mid-term-sol

mid-term-sol - Problem 3.35 The collar :1 slides on [he...

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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 TsinE-J — 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: “rind-3: 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 25-3 :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’VLZI-S}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 E-fi 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]. 1|What is the largest value of F3 fat which the tc]ler support at t." wi]l remain on the floor? Sflllltifl-ll: 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.“: = 1-5“ — rs; = T.z1_-1I|32i— 3j. The unit vector parallel to the rod is 61.1.3 = flflfl’ffli — H.3T5j. The location of B is 11.13 = ale-Ag = 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 6-5;: = Ill lfifl'i — fififlfiflj + D.T94I3k. The tension in the cable is T = |T|eBD. 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 —'|:|'.59l3-D 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.:LlI|32C'y-k = 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. Datum-mint, 1J1: hnriznntal fare: P calm-turd m A by tl'u: hydraulic piston and Th: Tactic-us. an tl'u: EJ'JCH-‘C-l HI C“. SDIIJHDII: The. baa-body 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} The-angled = manna — :11:be me theftuarb-Ddy diagram cnfjociutfi. F 1' win-obtain Ibe- equation EF =p+TEDIIIE= UH] Substituting H3:- Ei'mu infirmafion into Eq:. [Ll—[4] and halving. we. abhiu T =—19.EE|IIN. 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.

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mid-term-sol - Problem 3.35 The collar :1 slides on [he...

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