HW4 - *Q5.]3 ta] larger: the tension in A must accelerate...

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Unformatted text preview: *Q5.]3 ta] larger: the tension in A must accelerate two blocks and not just one. (h) equal. WheneverA moves by 1 cm. B moves by 1 cm. The two blocks have equal speeds at every instant and have equal accelerations. (c) yes. backward. equal. The fa rce of cord B on block 1 is the tension in the cord. “35.20 ta] Smaller. Block 2 is not in free fall. but pulled backwani by string tension. [b] The same. 'Wheneyer one block moves by 1 cm. the other block moves by 1 cm. The blocks have equal speeds at every instant and have equal accelerations. (c; The same. The liglu string exerts forces equal in magnitude on both blocks—the tension in the string. QSJ [a] The force due to gravity of the earth pulling down on the hall—the iieaction foiice is the force due to gravity of the ball pulling up on the earth. The force of the hand pushing up on the ball—reaction force is ball pushing down on the hand. [b] The only force acting on the ball in flee-fall is the ga'avity due to the earth—the iieaction foiice is the gravity due to the ball pulling on the earth. P52 For the same force F. acting on dif'feient masses F = mat and F = men: . . ml a2 1 [fl] —= = — at: all 3'- [bl F: [ml + m2 la: 4mg: = m] (3.00 malls2 a: 0.?50 rolls2 _ HL'i.I"h.1£ P5.“ EF=mfi leads (—2.00? +2naj+ snoi— moi—45.03) N= m[:?-.T5 ruffle where :3 represents the direction of 5 I_42-Ul—l-00ll N= m .335 n'u'ls2 a . .lJ I Z T'= “id-21]]2 +l:l.DIC|:l2 N at tan'J below the —x axis 142.0 ,1 E i": 42.0 N at 131°: mllTS n'u'lszjfi For the vectors to be equal. their magnitudes and their diiections must be equal. (a) Therefore a is at 181° counterclockwise from the x axis 410 N (b) m=—...= 11.2 kg. 3.75 in,-"s' td} ¥f=irf+fii=o+[3.75 infirs: at 131°}naa s so :-_,.= 37.5 mm at 131° ¥f=315mfs cos131°l+3151nfs sin131°i so T'J,= l—fiTSE—Dfigfljl Lay's I? “l: on? +0.39? in..-'s= 37.5 mfs d): J? P‘SJfi z-‘T=—= 10:. vY=—-= 9:: ' dr - dr ml d" (1,: -*=m_av=i=13r dr - dr A1 := 3.00 5:. q, = [DID Inf-5'2. a.T = 36.0 111fo 2 F = max: 3.00 kglilDD mfg-J: 3-0.0 N E F; = may: mo kg|j3m [HI-"'51] = mg N '— 2}": Eur: +53 = 112 N P52“ From CqL1i1111J'iL1111 of the suck: I; = Fg I; J ,‘I From 21".: D for 1110 kuol: TJ £11191 +II"2 5.1119: = FL, [2] From EFT = D for the knot: 1‘] m9, = T_. we: ('3) Eliminate T3 = T; = 1‘] c0591 fem-93 and solve for T, T F: unsnfi‘z F: cosé‘z J_ [un cor-93 --co.-91-msz)_ .~"1I:I.[(-3'L +92) T3=Fs=fi ' ~ 5251? . T = F “'—1= 2% N J Jamar; - T1 =TJI couQ‘J 1: Egg NI muffin I: 163 N ' $05.93 J. aco~¢25fl°x FIG. P520 “135.23 P526 fr F\ \J l 491w tn] [fiolflle either nines T + mg= mt: = CI |T|= lmgl The scale J'Efllilr': the teneion T. no r = mg= 5.00 kgtgee mr'szj= th} The solution to part tn} is: also the solution to (b). FIG. P5233} and {b} to] [fiolflte the [:iulleg,r i3 + 2?, = o T==2|TJ|=3mg= d T7='+'_I+ "=0 l ’1 E “ mg FIG.P5.23{c} Take the component along the incline HJ + T"Jr + mgJr = Cl 5". f1 . ,r or \ {Till-5:300“ CI+T—mg:;in3C|.D°=Cl i I 5.00 9.30 I" II":mg:;iri3Cl.|3I°=fl:(—1| 3‘ \ mm \_ E 2 = .l'-. x "t FIG. P5234111} Firm. COD‘SldBJ' the block moving along the horizontal. The onl}r force in the direction of movement is T. Thus. Efi=ma r=['5 kale t1: Next con sider the block that! mover; vertically. The fo rcer; on it are the tension Tend itr; weight. 