Homework#1 Solutions

Homework#1 Solutions - Problem 3.29 Two tow trucks lift...

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Unformatted text preview: Problem 3.29 Two tow trucks lift amotorcycle out of a ravine following an accident. If the IOU—kg motorcycle is in equilibrium in the position shown, what are the tensions in cables AB and AC? Solution: Weueedto findunit vectors 933 aodegg. Theo write TAB = TABQAB IdeAG ZTACQAG. Finally, “lite and solve the :3un ot'eqlfililm'um. Fordleringalfl.‘ medle knownlocatlomofpoinu A, 351111113, TAB _ TAG 9A0 — — ll'ABl ITACI GAB : r33 = —3i + 3_5j m |rAB| = £1.61 In 1'30 = 4i +4.5jn1 II'AcI = 6.05! m GAB = —l].fiBli + 0.?59-j BAG = fl‘fifixii + 034$ TAB = —l].fiSlTA3i + IEIJ'SM‘ABj TAG = fl‘fifixiTAci + DRAIWACj W = —mgj = —(lfll])(9-.Sl]j N Forequillbdmn, TAB + TAG +W = {I In compo-Ill form, we have F: 2 4.65m”; + 0.5641310 : n 1“,, = +0359?” + 034??“ 4931 = 0 Solving, we set TAB =658N, 1310:6115}: Amtfl 3am comm mg=um}(9.81)N Problem 3.33 The unstretched length of the spring AB is 660 mm, and the spring constant k = 1000 Mb. What is the mass of the suspended object? Solution: Use the lineal spilt-g fowe—ememium [elation to find the magnitude afthe temimlinapuingABt Isotatejmtetmefl. The tunes me the weight and the tensions in the cahlesh The angles me 35!] Lane: = (w) =U.5333, or: 301260; 35!] r. = — = 0.8175, = 412". an .13 (40”) J3 Themkhetweellthemnis mdmespuingflu. Thetensimlis TAB = ITA3|[iCCBfl+J-Bil1fl)_ Themlehetwemthernis andAC‘is (130—5). Thetensimis no = Incuimsusu — m +jsinusu — m) TAG = ITAGI(—i0053+jfiiflt3)1 Theweightis:W= Di—|W|j. Theequflihuium conditions: 2 F 2W +TAB + TAG = O. Suhatimte and collect like tam! _ Thetensionis |TAB| 2 m1. = (lDDIJ)(0.DB4-62) = 34.6N. 2F: = (ITABIGOSE‘_ ITAGIDDEJSJI =0, _ 0 _ O. ‘ ‘ ZFV = (lTaclsiImHTABIsinfi — Iwm‘ =0. F”— 3"” ”3—41'2’mm‘3‘1“ 0.948 Conn ' ; it}! W : — 34.6 243.62 N: SolveilTAcl= m.3}|TABIMdlw|=(lmm:3 )ITABI- ' ' (0.752% 1 The tension |TAB| is found from the linear spring force-deflectim relation. Thesprmg extenshmis The massis m. = 01%?) = {fit—gm} = 4.44'17 kg AL = ”R350? + (600)? — 660 = 694.65! — 660 = 34.62mm Problem 3.56 The system is in equilibrium. What are the coordinates of A? Solution: Determine fiomgeometrythe coottlinates I, y. Isolate the eahle junctule A. Since the frictionless pulleys do not change the magnitude of cable tension, and since eaeh cable is loaded with the same weight, aahiu‘at'ily setthisweightto unity, |W| = l. The angle hem-ewdlecableAB andthelznsitive 1: axis is or; the tensiooinAB is ITABI =imsa+jsium The angle between AC antlthe positive 3 axis is (180° — ,3]; the tensionis Tag 2 lTAC|(—icosl3+jfiinl3]. The weightis |W| = {Ii —jli "Illeeqnilibrinm conditicms ale ZF :TAB‘i'TAG‘i'w =0. Suttstit'nte milealIeuIfl-emms, 2F: 2 (coso—oosfifi :0, 2F, 2 (sina+sin,8— l].i =0. methefijstequatimoosa = 0053. Onthetealistieasstnnptim thatbothanglesateinthesameq‘nadrant, menu 2 fl Fromthe second equation sine; = {a or: or 2 SI]? “Fiththe angles known, geomeuy be used to determine the oomdioates :s, y. The migii] ofthe 2:, y coonliiiate system is at the pulley B, so that the oomdinate I ofthe point A is punitive. Define the positive distance r: as shown, so that h Similaaly, (h +E) = tenor. — m Reducetoobtaio I zit—hoota—Ecotcz. Suttstit'nte into the first equation to obtain 3 = (g) (b — hoot: or]. Multiplythis eqnationby bane: andnse : = thanato obtain a: (ta?) (b—kcom]. Thesipiofthe coordinate yisdetermioedasfollows: Sineetheoo- mdioate I is positite, the condition (it — hoot on) :- l] is tequitetl‘, lwith this inequality satisfied (as it nntst he, or the pmhlenl is itwalid}, s is also positive, as required. But the angle or is in the first quadrant, sothatthepointri is below-the pulley 3. Thus y = —E antlthe moniinates ofthe point A ate: 1 = (g) (b—hcot on], y: —%(btaua—h), or 230° Problem 3.64 Prior to its launch, a balloon canying a set of experiments to high altitude is held in place by groups of student volunteers holding the tethers at B, C, and D. The mass of the balloon, experiments package, and the gas it contains is 90 kg, and the buoyant“),I force on the balloon is 1000 N. The supervising professorcon— servatively estimates that each student can exert at least a 40—N tension on the tether for the necessary length of time. Based on this estimate, what minimum numbers of students are needed 313,0, and D? Solution: 2F, = 10-00 — (90)(9.31) — T = I] T = 11121 N am, 3, u; B[16,D,16) C(11), o, —12) D(—16,0,4) We need to write unit veetms 9,43, 9,40, and 9,11). 