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Unformatted text preview: P ROBLEM 4.89 IEGM For the frame and loading shown, determine the reaetions at C and D. SOLUTION Since member BB is acted upon by two forces, B and 1), they must be colinear, have the same magnitude, and
he opposite in direction for HD to be in equilibrium. The force 1?. acting at B of member ABC will be eqml in
magnitude but opposite in directianto force 3 acting on member BD. Member ABC is a threeforce body with
member forces intersecting at E. The ﬁb‘dfs ofmembers ABC and ED illusu'ate the above conditions. The force triangle for memberABCis also shown. The angJes a and ,8 are found from the member dimensions: a = tan—{m} = 215.565"
1.0111
= _1{1.5_m) = 5.5 3100
’6 \ 1.0m '
Applying the law of Sines to the force triangle for member ABC,
150N _ r: B
sinU? — a} sin{90° + a) sinl:90°  ,8)
at ﬁlm _ C _ 3
511129345" 5111116565" — sin33.690°
ISON ‘ . °
=( _}sm116 565 = 270.42}:
sm29.?45° and 0:3 150N}5In33.690 =IG?.TD4N sin 29345 ° or C = ZTON in 56.3“ '1 01' D =167JN 1: 26.6“‘ \._/I PROBLEM 4.70
For the flame and loading shown. dﬁ‘lc‘l'l'nine the reactions aid and C. Since member A3 is acted upon by two fDI'GES, A and B. the).r must be colineat, have the same magnjmde, and
be opposite in direction for AB to be in equilibrimn. The force B acting at B of member BGD will be equal in
magnitude but opposite in direction to force B acting on member AB. Member BCD is a three—fame body with
member forces kiwisearing at E. The ﬁb.d.’s of members AB and BCD illustrate the above conditions. The
force mangle for member BCD is also shnwn‘ The angle ,6 is found from the member dimensions: 60m ,6=ta'm"[1 00m J = 30964“ Applying of the law of Sines to the force triangle for member 3CD, 130N _ 3 _ C
sin[45°ﬂ) '— sinﬂ _ sinl35° 130N B C
or *._=—= ginl4ﬂ36° sin30.964° sin135° ' O
. A = B = (130 N)sm30.964
sinI4.036° = 275.78 N or A =ZTGN T 45.0% 130 N)sin135° and C = ( ‘ = 379.0214
511114.036“ or C=379N h. 59.0°§ PROBLEM 4.73. A ZOOlb cratcjs attached to the trolleybeam system shown. Knowing
that (1 =15 ﬂ. determine Ia) the tmsion in cable CD, (:5) the reaction
at B . .Frmn gwmety of forms _ —I J’ag
— tan
’6 [1.5 11]
ya: = 2'0 — yDE
= 2.0— l.5mn35°
= 0.94969 11
,3 = tan—(034969] = 32.339°
1.5
and a = 90°  ,6 = 90°— 32.339D = 57.66? 3 = ,8 + 35° = 32.339° + 35° =6_T.339° Applying the: law of aims to the force triangle. Zﬂﬂlb_ :— _ a
sing!P sine sinSS”
or (mills) _ ' T _ B '_
sin67.339° sinS'fJSﬁl" sin55°=
' ' {2001b}(sm57.551° .
a r: =183.1161b
0. 513167.339” .. '
or T =Isa.1rb «I
(b) 3= [20pr si.n.55°}=1?15361b
sin61339° or B = 1715 lb : 32.3”4 \.._/' ~_/ SOLUTION PROBLEM 4.38 A slender rod of laugh I, = mummis held in equilibrium as shown,
with (me and against a frictionless wall and tbs 0111:1 md attached to a
QOI‘doflcngih S = 300m Knowing thatthe mass ufthe rod is 1.5 kg, Hetermine (a) [he distance I}, (b) the tension in the cord. (c) the reaction
at B. .1...— From the f.b.d afﬁne threeforce member AB, forces must intersect at D.
Since the force Tintersects point D, directly above G. J’ss = h
For ml: ACE:
3’ 2(a): +011): (1)
For triangle ABE:
L2 = (AEY‘ +01)2 (2)
Subtracting Equation (2} from Equation (1)
$2  L2 = 3&2 I2_ 2
orh=1lS L
3 (a) ForL=200mm and S=3ﬁﬂmm
 2 2 . '
h =J 300 g 200) =129.099mm or £1 =129.11111111 (in) Have W = mg =[1.5kg)(9.81m22)=14.715 N 111111 a = sin'[E’i] : sin" __2(129.o99)
s 300
6 = 59.3910
From the force lrianglc
I = i _ ﬂ =110973N sins ' 511159.391“
or T = ”JUN“ (c) g=i=m=3J055N
tanéi Lan59.391° or B =8.?1N4—‘ ...
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 Spring '08
 Chang

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