Homework_Solutions___Week_2 - rn rn FEIEET Sample...

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Unformatted text preview: rn rn FEIEET Sample ExumInu‘linns 31' ml and 3?. The disuihuted 10ml Is replaced by 545M100?) mm}=2m:~q TIE K m The form is applied at {lfajafé fill—fiffi m from paint :1. EME:C| _ B I Jim—E T31)! [HEN] Nil—{ID THJRA =3 HA =146J§TN Answer is D. 38. Divide the area ink) recumgle R and triangle. T. :L'R '—‘ % ALI = 2 Err = 1 AT =fi a. : #111111 + MAT __ [23' + {13433} .4R AT 2 + E = 0.375 Answer is D. 3:]. First, find- thc react-1mm. REE-my Rugby: _ {mas N} + (mat) N] + {2}{25 N} _ 2 322.5 N — Emulation: inf Exam 'I—Muang Sealer: H tint the truss vertically through the eeeenrl seetjen From 4.3. The vertical component in BC is the same 3.5 the the left. reeetien at E. 25 N 5:: N' a 2 MA = ” [KM [T mute...) h (3 m}[23l'|fl N] -—. e I__-_[ REA, = 955.? N = .BGH, 2' 32.2.5 N K ADSWI-er is A. I I N m | t a l xx 3% I [ .__\1 44. N = WEBB e = [tale l‘x'jiees {10”} = 601.2 N ' ( 1F :' Ff... = Ne. =[ee1.2 N}{U.2} = 123.2 N .l' E L __ [ P;3 = {TEAS N}{etn4fl°] : 504.5 K . 15 :11 15 m F9 3 FL: The hle-el: ls. sliding. zen = 0 “(EU IIJJIF + (15 mflflfi N - 322.5 = U F = _223'125 N Anew-fer is C. FL“. 2 Np... = (5.11.2 NHDJE] u 9.11? N_ Answer 15 BI 45. The fieebody diagrmn nf the top Huck Fwd ' FEB 5 kg 4D. If there was a force in AH, there would be me other member at jnint A that. ctmlrl cancel a. vertical mmpo— _ nent. Member A33 is a zero-force member. FLE = Ill-'1'” _ FW- = #1739 ,_ ,_ x . m Answer is A. =te.1a](a kg} Keel 8—2] = “.35 N The freeberiy diagram of the lewer block 55 4]. F55 ' F Fm 20kg —Ifl-'-—F———i F = F}; 4- Flt-.45, . . _ u = tee N + [n.eetteu kg + 5 kg} —; a x =8 . N ZME = e U 9 {P Answer 15 D. L k3" --H|Bt1 -—{] 2 30; = LP’IJQH [mmpmsfiiom 4E. Since the fitruL'turt-z 15 eymmetrjeel. Angurer is D. REE = EEG Er. — n _ 3TH tees-1.1”] P = 0 414-1. = = N'fll PAC : Answer is D. " Answer is B. measslerwl PI.-l:|[terinrtsr Int. H FEKEIT Sample Emil“: w 1?. Draw the freshedy of the lower pulley and sprE-fi-dfl' her. All forms are the same because the puma-Jr's we fi-ietienle-as. F D N F = ' = 15.5? N 3 AIL-earner is E. El}. 11D)! LE 43. When she block begins to tip, it will] tip about the lower right corner. Since the herimn’sal mmpenent ef the reaction at this comer is: unknown. momenta must be taken about the corner to determine 3. Elfin“: rigid. cum-nu: "' D 4 m —z[2lf] N]. + {56 N) = 0 z = 5 Tn Answer is: D. . For n ideal as the. W and atrepic ically absu- 1t is when: rludic has a II] gt? — [1} {9.31 —) (13 a}: T1. 5‘ 2 5? 29m 0:: MI... Answer is C. 72. Impaci farce is normal L0 the wall in the 3 direction unly. $2.; = a v“ r — 0.5: w m 3 1:”: _|: ' JG) = 2% mg’s Answer is A. W K r -’.?V///_///L/r9'¥% Prnfessiunui Publimlium. lnz. fl WI hauim'linn: m This is e compound pendulum. The mass moment of inertia uf the Lube about its end [where it- is. Smipended] is 1 _ 1r . n _ '2 i ID = 371;; : flfifiliw = 0.01550 kg-m- The distance from the suspension paint to the center of the 1113.35 is L 0.3115 In 1 d E. = _2_.._ -. Learn The period of eecilletien of e compound pendulum is .' In T = 2?!" III m—yd 3 2” V" _ : : \J [0.5 kg) (9.51 {0.1525 In} =U.9':l4 5 Anewerr is D. _ 2-? _ {EJEED m} 74. a _ v—i ._ E E ma Ex I hr (1 km} been 5 =91] 5 Answer is D. _ 2 a ' __ [1?t + muting e {10;} [W335 — (eejfij + Tln [g] = 2.34 Answer is C. T5. Haring : Eprnjectiie .2 3 ,, e _ m" ikffl..-) — 2 __ . 