Fall 2008 Final Exam Soln

# Fall 2008 Final Exam Soln - wwm QMJ_‘ 7 Given the...

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Unformatted text preview: wwm QMJ _‘ 7 Given the fd'l'lowing syStem of linear equations: ' _ “ﬁg-wet:- X+z+2w=.0.. VZx—yw‘O ._ i 4x+3y7~0 ‘ a) write the SyStcm of linear e‘qﬁatiOrlsgs a matrix product of I ' 1‘ ' I " g1dentifymatricesA,X,B,(2); ' _ ' _ I: ' .. ‘ .' ' H b) u'sing‘t‘hc adeth formula calculate the'inverse. offmatrix A, (9)113; I _ “ I _; c)‘ deterrlriine'"cl-1‘6;valuesz'izistag-CraiwéfgruleT ('9) I , .' _ ,I .‘ '“Wh-‘en Calculating deteﬁni‘nant's;.31Eintérmedi-ﬁte_workmu\$tbéshQWn‘toreccive ﬁlll.‘ - ' .__;.crédit. ‘ H ‘ ' ‘ ' '7 ' ‘ " "' ‘ " "' Problem 2 (20 Points) Determine'ine ‘m’ for which then; is no motign in the ﬁgure shown‘ The coefﬁcient of friction bétwe‘éan mama's-"am ihe inclined surface is 0.35, while the coefﬁcient of ﬁction between the rope and the_ ﬁxed cylinder is 0.25. The contact angle for the cylinder, 9 is 30 (legrees. Mam mama.ng (‘P . «TL (mama: 5&9 @ T7,. W““’l“"\‘°/ % F~¢3D nglmmmw N053 Mr Mb“ Cow were; dodcmee m 5 c. Wﬁkﬁwakj G) 0&5‘0 (5* ‘fma 225% Ron Zmﬁf‘gw‘z 33%, .T q*& 320 2 mm WW4: q m T - \ ‘I 732' \ HEEMK ~é’juﬁ-2 (ﬁg; Q.Z§(11‘9m @651 2 @MV @aom» lkﬁi’b 2’ 72:8: =~> NU (as 23%»; = O ,\. * m \ YEN) \Ugsnaa “12% \3\$‘(‘%&>w\=o WKOS‘ wages) ~= Problem 3 (20 Points) For the beam shown a) Determine the resultant of the system of distributed loads (8) b) Locate the line of action with respect to support B (8) 0) Based on your aHSWer to a and b, determine the reaction at each of the tWo supports. (4) b) D‘isrmncg prom Emmi: MN“)- “m ~32 544‘? "PT" ‘_ “goo bishops; prom 3:: IQw—gehkﬁa .1: 5:56 C) LIL”%':~() ﬁrm «m C) - fgC‘ % W1 . M ' “W9 ,9f36 7% ML 1!» ID 81mm G) Reg”! ng €351 Him 30¢, fb it“ Q My? .9.) 4.9.th 5.0;, y <33 A»; z; t, om} mfg?“ ' “mummy” W Problem 4 (20 points) Block B of known weight W is suspended from a small frictionless pulley as shown. The weights of the pulley and the cable are negligible. 1f the distance between the supports is known and equal to L, determine the force P for which the sag s is equal to 12/2 when the system is in equilibrium. Solution From FBD for pulley B: it: + me 333 m T cos 8 w'? eee e ﬂ 4p0hﬁs Therefore, from the right triangle formed by s and BC: tanem s/(L/2) At 3 = L12, the angle is: /“"‘"‘---.