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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.?59j
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 compoIll 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 spiltg 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+JBil1ﬂ)_
Themlehetwemthernis andAC‘is (130—5). Thetensimis no = Incuimsusu — m +jsinusu — m) TAG = ITAGI(—i0053+jﬁiﬂt3)1 Theweightis:W= Di—Wj. Theequﬂihuium conditions: 2 F 2W +TAB + TAG = O. Suhatimte and collect like tam! _ Thetensionis TAB 2 m1. = (lDDIJ)(0.DB462) = 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 forcedeﬂectim
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
hemewdlecableAB 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 milealIeuIﬂemms, 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 belowthe 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, = 1000 — (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]' — DABSFAcj + 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 nipnine 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 cablelaw 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_
Ireml 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.6092T,qgl — 0.7422Tw
—0.9487TAO]'1 = 0 2F” 2 (0.1644(50) +0.6092TA0 +0.2401Tw
—O.3162Tso)j = 0 2 Fl 2 (+0.5077TA0 — 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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 Fall '08
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