Beer, Johnston, Eisenberg Vector Mechanics for Engineers – Statics 8 ed Ch2.1-6

Vector Mechanics for Engineers: Statics w/CD-ROM

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Unformatted text preview: 0 mtworK . § “\\N\\WM\V~“\VLN Wmflmmmwaflafiwm‘wi‘m‘m““Mw‘mxwmw“\\\‘\\m\w\»\w\w\«WNW» Two forces are applied to an eye bolt fastened to a beam. Deter- " graphically the magnitude and direction of their resultant using (a) the V ‘ elogram law, (b) the triangle rule. l V\C\.§ CAI/3‘5. k: K15 KM °(: 32.5" T E l l l 3 Network l l E l i i 4 ”mm“, “away“,WWMWWW”WWW/WA.WMMW.,W,,WMM“WWWW.”“MWmywymwgmwmaw “WM/Wig“!«WWMMWMWWW/lmuwwww,wwwaonwmmywmmwwaflmuu/w \\N\“‘.\N\\m‘mmwm‘. E E E E E E E H o m ewe r‘ K E \\ MHWNWEZDJ‘EQ \\ a t r “WNW“NWNWMWE““W”“W”‘“”m“W"“WW“‘W“““WM“““““““‘““‘““““ “W“ E E E E E E E T he ZOO—N force is to be resolved into components along lines a—a’ E 4. ) Determine the angle at using trigonometry knowing that the E t along a—a’ is to be 150 N. (b) What is the corresponding value ponent along b-b’? Fig. P2.5‘ E E E E E Ea) E E E EEW. ; F5- 5‘ E a E \50 N E E E T— : 2o 0 N E E E E E E . . E Us'mcl +Y‘Roumc\\€, (Q\t— OVA Law 0% $.ne3 E E E E 200 M _, $urx LE 5 ° E E \50 M _ S“ n Ev E F) E ~ E E E 3;“ : o, s 3 E 193 E E : . a E fl 52 O E E E <>< 4: + LE 3 ° a No" E ”6 E E l <>< = I o 3 ° E E E E E9 U 51 “3 W lam} 0Q ‘l‘imflg E E E FM; : 3“ V‘ " 073° E E 200 3% “5° E _ Q E E E Fro»; , t; e M E E E E E E E E E E E E E E ,a 11/ [WM ““1 “mm w M. WW“ ‘1 ma “w ‘1“ «NM ‘1 \t N “W :Pl‘ mlt‘f$13.:ngWWW‘MWWWENMWWMW : l 2. 9 /l Two control rods are attached at A to lever AB. Using trlgono try and knowing that the force 1n the right— hand rod IS F 2 =20 lb determ‘” (a) the required force F1 in the left- hand rod if the resultant R of the foré exerted by the rods on the lever 15 to be vertical, ([9) the corresponding nitude of R. " Flg. P2.8 and P2.9 t 3 § § § 3 l s g g E M E 2 ea: ‘10“ 10°: %0° E E ‘ :1 ‘2 —38’: o l V E J5 °<o o (a; E s E i E Ugly“) W lam) O‘Q' gIVLLS E ‘ l l ____2_Q: Sin (3 : swxeao g g 2 Slnto $1W38° E l R: Exam 3 : § § 5 : E i E § § v E 1 l i s l g i i 1 1 W m N txxxlxs‘fixxxwak‘\sS‘iiw3x33»§\§\§§\§»\\§3§§\!§§§i§x§~i§i§§i§x xx \ x»»3\§\\\\53§§§§3§~x§$§§$§i§§§§$§k\xsiigsixsaii‘xkxxksixxii3§§§S§3§333%\«5!S§I\»«iflfigxxxxxxwkgxxxxxs 3333 5‘ gigxiiug‘»\\§§\\§3\§3h§§§§i$§§§§§§§§x5§§\gixxskfis x? Musg—fifzmmm“ ’ \ IOS" “w““mW“«N‘Nmmm\\\\W\&w\»N“w\mwm““\MVNWNI. (SKmH (3. «wk 2 (BKMYEKW\Q63 \oS" 3‘30 2.17\: Solve Prob. 2.3 using trigonometry. : |%O°“95°—500 cg R7. US‘W,‘ ‘We, lav.) a"? Cos‘mfli‘; Hommgorvw as:335.3x§§»§§§§§§€§x§§$§3§§§§§§3§§g33§§§§§§V§§i§§§§x§§§§~k§§§§ o V V i S i § \ : mm MWVWmmtchgmgmwmMWWWgWWW.cmWtWvcmm.Wmmm Ehgg‘méi‘émmmwm We“ E 6% Two structural members A and B are bolted to a bracket as shown. Knowing that both members are in compression and that the force is 30 RN in member A and 20 kN in member B, determine, using trigonometry, the \x magnitude and direction of the resultant of the forces applied to the bracket by members A and B. E g Fig.P2.19 3 g a i E % i i 3 i t i E E i 3 : \\o° i E U%‘. lewd OQ Qofii m8. 3 g L E 117‘: @0ka t (20 KMYL - 2(SOKN\C?’°\<”\ “END. 3 § E lRfl—hAKUl § i E Umfl 44% {and 0% Sets g E ‘ tun KN _ Sin \\o° g % .ZOKN — Slh°< g E g °< —. a?” i i W é ‘ . : 1} ° ) é a 1/5 a i" i E 3 i g g § % ...
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