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Unformatted text preview: Problem 3.35 The collar :1 slides on [he smnolh ver—
lical bar. The masses mg = 2U kg and mg = ll]I kg.
When h. = [Ll In. [he spring is unslrclehcd. When [he
syslem is in equilibrium. l1 = (1.3 In. Delerlnine the
spring eonslanl 3:. Solution: The triangles formed by the rope segments and the hor
izontal line level with A can he used to determine the lengths Lu and
L3. The equations are Lu = @125)? + (0.1)2 and L3 = V"(tlasn +013)? The stretch in the spring when in equilibrium is given by (5 = L3 —Lu.
Carrying out the calculations, we get Lu = 0.269111, L3 = H.391 tn,
and 6 2 [L121 in. The angle, 6, between the rope at A and the
horizontal when the system is in equilibrium is given by tanﬂ =
03,5025, or E? = 50.2”. From the free body diagram for mass A, we
get two equilibrium equations. They are E Fr 2 —_'\"_4 + Tens 6' = CI
and 21:13. 2 TsinEJ — may 2 (I.
We have two equations in two unknowns and can solve. We get NA = 163.5 N and T = 255.4 N. Now we go to the free body diagram for
B, where the equation ofequilibriuni is T — mBg — k6 = D. This equation has only one unknown. Solving. we get l: = 129? Ni'tn r: l; W mBg 5P3: “rind3: 3” HQ)
“T i'.\ 3.:
:fa' 3' 9 W "— Ca)
, 6,, CA W C,
i Mp. " W 3 T Q (‘3) FED “23 1101‘:
1’. ‘ k
was«5 L3) 2:; [.23 m 14.41“ mm?» 8% “L . k ( f
N f' l "—0 f. {Ed‘1":
‘eﬁ ﬁrm‘en (53> .' . g3 Tlﬂéhﬁ z A if)
2. Fm 1 She:— mjm9:.§_ pz—xf—J S g a
73 (13:33 ‘ g k :3 h+b3 253 :5 ht]:
S 2. " g M” Cw W also 133 okrch ._._ ‘ . Q2 a (NEW 02 $ “"l mg
L; J Lu. Ar 0‘2 c h?— M32
1—H” 2 2
S PL
h ;
'2— __ L : SL53— _—_) K
1:3, 3 L x 3 am a 1: jé’VLZIS}O .3 {é 2.L far (1 "I33 131‘.) “rep—O ﬁﬂi?b‘ﬁuw bqﬁdi'qw :6 S) 2L I‘I'uhlem 3. ES plat The Eﬁ bar is supported by a ball and socket support at A. the cable ED. and a
rol]et support at C. The force F : ij — {Hills Uh]. 1What is the largest value of F3 fat which the tc]ler
support at t." wi]l remain on the ﬂoor? Sﬂllltiﬂll: The strategyr is to determine the sum of the moments
about A, which will involve the unknown reactions at B and (3'. This will require the tmit vectors parallel to the rod and parallel to the cable.
The angle formed by the rod is 3 . 1 . . o
 — — I
Ct 8111 ( ) The vector positions are:
121 = 3L
I'D = Eli + 2k
and r3 = [:8 cos 22°:Ji = idlﬁﬂi. The vector parallel to the rod is
11.“: = 15“ — rs; = T.z1_1I32i— 3j.
The unit vector parallel to the rod is
61.1.3 = ﬂﬂﬂ’fﬂi — H.3T5j. The location of B is 11.13 = aleAg = 3.7D81i — 1.5j.
The vector parallel to the cable is
I'BD = I‘D — lira1 + rag} = 0.2919i — 1.5j + 2k.
The unit vector parallel to the cable is 65;: = Ill lﬁﬂ'i — ﬁﬁﬂﬁﬂj + D.T94I3k. The tension in the cable is T = TeBD. The reaction at the roller
support (3' is normal to the m — a plane. The sum of the moments about A i j k
2 :MA = 3:031 _1.5 n
—5£+ I '3 FY
i j k
+T ‘ 3.7031 _ .5 o ‘
D.ll'3l:l —':'.59l3D 0.7946 
i j k T4162 —3 D‘=IJ
 D Cy HI = 75i + 135.4j + 3.?D81Fyk
+T [—1.192i — 2.94am — 2.13313kj + +T.:LlI32C'yk = D, from which, T = 13552 = 132.92 lb. Collecting terms in k, 3.7D81Fy + 2.334lTl — T.4162C—'Y = CI. FOI'C'y = 1), FY = % = 34.54 lb Problem 6.95 Th: dimmaion: an: a = QED mm. b =
300mm_c=2::mm..d=150mm.e=3[lﬂrnm.
and I = EEEI mm. Tut, ground cantata a mical form
F = TIIIEI N on Thu shovel. Th: mass of Th: shovel is
9'3 kg and it: wcjght acts. at £3. Th: weights. of th: links .43 and AD an: nagligiblc. Datummint, 1J1: hnriznntal
fare: P calmturd m A by tl'u: hydraulic piston and Th: Tacticus. an tl'u: EJ'JCH‘Cl HI C“. SDIIJHDII: The. baabody diagmrn aftharhmli: from which we
ablaiu the:quaﬁon: Es", =CJ—chud=2. (1:.
if}.I = I3“, +Tucind+F — mg: III.ij
E HMS? = 31" — emg + [a — chaLnﬁ'
+dT cmﬁ = n. {a}
Theangled = manna — :11:be me theftuarbDdy diagram cnfjociutﬁ. F 1' winobtain Ibe equation
EF =p+TEDIIIE= UH] Substituting H3: Ei'mu inﬁrmaﬁon into Eq:. [Ll—[4] and halving. we.
abhiu T =—19.EEIIN.
P =15.I!EIN.
L7: = —lﬂ.IIEIIIN. and I?” = 513 N. Show] ...
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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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