9,43 = D.66'i’i — ossaj + 066?]: em = 0.57111— 0.4565 — 0.684]: em = 41.31731— 0.436j +0318]: We now write the fmees. in mm: ofmagm'tudes and unit vectors FAB =CIfiE'TFABi—0.333F33j+0_flfi?FA3k FAG = 0.57UFA01' — UASEFch — UlfiWAck FAD : —0'.S73F'AD]' — DABSFAc-j + 0.213FAGk T = 117.1] (N) The equations of equilibuium ate 2F: 2 0.667FAB +0.5?DFAG — 0.313an = 0 EF, = 4.3331343 — 0.45650 — ammo + 11171 = o E 1ra = 0.6671343 — dogma + 0.2131110 = n Solving, we get FAB 264.3NN251MEI1L5 FAG 290.8Nm351udmts FAD 2114.8Nm33mdems c [lD,tl,—12.) m (4610,4111! x a (16,0,16]m A (113,011.: c [lD.D,El2]|:n I a (1110, 16) m [BOON (90) g [— 16. D, 4] a a (16, -:1 [6) m Problem 3.76 The cable AB keeps the S—kg collar A in place on the smooth be: CD. The 3; axis points upward. What is the tension in the cable? Solution: The coordinates nip-nine C and D are C (0.4, 03, fl), and D (02, 0, 0.2.5). The unit vector fmrn C towmd D is given by 905‘ : Qcpzl +903” + Bcpak : —O.456l — 0.634] +0.57flk. The locsh'orn ofpointA is given by EA = 1:0 + (103903,, with simflaa equations font y}; and 3A~ Floatthefigme, dag = 0.2 m. me this, we find the mountinates of}! 3344. (0.309, 0.162, 0.114). me the fignne, the coonfinntes of B me 3 (0, 05, 0.1.5). The unit vectou fiomA towmd B is then given by z 3,43 =9”; : 933,5 : ”mtg: 0.6m : 0.73% : hawk. Thetenaioulfome inthe cable-law hemittenns TAB = —U.G74TABl + 0.735TABJ' + OfiTQTABk. me the flee body diagram, the equilihtium equations me: FN: +TABQABz = 0, FNy +TABBABy - my = U, Ellll FN; +TABQA33 : D. We hate three equation in font mknowns, We get anothel equation Eton:- the condition that the bar CD is smooth. This meant! that the normal fume has no component parallel to CD. Mathematically, this embestatedas FN -QGD =0.E1pandingthis,weget FNIEGD: + FNyE‘ch + FNaQCDz = '1 We now hue foul equatimn in our founmknowns. Substituting inthe nomhets and solving, we get TAB = 57: N, FN= = 3319 N, PM, = 361 N, and PM = —4_53 N. Problem 3.8? Cable AB is attached to the top of the vertical 3—m post, and its tension is SO kN. What are the tensions in cables A0, AC, and AD? Solution: Get the unit vectors parallel to the cables using the coordinates of the end points. Express the tensions in terms of these unit vectou's, and solve the equilibrium conditions. The cour- dimstes of pointsA, B, G, D, 0 are found Eromthe problem sketch: The coordinates ofthe points are AGE, 2, 0], 3(12, 3, 0], CH), 3, 5], D(0,4, —E), 0(D,fl,0]. The vector locations ofthese points are: rs=3i+2i+fllh rB=12i+3j+0ks Pc=fli+aj+51h EDZDi+4j—5k1 rozfli+flj+flk. The unit vectorpsrsllel to the tension acting bemeen the points A, B inthediiectimofBisbyrlefiaition 9A3 =fl_ Ire-ml Performthisforendlofthetfltvectows a” = +0.9864i —— 0.1644j + 0]: am : —0.0002i + 0.00025 +0.50‘rrk 3,45. = —0.7440i —— 0.24315 — 03202;: em = —0.0487i — 031023 +01: The tensions in the cables are expressed in terms ofthe unit vectors, TAB = ITABIGAB = 509.43. TAG = ITAGIQAGa TAD = ITADIBAB. TAO = ITsoleso- Theequflihu'ium conditions are 2F =0 = TAB +TAG +TAD +Tso = 0- Suhstit'n‘te and collect [the terms, 2 F: = (0.0304(50) — D.6092|T,qgl — 0.7422|Tw| —0.9487|TAO|]'1 = 0 2F” 2 (0.1644(50) +0.6092|TA0| +0.2401|Tw| —O.3162|Tso|)j = 0 2 Fl 2 (+0.5077|TA0| — 0.6202lTADDk = 01 This set of sinmltsneous equations in the tmlmown forces may be solved using my ofseveml standard algorithms. The results are: ITAOI =43.31’N, ITACI 23.3]:N, ITADI =5.EkN. ...
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