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Unformatted text preview: Problem 3.29 Two tow trucks lift a motorcycle out of a
ravine following an accident. It the IOOkg motorcycle
is in equilibrium in the position shown. what are the
tensions in cables AB and AC"? SOIlItIOll: We need to ﬁnd unit vectors 6.49 and eAc. Then write
TAB = TA 36.43 and TAC = TA 1:9.4C'. Finally. Write and solve
the equations of equilibrium. For the ring at A. From the known lorations of points A. B. and C'. TAB [AC
9.43 = C‘AC =
“148' IFACI
ma = 3i + 355 "1 II‘ABI = 4131 m ['30 ll'Acl =G.02m
e“; = —0.651i+ 0.759j
QAC = + T” = —o.051ngi + [17591.35
T.“ = 0.661T.4ci + ommw j
w = —mgj = —(100)(9.81)j N Forequilibrium.
TAB + TAC +“tr = 0 In component form. we have 2 F, = 4mm“; + 0.6mmC = 0
Z a, = +0.70%” + 0.7471“~ — 981 = o Solving. we get mg=(100)(9.8l)N TAB = 658 N. TAC = 645 N Problem 3.33 The unstretched length of the spring
AB is 660 mm. and the spring constant k = 1000 N/m.
What is the mass of the suspended object? Solution: Use the litrnr spring forceextension relation to ﬁnd
the magnitude of the tension in spring AB. Isolate juncture A. The
forces are the weight and the tensions in the cables. The angles m are
tana = ($) = 0.5333. 0 = 3026‘. 350 m) = 0.875. a = 41.2°. tan:3=( The angle between the 1' axis and the spring is a. The tension is
T33 = ITAgIU coso + j sina J.
The angle between the 1' axis and AC is (180 — :3). The tension is TAC = T_4c(icos(180 _ a) +js‘n(180 _ 3))
TM: = HACK—“063 +58in 3) The weight is: VV = 0i — The equilibrium conditions: 2 F = ‘V + TAB + TAC = 0. Substitute and collat like terms
The tension is TAB = IcAL = (1000)(0.03462) = 34.6 N.
ZF, = (T_“;coso — T_4c cos :3.)i = 0. Z P” = (ITACI sine + ITABISin 3 _ “VI” = 0. Fora = 30.20: and .‘3 = 4112"“. the weight is 0.918
a ' o .3 ‘V = 34.6 = 43.62 N;
sow: Incl = (2:3) ITAtaland [WI = (%$4) ITABI ' ' (0.702% ’
The tension ITAEI is found from the linear spring fortedeﬂection
relation. The spring extension is The mass is m = ( AL = «(350)? + (000)2 — 060 = 694.02  660 = 34.62 mm Problem 3.50 The system is in equilibrium. Whatare
the coordinates of A? Solutlon: Determine from geometry the coordinates I. y. Isolate
the cable juncture A, Since the frictionless pulleys do not change the
magnitude of cable tension. and since each cable is loaded with the
same weight. arbitrarily set this weight to unity. VV = 1. The angle
between the cable AB and the positiye 1‘ axis is a: the tension in A Bis ITABI = icoscr + j sine. The angle between AC and the positive 1' axis is (180° — :3): the
tension is TAG = T,4c(—icos.'3 +jsin 3). The weight is VV = 0i — jl. The equilibrium conditions are ZF =TAB+TAC+VV=U Substitute and collect like terms. 2 F, = (case — cos Sill = 0. 21"” =(sino+sin.‘3—l)j=0. From the ﬁrst equation eoso = cos :3. 0n the realistic assumption
that both angles are in the same quadrant. then a = :3. From the
second equation sin a = oro = 30’. With the angles known.
geometry can be used to determine the coordinates .r. y. The origin
of the .l‘. y coordinate system is at the pulley B. so that the coordinate
a of the point A is positive. Deﬁne the positive distance 5 as shown. sothat
( ) =tano. h
Similarly. (b + ) =tana.
—I H In. It. Reduce to obtain
1' = b—hcota — acotct. Substitute into the ﬁrst equation to obtain 1‘: (b—hcota). Multiply this equation by tano and use 5 = .rtana to obtain t
s =( ago) (b— hcoto‘). The sign of the coordinate y is determined 15 follows: Since the co
ordinate : is positive. the condition (b — hcoto) > 0 is required;
with this inequality satisﬁed (as it must be. or the problem is invalid}.
sis also positive. as required. Butthe angle a is in the ﬁrst quadrant.