\ 6 = arctan (1) m 45" ® 3 6» (ms 8‘ :2 nee at! e points for the FBI) M" . \s: T 3 “'11—” 1 oint t 2 sin 6 I) “a. L‘ m T \W ‘3’“:- I”: The cable is continuous, so: ‘ N ' i l .7; : W ‘ VQ‘V P T 2 311145” 4 9 ‘ P: W/V'f: 0.707w Moment About a line problem: For the system shown, the bent bar is securely anchored to the wall at point 0. a) Determine the scalar component of the moment produced by the 750-N and 300-N forces indicated about the line 0C. (15 pts) b) Determine all the reactions which must exist at O for this system to maintain equilibrium. Express answers in Cartesian form (10 points) 1 K - 4 ~.. A A A [80 A A Q“ : g : O-L/Uél ‘lr 049138K l/(ao)2+ use)2 80 mm i "' A ,, A 3 g : V ‘ 1 ----- mm“ 1602 ~ IBDK 3‘ "A'— I 0.60M?» 0-74742 : e ' (l(,o)‘+(-IBL’)‘L [801mm ( EKUmm _, a 3 A El éAc: 80' _ZSOJ to.3ouai ~ 03521;! . it": : I We on +c- zso )1 /\ X A I 750 Nf - y J. x ,w" Hit) mm B ""“M «1" éAL H’s/0 *OA —v “Ag A .\ fu14‘"\——«_——r-——;: 1‘ 4 M = r Y (FA: F ): (0-2501-o.pgux)x 300(0301/51- 0.9SZ‘1J)+ 7§til(o466’-I‘II-Q?L/7VK)—( "l' N I [0293' WW?” [ ( ‘YI-WIA ~285.?23)+ (4983.3 560.6 2)1 2“ A A a A : (0.250j_o.:aox)><i58q‘7qi_28g,;J ~560.(,K] w i J ,2 ] ‘1 A _ ’88 7“? +,06.I§J' ‘1‘47.L/5'Z [Mm] -7 FRisuuPN ’ n O 9 C l o (E C l A A 11% MAO/o. éoc : [_88.7,,~ + (06,”;3 wit-{7.LIZKJ‘LO-‘406‘I +0.7138K] Me“ LI :"?O.7g Nlm A or Could have ULPJ Sea/Ar-trl-P’? PVOt/“cf 12‘ 56: l / I I‘ M E _» A 1‘ “ -" 2 2 F: [g89.7‘/ + szi + [-23 5‘72 +Ry:l~l +[-§60:6 Will-‘0 My K M2 FAI, \\_ \J y \\ EMU = [‘88]! +Mx]f4[wg.;5+ mﬂff [-unW I": J2 [” "11:5 22” " I I / 1 it :7 R;:-5m.7y~, Elysium R°z= 560.6N M; : 89.71Nim, MI”: -;og.r§N-m, m;:143.¢IN-:n Problem #6: The 6-m pole ABC is acted upon by a 45 S-N force as shown. The pole is heEd by a bail—and—socket joint at A and by two cables BD and BE. For a x 3m, determine the tension in each cable and the reaction at A. i ! : Soiution £20 goints totai}: FBD of pole (3) r ( ;3,—6,2\ F m 455k J = 65 W3,w~6,2)N (2) r 1.5 —- w T 3J=-§£<L-2,—2)N m r { 1.5,—3,3) T x. TEE : TBEL 4 5 J = €L09~292)N (2) g" F 'r F r I' Z MAerIAXF + rBJAXTBD + {BIAXTBE : 0 (2) ijk ijk ijk T 65060+~§£030+m§5~03020 (1) m3w~62 1—2—2 1—22 Using cofactor expansion over the 2nd row, , T , T . 65(6)(21+3k)+ \$30.21 w 1<)+ \$121— k)m 0 2M,“ = 0 780 ~ 2TBD + 2TBE = 0 (a) (1) ZMAZ = 0 1,170—TBD—TBE m0 (b) (1) Multiplying (b) by 2 and adding to (a), 41339 = 3,120 —> Tm) = 780N m (2601 ~ 520 j~ 520k)N TEE = 390N m (1391—260j+ 260k)N (2) Z};=0w~>qu195+260+130=0anw—195N (1) ZFy=0—>Ay—390—520~260:0~—> Ay=1,170N (1) ZFZ=0~«>Az+130w520+260=0~>Azm130N (1) A=(-~19Si+1,170j+130k)N (1) ...
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Fall 2008 Final Exam Soln - wwm QMJ_‘ 7 Given the...

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