgfl kWO aqua/Vow ‘ LZ~ ~1=C3-L32-b7‘ J} _ cm? = m — mu)?— (sis—My '5! ‘erL Solution (5 _ ~, _ _ 33.1 VERY swath/a. +0 (9.35 (pH HOL HI. hawk-er of; ficgw‘cé‘m‘é L = 2- 53:41 aafifig cwité +0 Medit— 4%. “Wale.* b - 5'45““ - 4 33“ C48} ‘ (%'—L) 5.495% fl: " ‘i 2...? 0 2— ~. M 64 :- qfl43 £36530 2 ‘ TAELQCQK ‘- 35NLO$°< +Tng¢ Cos/5 30h; 9m" TFEL . ‘TMC {cos}, -Qos at) : 35N<Los °< NC 5o< TRBC. 35 O Cos #‘ Qe$°4 ‘P Ramework 3,53 can-G ,‘far‘c: 2:25” €:é): O 1 TRBL $14K 4- 35k) Sifi°< + T“3Q anfi— Uh) So\v<,. ~9e>r L0 \ U -. 13% N ...
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