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Unformatted text preview: Problem 2.58 The cables A, B, and 0 help support a
pillar that forms part of the supports of a structure The
magnitudes of the forces exerted by the cables are equal:
IFAi— — iFgl— — §Fg. The _magnjtude of the vectot: sum MM “yew of the three forces 13 200 kN What' IS FA? Solution: _ Use the angles and magnitudes to determine the vector
components, take the sum, and solve for the unknown. The angles
between each cable and the pillar are: Measure the angles counterclockwise form the maxis. The force vec
tors acting along the cables are: FA : IFAIGCOS 303.70 +jsin 303.70) = 0.554SIFAIl — 0.8319iFAij
: F3(ico:3323.1° +jsin323.1”) z 0.7997'F5Ei — 0.6004313}; L3
= FC(i cos 333.4o +jsin 333.41") = 0.8944IFgi—0.4472ingj The sum of the forces are, noting that each is equal in magnitude, is 2F: (2. 2489 FAEi — 1. 8795mm). The magnitude of the sum is given by the problem: 200 : in; (2.2439)2 + (1.8795)? x 2.931}FA!, from which IFAi = 68.24 kN ‘
i
l
i
w
l
w
L ‘ Problem 2.97 The 70m—tall tower is supported by
1 three cables that exert forces F A B, F Ag, and F A D on it.
The magnitude of each force is 2 kN. Express the total
force exerted on the tower by the three cables in terms
of scalar components. Solution: The coordinates of the points are A (0, 70, 0), B (40, The resultant force exerted on the tower by the cables is:
O, 0), C (—40. 0, 40) D (—60, 0, "60)
The position vectors corresponding to the cables are: x (750 , OJi + (o H 70)j + (—60 — 0)k
= —60i — 70k — 60k
= (—40 — 0)i + (0 — 70)j + (40 — 0)k
= —4Di — 70j +40k " = (40 —0)i+ (o — 70)j+ (n —0)k
:40i770j+0k _. Z 3 The unit vectors corresponding to these position vectors are: rAD —60_ 7o, 60 lgmpl 110 110 110
= “0.545511— 0.6364j — 0.5455k I'AC . 70, 40 “AC 21ml: R" gs” e : ~0.4444i — 0.7778j + 0.4444k rAB 40 , 70 , .
: = m _ — +0k=0.49631—0.8685' +Uk
"‘43 mg: 80.61 80.5J J —l——.l_’—‘ “AD The forces are: FAB : [FABIUAB = 0.9926i — 1.737j +014: FAG = fFAcuAC : —D.8888i i 1.5556j “1 0.8888
FAD = iFADillAD = —1.0910i— 1.2728j — 1.09101: Problem 2.122 In Problem 2.121, determine the vec—
tor component of T normal to the bar OD. Problem 2.123 The disk A is at the midpoint of the
sloped surface. The string from A to B exerts a 02
lb force F on the disk. If you resolve F into vector
components parallel and normal to the sloped surface,
what is the component normal to the surface? _.,_._ .1 4)
if: ‘15s} Moi (3:9,, Solution: Considera line on t e sloped surface from Aperpendic—
ular to the surface. (see the diagram above) By SIMILAR triangles we
see that one such vector is r N : Sj + 2k. Let us ﬁnd the component
ofF parallel to this line. The unit vector in the direction normaE to the surface is
my 8j + 2k _
e : —— : 1i 2 0.970 +0243}:
N Ier x/s? + 22 J The unit vector 6113 can be found by
(we , mli + (as  aAh’ + (213 — ZANI
(we — “)2 + (as *yaiz + (Zn  ZAP Point B is at (0, 6, 0} (ft) and A is at (5, I, 4) (ft).
Substituting, we get 3:113 = 9A3 : “0.5154” 0.515j e 0.492k
NOWF = iFieAB =(O2)3AB _
F : 70.123i “t“ 0.123.; — 0.0984k (lb) The component of F normal to the surface is the component parallel
to the unit vector eN. ’ ' FNORMAL : (F'QN)6N £(0.955)eN
FNQRMAL = 0i + 0.0927j + 652351511. Problem 2.124 In Problem 2.123, what is the vector
component of F parallel to the surface? Solution: From the solution to Problem 2.123,
F E —0.123i + 0.123j  0.0984k (lb) and
FNQRMAL : 0i + 0.0927j Jr 0.023211: (lb) The component parallel to the surface and the component normal to
the surface add to give F(F : FNORMAL + Fpmallel) Solution: From the solution of Problem 2.12:, T = an N, and the com
ponent of T parallel to bar CD is Tparanel : “7.52 N. The component of T
normal to bar CD is given by Triormal : V iTi T (Tperallel) : 494 N Thus Fparallei = F — FNORMAL Substituting, we get Fpmncl = 41.12311 + 0.0304j — 0.121614: lb Problem 2.141 The cable BC shown in Prob lem 2.140 exerts a BOOlb force F on the hook at B. (a) Determine rAB x F and rAC x F. (b) Use the deﬁnition of the cross product to explain
why the result of (a) are equal. Solution: (a) From Problem 2.140, the unit vector 9130 = 0i — 0.6j + 0.3k, and rAB = —12i+ 6j — 12k Thus F ﬂ FeBo E 0i — 180j + 240k, and the cross product is
i j k I'AB X F x 712 6 —12 m —7‘20i + 28305 + 2160k(ftlb)
0 —180 240 The vector TAG : (4 e 16)i+ 0j + (8 —12)k = —12i+ Oj — 4k.