so that the point A is below the pulley 3. Thus 5/ = —s and the
coordinates of the point A are: l l
.r = (b— hcotov). y= —§(btana — h). a = 30° Problem 3.64 Prior to its launch. a balloon carrying
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 buoyancy force
on the balloon is 1000 N. The supervising professor con sen'atively estimates that each student can exert at least
a 40N tension on the tether for the necessary length of
time. Based on this estimate. what minimum numbers of students are needed at B. C. and D? Solution:
21“, =1000—(90)(9.81)—T =0
T = 117.1 N A(0,8,0)
8(16. 0. is)
000.0, —12)
D(—16,0.d) We need to write unit vectors e33, 9A0. and 6A0. aw = 0.667i — 0.3335 + 0.6671:
GAG = 0.570i — 0.456j — 0.68:1]:
BAD = —0.873i — 0.436j + 0.218]: We now write the forces in terms of magnitudes and unit vectors FAB = 0.607FAgi—0.333FABJ+0.667FABR
PAC = 0.570FAci — 0.456}:ch — 0.684FAck
FAD = —0.873FADi — + 0.218FAck T = 117.1] (N)
The equations of equilibrium are 2 F, = 0.657519 + 0.5mm: — 0.8mm; = 0
X: F” = 4.3331233 _ 0.4501?“ _ 0.4361?“ +117.1= 11 Zr. = 0.6671?” — 0.93.11?“ + 0.2131?“ = 0
Solving. we 3d FAB = 64.8 N ~ 2students
FAC = 99.8 N ~ 3 students
FAD =114.6N ~ 3 students C (I0.0.12)m (46.0.4) 111 x
B “6.0. I6) m C(l0.0.Dl2)In B (16.0.l6) m HIDN 190) g B(l6.0.l6)rn N Problem 3.76 The cable AB keeps the 8kg collar
.4 in place on the smooth bar CD. The y axis points
upward. What is the tension in the cable? Solution: The coordinates of points C and D are C (0.4. 0.3. 0).
and D (0.2. 0. 0.25). The unit vector from C toward D is given by 9CD = ecp,i + ecpyj + acmk = —0.456i — now + 0.570k. The location of point A is given by r}. = 1C + dCAeCD, with
similar equations for 515 and :A. From the ﬁgure. dCA = 0.2 m.
From this. we ﬁnd the coordinmes of A are A (0.309. 0.162. 0.1l4).
From the ﬁgure. the coordinates of B are B (0. 0.5. 0.15). The unit
vector from A toward Bis then given by (I 9A3 = 9,43,] + eAgvj + eAB:k = —0.674i + 0.735j + 0.079k.
The tension force in the cable can now be written as TAa = —0.674TAai + 0.735TAaj + 0.079TABk. From the free body diagram. the equilibrium equations are: FN: +TABeAB: = 0 FNy + 73486.43”  m9 = 0.
and FN: +TAB°A8: = 0
We haw: three equmion in four unknowns. We get another equation
from the condition that the bar CD is smooth. This means that the normal force has no component parallel to CD. Mathematically. this
can be stated as FN  ecD = 0. Expanding this. we get FNreCDx + FNyeC'Dy + FNeeC‘D: = 0 We now have four equations in our four unknowns. Substituting in the
numbers and solving. we get :3 to
u
L1 7.7 N. FM, = 38.9 N.
FNV = 36.1 N. and FN; = —~‘1.53 N. Problem 3.87 Cable AB is attached to the top of the
vertical 3m post. and its tension is 50 kN. What are the
tensions in cables A0. AC. and AD? Solution: Get the unit vectors pnmllel to the cables using the
coordinates of the end points. Expres the tensions in terms of
these unit vectors. and solve the equilibrium conditions The coor
dinates of points A. B. C. D. O are found from the problem sketch:
The coordinates ofthe points are A(6, ‘2, 0). 3(12, 3, 0). C(0. 8, 5).
D(0. 4, —5). 0(0. 0. 0). The vector locations of these pains one: I" =6i+2j+0k, r5=12i+3j+0k, rc=0i+85+5k,
ro=0i+4i—5k, ro=0i+0j+0k. The unit vector parallel tothe tension acting between the points A. B in the direction ofB isby deﬁnition
a” = &_
Ira  I'Al Perform this for each of the unit vectots a“; = +0.9864i + 0.164“ + 0k
eAc, = —0.6092i + 0.60% + 0.5077k
9,”; = —0.7442i + 0.2481j — 0.62021:
er = 0.94871 0.3162] + 0k The tensions in the cdtles are atplessed in terms of the unit vectots. TAB = ITABIGAB = 508,43. TAC = ITACIGACs TAD = ITADIBADe TAO = ITAoler
The equilibrium conditions are ZF=0= TAB+TAc+TAD+TAo =0
Substitutc and collect lilac terms. 2 F. = (0.9864(50) — 0.6092ITACI — 0.7422T,.D
—0.9<187]TA0)i = 0 Z P, = (0.1644(50) + 0.6092TAc + 0.2481TAD
—0.3162TA0)j = 0 2 F: = (+0.5077TAC — 0.6202T,.D)k = o. This set of simultaneous equnions in the unknown forces may be
solsed using any of several standard algorithms. The results are: iTAOi = 433 kN. ITACI = 6.8 IcN. ITAD = 5.5 m. ...
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This note was uploaded on 04/26/2009 for the course MAE 101 taught by Professor Orient during the Spring '08 term at UCLA.
 Spring '08
 ORIENT

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