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Test_1_Solutions - Name ID Section number Michigan State...

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Unformatted text preview: Name ID Section number Michigan State University Department of Civil and Environmental Engineering CE 221 Statics (3 credits) Fall 2010 Test 1 Maximum time is 50 minutes Note: — This is a closed book and notes test. You are prohibited from using formulae and/or problem solutions stored in the memory of your calculator. To facilitate grading, your “answers” must be clearly marked in boxes or underlined. Make sure that the appropriate Scientific units are clearly stated by the answers. All work leading to the answers should be attached and sufficiently complete for the grader to 'ude artial credit and reasons for error. Name ID Section number Problem 1 - 20 points For each statement in the table below, check under T for true and under F for false. Statement If r is the position vector from A to the force vector F, then the moment vector M of the force F about A can be obtained from F cross roduct r. M: r x l:- L The position vector from A to B can be expressed by the coordinates of A minus the coordinates of B. 3 Dot product of two vectors can be used to determine the angle between the vectors if the magnitudes of the \/ vectors are unknown. A- (5 : ‘1 34059:» case; fif—Bm Cross product of two vectors produces a third vector at nine de rees to the lane of the two vectors. \/ 5’ If two directional angles of a vector are 40 and 60 degrees then the third directional angle is 66.2 ‘/~ degrees. (103140 + £04,th + cod/keg. L .2 g Q The dot product of vector F with the vector F is zero. \/ If r is the position vector from A to the force vector F, then the magnitude of the moment M of the force F about A can be calculated as the magnitude of r times the magnitude of F9 time the cosine of the angle V between them. M: ii'r P : rF sane} L g 5 + 2 - 2 = 5 Wthe position vector (1') from point A to the force “1 vectorF[F={10i+20j+30k}N]isr={1i+2j} m, then the moment vector M about A is \/ 7 M = (40i — 30flk) N—m L m A moment of a force F about point A in space can be calculated as the magnitude of the force vector F times the magnitude of any position vector drawn 1/ ’ ifromAtotheforce. a: vxF :rF—u'ne _j q) 5A? 601?? p " k :1 J .o 2 e 9‘ l o 9 Z i 2 o l ‘ “d 'H“ 'LO 9’0 \D 30 IO 21‘ i0 90 30 2 0 a Name ID Section number Problem 2 - 30 points Determine the moment of the given 20 1b force about the hinge using vector method Copcrdt'oaigl a; A (,3, ol- (4). CoxcnrdrndiZA [q B ((3’wa ) gun 9/0). a ——_ . YBA : (3’0): + (o‘»-3C031o)§ +£q£3$1nw>l<o :- 3? _ 28qu 7+ 244a K.~1Lt' m] «a a U " YBA Bio «98qu +9-4;qk ex» ,, : _/ 1 Z rm, 31+ L” 22814!) + (2.61414) a. . 0.5q011»? 