Thus the cross product is
i j k
rag x F r #12 0 M4 = —720i + 2880j + 2160k (ﬂuib)
0 —180 240 (b) The deﬁnition of the cross product is r x F = ErHFlsinBe.
Since the two cross products above are equal, #343 IiFI sin 913 =
rA0§Fisin92a Note that rAc = FAR + 1‘30 from Proh—
leni 2.1l6. hence mm; X F m I'AB X F + r50 x F =
rABHFsin91e + chEFsin08 : IrABHFlsinﬂle, since
1'30 and F are parallel. Thus the two results are equal. H Problem 2.142 The rope AB exerts a SON force T on
the collar at A. Let rCA be the position vector from point
C to point A. Determine the cross product roA x T. Solution: The vector from O to D is r01) = (:L‘D — mcﬁ +
{3,113 — yc}j + (23 w 2:0)k. The magnitude ofthe vector Ircnl = V (we — $002 + (to) — yo)? + (213 — ZCP i The components of the unit vector along CD are given by 11.03,; =
i (mp —mc)/jrcpl,ucpy : (yp iyc)/IrgD,etc. Numerical . ;
1 values are [ram = 0.439 m, ucpx = 41.456, Hcpy = 41.684, . I
i and etcDz : 0.570. The coordinates of point A are given by wA : ! rec + iI'C'Ai'UCbma 3,1,; = my + rcAuch, etc. The coordinates
of point A are (0.309, 0.162, 0.114) in. The vector I‘CA is given by
ray; =(:1:A ~mg)i+(yA—yg)j+(zA—zc}k. ThevectorrCA is
rCA : (0.091)i + (70.137}j + (0.114)k m. The vector from A
to B and the corresponding unit vector are found in the same manner
as from C to D above. The results are ITABI = 0.458 m, MAE” =
—0.674, HA3!) : 0.735, and UABz : 0.079. The force T is given
‘ by T = iTiUAB The result is T = —33.7i + 36.7j + 3.931: N.
The cross product rCA x T can now be calculated.
i j k
I'CA X T m —0.091 —0.138 0.114
—33.7 35.7 3.93 : (W4.65)i + (»3.53)j + (—7.98)k Nm Problem 2.158 Suppose that the forces F A and F 3
shown in Problem 2.163 have the same magnitude and
FA .FB z 600 N2. What are FA and F3? Solution: From Problem 2.163, the forces are: The dot product: FA  F3 : FAfF3(U.6233) : 600 N2, from _ i FA =EFAE{icos40“ sin50“+jsin40“+kcos40“c0550°) W I lF i 600 3103N
A i B : W x . ’ n {F A!(0.58681 + 0.6428j + 0.4924k) 0.6233 FB = FB (—icos 60a sin 30° +jsin 60° + kcos 60“ cos 30°) and
: IFBIHJ 25i+0 Sﬁﬁj +0 433k) FA = 18.21i + 19.95j + 15.28k (N) i F5 : ”7.7m + 26.87j + 13.441: (N) H Problem 2.159 The rope CE exerts awspoﬁ forcejl‘
on the door ABC'D. Determine theé ctoi: componenft‘fw
of T in the direction parallel to the liné"f
point B. '“poihtit'tn (0,0.2.0) m ' (0.4,0.25,—0.I) to i i ‘ .. . A(0.5,0,0)m x
i " (0353,02) :‘n Solution: Two vectors are needed, to}; and I‘AB. The end
points of these vectors are given in the ﬁgure. Thus, rc E :
($13 — wgﬂ + (yg — ycﬁ + (21; —— zc)k and a similarform holds
for 1343. Calculating these vectors, we get 13;; = 0.4i + 0.05j — 0.11: m and 1343 : —0.15i+ Clj + 0.2k In. E (0’4’0'25’411) m The unit vector along CE is 80E = 0.963i + 0.12Gj — 0.241k C and the force T, is T : ITIBCE. Hence, T : 500(0.963i + (0910)!“ A (05 0mm 5 ,
0.120j —— 0.241k) = 482i + 60.23 — 120k N. The unit vector along ' ’ ' x . J i'
AB is given by mm a —O.6i + Oj + 0.8k and the component of  5
T parallel to AB is given by TAB i T I GAB‘ Thus, TAB : B
{482)(—0.6) + (60.2)(0) + (—120}(0.8) = —385.2 N (035,0,02) In Problem 2.160 in Problem 2.169, let r 30 be the po—
sition vector from point B to point 0. Determine the cross product r30 x ’1‘. i Solution: The vector from B to C is i
5 r30 = (30 W meli + (:90 e yell + (20 " ZB)k
z —o.35i + 0.2j w 0.2k m. The vector T is T : 482i + 60.2j — 1201: N. The cross product of
these vectors is given by i j k ' 1
mm x '1‘ : —o.35 0.2 “0.2 : ——12.oi m l38j m 117kNm i
482 60.2 —120 . Problem 3.35 The collar A slides on the smooth ver
tical bar. The masses mg : 20 kg and m3 = 10 kg,
When it : 0.1 In, the spring is unstretched. When the
system is in equilibrium, h = 0.3 In. Determine the
spring constant 14:. Solution: The triangles formed by the rope segments and the hor
izontal line level with A can be used to determine the lengths Lu and
L3. The equations are ‘ Ln = #0125)2 + (0.1}2 and L3 = #0125)? + (0.3)2. The stretch in the spring when in equilibrium is given by 5 = L 5 — Lu.