3 cm; :11 u 1 kg 1111 ‘5. «5 u — x [ ("11] (lm Em) _= LEE mffi Answer is E. "-H'. _ “lu- _ austere—e) W P_ Lit _ e: {flkgliemozflf~em§fl =122.e x 1e” W (12?. GW} Answer is B. 7'3. 3'! / [Vssne— yam = —e a vein 3e: “- égfiflflz = —h Cheese the positive root. . .-' _ a: = [Vfilnfl' + 1...: V3511:2 H + Egh.) Answer is D. 79. The initiai kinetic energy is mara Bk = -—- 2 I m‘ 1’ _ {32 kg] (35m 1 km W 2 105 Fig-mi 32 "r 2 2i}? PET-I After launch. some of the kinetic energy will he trams- formed into potential energy; but the total energy will he unchanged. Answer is C. Pmififlflml il'fli. w unc— d be mod-- .16 .ngle IE: of .o to 19. The walls are frictionless, 5|: there are no vertical wall components to help support the Spheres. The total weight of 50 N must be roiistcd by the floor forum:l F. Answer is: D. 20. B} itself1 sphere B would require no horizontal sup- _ port. Homeven the horizontal component of sphere A‘s tum weigh: must. be resisted. On sphoro E: ‘+ WA 1131 N ‘0' 2W =-—-—:n—-—--=14T"'.32'.\.T E} I E M” tam? Leufifl" Answer is C. Eullfinns hr Exam I—Aflerlm Section 33 21. Amumc 13'z = D. 2.1;... = —{3.?5}{fifll} _ [7.5J(60D}+ [again-{g} n (3.?ajfauflj (g) + 25;}. = 0 F]; = 245.] .3 12-0 the right- Ariswar is D. 24. DiVidL-J the Lanai area. into three shapes: III‘JE'TBU- mutangle, .55., and two fimaJi interim" rectangles. B rimi C. I... ism-“i .—. (it:E [4}[i213-_ 51's Zia. -_ Ag. — SCI — 60C — {300] If —. (i Um: Lin: paralicl LLXiS ihmmm to find the mmimrlt nf , i inertia uf the SIIlliHBI rectangle about the LIL-axis. .411” = 53] [upwade EMA = {335M501 — {151(st — [imam] 4’3 It? — #313 “ME +(2H300Hfi'1-[3.’«'fi}[30rij{g} = a = (1‘1?) {EMF + {and}: E: 1220.5 {1.4: the right] = 32-57 JT 7- L1 —I:1T5 -5- f3} 3' 5?fi-— [EHEQfi‘fl 2F; = A... +1220.6+[3nm[§]=-3 4 MT : ] _ .4511: = ~14fififl = ldfirflfi to the left _ Ana'wer is C. Ansmr is B. 22. AL point E. ?_l' 212:me +Ei=0 15“. F113. + 1220.15 0' FED L —13b}3.3 [oumprewiun]: ‘2 25'” = Fan (4'25) + Em = 0 F -— 4—1133 3) 2 “E _ ‘ ' 4.25 = I511] (minim) A1: puim fl; 2 2E) = —FAE -' FAD-f1 A“. = '0 4.35 FAD — (330 — 55]] 2—) = 339.111 {temion} (Mamba: ABC is horizontal and has no vertical fume component} Answer is C. 23. A horizontal fume to the right will mutiler the othcr moments. -r _ i. —— Sulul'ium in: Emu l—Afl'emuol Sulfa: 85 — 1'01]. 312:. v=535=12fi+4h1 1’11. dim {d 2 =—---= — 12 4x—1.—24:+4 ” m2 Jim): 3+ 3' Al. x = 2. a = [24}{2} + 4 = 52 Answer is C. 3?. The distance traveled, not. the change in pusitium is rmquired, so check fur changes in the Sign fir vclflcfly. If there are any rlwcrsalfi in position. then 1.? must pass through zero. E = =1212+-1t—1 V d: v is 23m whan The pflsitinns at thu intnrval endpoints :11": III-21:31}??? —' {EH—2}“ — [-2] + 3 = —19 zt—%}=I:4H—af+[2n—%f— —&}+3=-3-s z [+5 = {4: ref T1121 [-a-‘f — w + a = 2—90? 3 (+2} = (4}[233 H2319}? — 5: + 3 = 41 The difitanne travnifld E3 |3.5 ~{—19)1+|2.9n? — 3.5| +i41 — 2.91m -_ 51.: Answer is B. 38. Inhagrate using the average force fiver math 1:5an interval. _ '[fl N5 + 45 m +1335 N-s _ _ . __. .4 kg = 13.2 mfs Amar is. L“. 39. From the impulse—momentum principle, a impulsae -— El momentum Fflf.=mflv 3-5 n x 1' ' WI +(2flN1—25- >[253J 2 =55.U' N-5 The mass is irreie'lrant. Answer is B. 4121. Using the impulse-momentum principle and the re— 51.1135 UT Prob. 39, 3.5 .3 3.5- mm: 3;] Fcii=j th+ F'dt El CI 3 2F T - N :55 351,5 _lu. {9.5 S} -— {SEES N-s Answer is D. # FI'IJIhfifDllDI PUbe-Elliflflir IflL ...
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Homework_Solutions___Week_2 - rn rn FEIEET Sample...

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