00:555IJ4—O‘5856K- 3 Name ID Section number Work Sheet ’7 > Z .7 F - r J M ’ Uhinja C 08% EM 3flf/ 2/ ./ /// ‘7 211 $3 / / Am°""~=- V a” : u u If 0- ’M'igil’flom @A FC 8*) i Q : oh (05610.44 0 —— o 5'55” +0 583%) Wis! E — H Ioj+ H 1H4 : 1 00 a, a fig/Kmegg L A 03 K. __ 033 .4, 639010 " ‘_57 e . A Y ~ Co~DDL + (3‘03”) 08 P I O ”09.1) K' ._ 02+ 3&9st + 55* o »- 05+: 81J~+|03K (H p 0 K T? x 2 - L J “‘8' -——H IO “‘71" 87, o 2 ‘03 ‘05 G O ‘ +K t 281 ['4' “J we: IN“ \‘nxss ””0 : «we :1 ‘ ¢ a 444g? + .2- \83 — 333K. LION“ F“ ';~: M : Uhfnfic I (YOBX BA) 9 9 _ e .; 4: . (4,4.4SL +11~I84 ~33-3K) : 4445" Ubfit‘ Name ID Section number Problem 3 - 10 points The coordinates at which a person is standing on the golf course are (10, 10, 10) ft. His golf ball landed at the coordinates (200, 250, 10) ft. Calculate the distance between the person and the ball. Work Sheet ' , fl \ A"? kou‘lvRoo Mm: 644 / / \ Pugvh (,5 uc«;wdt‘n3 3/ ‘ ‘- [73.3 [Jana «Hi/rare 441a "M—gr?‘ - , A t; B 30L} (gall (Clndflfiiv AB 00.10.10) (loo/mum) A . rm -: (loo-I0): + (WHOM, ~+(Io~—to)K. : I‘IO(Z + 24403 + oK. gir‘ dx/s'lfinCe, balk/czar) [Jot‘n‘lé A and B LLA ‘- "7 (A43 : Haam‘luoie 51 rm 1 L 7.7 , x @510) + (LL10) +(D3 J:- 306‘ | 45‘ 527‘ Name ID Section number Problem 4 - 20 points Force F2 is located in the y—z plane at 300 above the y axis. The directional angles of F1 are given. Assume a weightless hook and calculate the resultant force acting on it in vector form and the directional cosines of the resultant force. Z F2 = © 2007 by R. C. Hibbeler. To be published by Pearson Prentice Hall, Pearson Education, Inc._ Upper Saddle River, New Jersey. All rights reserved WORK SHEET q F :- ECU). I ,-. c r 0 dficcii‘on magic/C =r oz ; ugo I {l : 60 I «Z» 2 M0 d ”6‘“ [>052 Name ID Section number F2 = 1001b F x 1: 2001b (3) W 2007 by R. C. Hibbeler To be published by Pearson Pvermoe Hail. Pearsun Educahon‘ Inc.. Upper Saddle Riven New Jersey. A“ rights reserved. W7. \ltcTo—v TF2. (42.5 {n 515-"? Wang. A F) 1: P>%f+g'j3 «f ELK o. _ * @(L’ + EmgoJ +F1$J.n‘30 K P 0 0 31030 K. 0L+HDOCDSBDJ «H00 0 : 05+ EeéJ +€OKU> «:7 Jr —7. FR: Ira—kg, O H EcALLH-afli’ aflce‘ ‘ ! — QUI‘U l —t ‘00 ET : .— 3(a).) a 7‘? , . : 200 Coqu-‘H +o-SJ .06 K) o a : tut-UL +100J« [ODKaUD (J j _ )00K).+ C70: «+96% + 530") nuuui+1~ee ej ,sokwb- :JLu. u) +Qge 6) +502 7 «50K Umwc'w M) a - [W FR ’ 230! H dlrcclt’on WV”) 31' F‘ r A ~ _ 7 ’ ‘V cm: “it “Lu 2 05m, arsfg— fl = 6? zed-4‘? In 3 934'” FR 256"” 7 ‘2 5 F27“ ~ "50 : ~0‘lOC'i C55. 7: ' 234M Name ID Section number Problem 5 - 20 points The light fixture is suspended from 2 springs each having an unstretched length d of 4 ft and stiffness k = 5 lb/ft. Determine the weight of the fixture at equilibrium if the angle (9 is equal to 55°. WorkSheet bABD {A Axmi'ar l?) A {5‘43“ Cm b kBD 50 em C433 45 ; M3 M i E A W :> to: at“; «W? E 7 If? Uh grlwvlcl/xed long [It 2 U 'H dongallro‘r) a1. $b7i‘na <5, = force in "Hm Slm‘nj F5 2- KKci" Name ID Section number i], “A A: sum: * o I , ‘ d % F50, (10$ 350 4- PSI C,D$ 85; 2 9k] ' CJcnflclbL-“ng 5" . fl 0 A 0 Sam :7 F5 craggs + {5 cases 2 lad“ O :3 213509355 : \od :3 £115” “MRS- : W / ...
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