Carrying out the calculations, we get Lu, = 0.269 m, L5 2 0.391 m,
and 6 : 0.121 m. The angle, 9, between the rope at A and the
horizontal when the system is in equilibrium is given by tame =
0.3/0.25, or 3 = 50.2°. From the free body diagram for mass A, we
get two equilibrium equations. They are ZFI =—NA+Tcosﬁ':0 and EFF" : TsinﬁimAg:0. We have two equations in two unknowns and ban solve. We get N A =
163.5 N and T = 255.4 N. Now we go to the free body diagram for
B, where the equation of equilibrium is T — m B y — Fat? 2 O. This equation has onty one unknown. Solving, we get k: = 1297 Nlm Problem 3.48 The SOIb cylinder rests on two smooth surfaces. (3) Draw the freebody diagram of the cylinder. (b) If a : 30°, what are the magnitudes of the forces
exerted on the cylinder by the left and right surfaces? Solution: Isolate the cylinder. (a) The free body diagram of the
isolated cylinder is shown. (b) The forces acting are the weight and the
normai forces exerted by the surfaces. The angle between the normal
force on the right and the :1: axis is (90 + E). The normal force is NR : NR(i 003(90 + 19) +jsin(90 + ,6»
NR = NRI(—isin,6 +j (205,6).
The angle between the positive :1: axis and the left hand force is normal (90 7 or); the normal force is NI, : INLKisinoz +jcos CE). The
weight is W = 0i — L“?! j. The equilibrium conditions are 2F =W+NR+NL =0.
Substitute and collect like terms, ZFE = {—iNﬂgsmn+ INLslnoz)i = 0, SEE : (NRcosﬁ+ lelcoso:  W}j : 0. Solve: lNRl z (sing) lNLI:
em W sinﬁ )
sin(a + ,9) I and iNLi = ( For W = 50 lb, and 0: = 30°, [3 2 45°, the normal forces are NL =36.6lb, [NR 325.911: Problem 3.49 For the 50—11) cylinder in Problem 3.48,
obtain an equation for the force exerted on the cylinder
by the left surface in terms of the angle 0: in two ways:
(it) using a coordinate system with the y axis vertical,
(b) using a coordinate system with the y axis parallel to
the right surface. 1 Solution: The solution for Part (a) is given in Problem 3.48 (See
free body diagram).
\
i
i
l () <W—> Part (b): The isolated cylinder with the coordinate system is shown.
