me211solution1 - 2-9 ‘ W eAB =‘{c.o’s45isin45° = eAC...

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Unformatted text preview: 2-9 ‘ W}. eAB =‘{c.o’s45isin45°} = eAC =.{_cés 30°, —sin:30?} ¥ We can write F = aeAB‘+ beAC, where a and b are C0nstant. 22'22 '[Z—a #0; ~fl'aél‘b: —500 2 2 V ', ‘2 ‘2 ' Solve the equ‘atio‘ns for a and b a 7 —448,b = 366 ‘ I The magnitude of two components is" FAG = 366N <{AmueV wauoofl Pym/{me “We, @W'FOWM9 FAB avid Fae. Fm; : (3094?, .F/WQMQ’S") Eflc : ($750609ng ‘Ffio7m300> 317,45“? EM 1 E (TM; 051,450+ “FACCOE ?0° ,0 ‘FAGQ'M¢§v*FAc4\M?C‘> 2 (0 I _é7m> FAD, Cfi¢§0+FAc C0230“ 3: 0. FM; 4M4? 1 Ewan/30° i «3‘00. golUQ ‘fiJJO Q%Owl+t2¢/\$. FAQ 6 Wficc ) LMefidJfive am wean; MMQ ohmm 04f,” T} OFFMHN, f0 0-? i-vx7+Fa/Ha MCéb‘YVWDQ W. 2-20 ‘ 1:2 {—6255in30°,4625¢os30°} F3 {~7505in45°,750cos45°} .‘;FR'=F]+F2+F3 _ _ Y 7‘: {(680— 625siri 30° — 7505111450), (—510 —' 625605 30° + 75000s45‘?)} i ={~162.8,—521} ' ' The magnitude of the resultant force is ‘ _ .FR'-="/(—.162.8)2+.(—521)2 545.8N » ' ¢=tan-F[ ‘521 j=73° , —162.8 - '_ The angle of the resultant force measured from the positive x is ' . 9 : 180° + 7341:2'53o 2-48 Fl {0,350sin40°,350cos40°} {0,225,268} F2 F3 '{250008 60°,—25000$45°,250c0s 60°} {125,—177,125} R=E+E+E, ' = {(71 +125),(225 +50—177),(268—50+125)} = {196,98,343} {100cos45°,100c0s60°,100cos120°} {71,50,-—50} The different notation using i, j and k is also possible F1 = 350{sin 40°j + cos 40°k}’= {225 j + 268k} ' "F2 =100{c'os45°i+ cos60°j + cos120°k} = {71i+ 50j — 50k} ~ 2F3 '= 250{cos 60°i .—-cos 45° j +“c_os 609k} = {125i —_17_7 j + 125k} FR; F, + 1F; 4 F3 = (71 {r 125)ii+ (225 + 50—177)j+ (268—— 50+ 1.25)k = 196i + 98 j + 343k ' ' ‘The magnitude of the resultant force is 2 FR = JFRj + FR; + FR] = x/1962 +982 + 3432 407N " The coordinate direction angles are - FR 407 : .' - F '52:" c'o_s“,6'= —R.y_ '= , : FR" 407 0087 = FRI egg, 7_=32.6° 2—70 - A ' ‘. A r“, :5 frA 2 {4000} l— {125’sir-125°,125cos25°}_= {(40041255in25’°),(—12Spos25.°)}2 I ’ ‘ ':_{452{8,..—113.3}"1 ' ” " " " 'or . {AB = rBI.— IA 2 4001.4 (125 sin 259i + 125 cos-25°j) = 452.8i—113.3j ‘ _ The magnitude. of the pogition vector from'A to B_ is ' erB|¥‘/452.82+(—113.3)2 , 2-89 __ .__ u The component of F along AC is the projection of F along AC-(FI‘I) " The projection can be found by dot product with_the'unit vector along AC ’ #175930 (1) Express F as a Cartesian vector F=|F|eBD Vector F can be expressed by the magnitude multiplied by the unit vector, which tells us the direction of the vector. The unit vector along BD can be calculated from a position vector along BD. _ l.BD _.eBD. _ erDl. r '="rOD —r ' BD 01; :Although I‘OD can be'fOun’d directly from the picture; (08 is not simple. We need to use i "information about point B given in the problem “point B is located 3m along the rod- from end .I '- .which means the length of BC is 3. 76 rOB:r0C+rCB r _ _ CA ' rCBferBleCA _erBl I . ‘ Ira-l. ,. I "m ;.r0;.—yoc 5 {0,03} _—{—3,4‘,0} = {3,—.4,4} Therefbre, rCB = [ram CA )2 (aw = {1.4056,—1.8741,1.8741} 1' lrcAl J32 + (—4)2 + 42 r05.={—3,4,Q}+{1.4056,—1.8'741,1.8741}'={—1.5944,2.1259,1.8741} ‘ ' rBD.:{4',6,O}+.{f1.5944,‘2.1259§1.8741}={5.5944,3;8741,—1.8741} 7 .I " {5.5944,3.8741,—1.8741} ‘/5.59442 + 3.87412 + (—1.8741)2 F: (600) = {475.5659,329.3275,—159.3125} rAc fie; {—1474} W={e0.4685,0.6247,416247} (2) e. ;=—= --".C- Incl Ire/4| '.\/(—13)2+,42+(—4)2 ' From (1) and (2), FN = F-eAC = {475.5659,329.3275,—159.3125}-{—0.4685,0.6247,~0.6247} 82.45N The compc‘ment of F perpendicular to the rod is Fi Ff + Ff = F2 = 6002 Fl = «6002 —82.452 594.3 N 2-91 1 ’PYDJéch‘fiA a"? Fl A attng m poke, .The projection can be found by dot product with the unit vector along the pole Hfiflfim ' Unitvector along the pole: _ rOA _ {23234-1}1 2 '2" 1 %“t|”?32f7‘??7 0A ( 1) . “Therefore, perF: F” = {2,4,10}-{§,-§—,7%} 0.667kN ...
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me211solution1 - 2-9 ‘ W eAB =‘{c.o’s45isin45° = eAC...

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