The angle between the right hand normal force and the positive :1: axis
is 180°. The normal force: N g = —NRi + DJ. The angle between
the left hand normal force and the positive a: is 180 w (or + ,8). The
normal force is NL : INL§(—icos{o: + [3) +jsi11(a + ,6». l _ , . . . Substitute and collect like terms,
The angle between the weight vector and the posttive 3 arms is 43. The weight vector is W = iW(i (308,3 — j sin 13). 2 Fa; = (—NR — lNLl cos(or + ,6) + lW] cos ,8)i 2 0, The equilibrium conditions are
Ea}, = (NL sin(a + ,9) — W sinﬁ)j : 0. ZF:W+NR+NL:O. Solve: INLi =( W sinﬁl )
sin(rx + ,6) Problem 3.56 The system is in equilibrium. What are
the coordinates of A? Solution: Determine from geometry the coordinates as, y. Isolate
the cable juncture A. Since the frictionless pulleys do not change the
magnitude of cable tension, and since each cable is anded with the
same weight, arbitrarily set this weight to unity, [W = 1. The angle
between the cabie AB and the positive :1: axis is 0:; the tension in AB is ITABl = iCOSCt +jsina. The angle between A0 and the positive .1: axis is (180°  It3); the
tension is TAG : ITacchosﬁ +jsin a), The weight is lW : Oi 7 ji. The equiiibrium conditions are 2F =TAB+TA0+WEQ Substitute and collect like terms, ZR; = (0056! — cosmi = 0, SF. = (sina+sinﬂ— 115:0. From the ﬁrst equation cos a = cos ,3. On the realistic assumption that both angles are in the same quadrant, then or = ,6. From the
second equation sin 0: = (é) or or : 30°. With the angles known, geometry can be used to determine the coordinates 2:, y. The origin of the :12, y coordinate system is at the pulley B, so that the coordinate . :1: of the point A is positive. Deﬁne the positive distance 5 as shown,
so that h+e
b—m Similarly, ( Reduce to obtain
:0 =b— hector —£cotcu. Substitute into the ﬁrst equation to obtain a: r (in) (1).. hector). Multiply this equation by tan a and use 5 = m tan 0: to obtain 5 = (toga) {hm heota}. The sign of the coordinate y is determined as follows: Since the co
ordinate m is positive, the condition (I) —— hoot. a) > U is required;
with this inequality satisﬁed (as it must be. or the problem is invalid),
e is also positive, as required. But the angle a: is in the ﬁrst quadrant,
so that the point A is below the pulley B. Thus 3; = —e and the
coordinates of the point A are: 1
m:(%)(behcotoz), y:—§(btana—h). 3:300 E
Z
i
.
l
l
i
E
l
: Problem 3.76 The cable AB keeps the 8kg collar
A in place on the smooth bar CD. The y axis points
upward. What is the tension in the cable? Solution: The coordinates of points 0' and D are C (0.4, 0.3, 0),
and D (0.2, 0, 0.25). The unit vector from 0 toward D is given by 9019 = 9013314 BCDyj + eaDzk 2 —U.4561 m 0.68% + 0.570k. The location of point A is given by an; = mg + dCAeCDzs with
similar equations for yA and ZA From the ﬁgure, dCA = 0.2 m.
From this, we ﬁnd the coordinates of A are A (0.309, 0.162, 0.114).
From the ﬁgure, the coordinates of B are B (0, 0.5, 0.15). The unit
vector from A toward B is then given by 3A3 = eABmi + BAByj + 614sz : 70.57414» 0.735j + 0.079k.
The tension force in the cable can now be written as
TAB : —U.574TABi + 0.735TABj + 0.079TABR. From the free bod)r diagram, the equilibrium equations are: FNa: + TABBABm = 0, FNy + TABBABy " my : 0.
and FNz +TAB‘3AB2 = 0.
We have three equation 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 F N  60D x 0. Expanding this, we get FNweCDm + FNyecny + FNzeCDz E 0 We now have four equations in our four unknowns. Substituting in the
numbers and solving, we get ‘ ’B’S'Tr'v' To Swim Fm. = 38.9 N.
FM : 36.1 N, and FNZ : —4.53 N, Problem 3.77 In Problem 3.76, determine the magni
tude of the normal force exerted on the collar A by the
smooth bar. 80111151011: The solution to Problem 3.76 above provides the mag
nitudes of the components of the normal force exerted on the collar
at A. IFNi : V (1?pr + (FNy)2 ‘i" (FNz)2o Substituting in the values found in Problem 3.77, we get gFN: = 53.2 N. Problem 3.84 The mass of the suspended object A is
111,; and the masses of the pulleys are negligible. De
termine the force T necessary for the system to be in
equilibrium. Solution: Break the system into four free body diagrams as
shown. Carefully label the forces to ensure that the tension in any
single cord is uniform. The equations of equilibrium for the four ob
jects, starting with the leftmost pulley and moving clockwise, are: 373T=0, RA3S=0, F—SRrO,
and 2T+2S+2R—mAg:0. We want to eliminate S, R, and F from our result and ﬁnd T in terms
of WA and 9. From the ﬁrst two equations, we get S = ST, and
R : BS 2 91". Substituting these into the last equilibrium equation
results in 2T «1— 2(3T} + 2(9T) = mAg. Solving, we get T : mAg/26 . .9 Note: We did not have to solve for F to ﬁnd the appropriate value of T. The
ﬁnal equation would give us the value of F in terms of mm and 9. We would get
F = 27mA 9/26. If we then drew a free body diagram of the entire assembly,
the equation of equilibrium would be F  T — m A g = 0. Substituting in the
known values for T and I", we see that this equation is also satisﬁed. Checking
the equilibrium solution by using the “extra" free body diagram is often a good
procedure. ...
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 Fall '08
 CHATTERJEE

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