88.2 N. FIG. PSJE We have = ma 38.2 N—T=l[9 leg-hi (2; Note that both blocks must have the same magnitude of acceleration. Equations: [1] and [2] can be added to give 33.2 N=l14 kgln. Then a: 630111,."5‘ and T = 3-1.5 N P529 After it leaves: your hand. the block‘s: speed changes: onl}r because of one component of its weight: 2 F; = maJ — mg sin 30.0El Nit! 1! II ' 't'3='t'7+2a[L—x.l .I r ..I ‘; Taking I-‘r = D. I-‘é = 5.00 mfs. and a = —gi~"tn|:20.0°) gives D:(5.00)}—2('§|.E0}Is:ln[20.0°][:.rf—0] i 25.0 Jr=—_ = 3.?31n ' 2(9.30]s:1n(20.0°) P533 Finn. we will compute the needed accelerations: [III BefoJe it 5tm‘teto1nove: a} = 0 Elf—1"; 1.20 [2) During. the firr-‘t 0.30m; a}=—l——=$ .t 0.300 e =1.SD 111,55! [:3] 1l-‘tr'h'tle moving at constant velocity: 1:} = 0 Inc—"1'3- 0—120 tn.-".~; 1 1.50 5: 41300 nuts” [4] During the [net 1.50 e: a: = FIG. P533 Newton's second law in: EF, = mo} +S—{72.D kg']|{9.se 111M}: (no kgln} s: 706 N+If71n kgla, tn] When n} =0. 5:. {In} When a} = 1.50 111,552. 5:. to} When a} 0. 5:. to]; When a} 41300 mien. S=. P5. 39 m = 3.00 kg. 6': E'-0.0""~ x: 2.00 m. r: 1.50 5: . 3_ (in x— Ea: . 2m m= intuit: 532 2 4.00 -. Along}: 0— f +mg5i1130.0°= ma f= m['gs:l1130.0°—al] Along y: H+ 0— mgcos: 300°: 0 H = mg car; 500“ (b; I“. = i: W_ I“. = tw130.0°—;= 0.363 H mg cm; 30.0° gc0$300° (c1 f=m{gam3o.o°—a]. f:3.m(9.305jn30.0°—1.73]= 9.3-?N (d; If. = I-‘E +2a[:.r_|, —.r[. whale x]; — x,- = 2.00 m .-}=o+2(1.?8){200)=?-11 “1335’ VI = 1“Hill mzj's: = 1” FIG. P139 I-nl P542 Let a J'CPJ'CFCL'JI the pniitive magnitude of the acceleration —n:i of m]. et'tne neeeierntinn—ni 01' rug. and ot' the neeelei‘ntion +nj of m]. Cali i"J2 the tension in the left rope and I":J the Munich in the cord on the right. For ml. 2 F} = ma} +113 — mlg = —ml.r.' For m3. 2}}: ram: —]"Jz+luin+T33=—m_.a and 2 F} = we} n— mlg = D for 033. E F} = we} TH— 0333: +m3c.I we have three simultaneous equations i-‘l'J th} —Tl_. + .19.:- N = {4.00 kgln +11: — 0.350t930 NJ— T5 = {1.00 kg]a +TJJ—J90 N = (2.00 kgia Add them up: +392 N— 3.4?- N— [9.6 N = [$.00 legit: m. g a: 2.3] till-"5'. down for m1. left for m1. and up for ma Now —I,_, +39: M: [4.00 kg)[2.31 mix-3] ru= 30.0 N 0nd IFS—19.6 N = [2.00 kglpai mfsz) ' Tu =24:- N . P5.-1-2 103g H54 (M am ('31 [d] ‘i-Ve write EF‘. = mat for each object. 13 N—P=|{2 kgla' P—o=t3 kgia Q=Hkfla Adding gives 13 N= ['9 kgja so Q: 4 1(ng mfszl= 8.01] N net force on the 4 kg P— 8 N: 3 kgl:2 [HI-F52 = 6.03 N net force on the 3 kg aitd P= [4 N 13 N— 14- N= 2 kg(2 m..-"s3}|= 4.00 N net force on the 2 kg continued on next page From above. 42: and P = The 3—kg block models the heavy block of 1.v-rood. The contact force on your back is represented by Q. which is much less than the force F. The difference between F and Q is the net force causing acceleration of the S-kg pair of objects. The acceleration is real and nonzero. but lasts for so short a time that it never is associated with a large velocity. The frame of the building and your legs exert forces. small relative to the hannrter blow. to bring the partition. block. and you to rest again over a time large relative to the ham— irter blow. This problem lends itself to interesting lecture demonstrations. lCine person can hold a lead brick in one hand while anotherhits the brick with a hammer. ...
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HW4 - *Q5.]3 ta] larger: the tension in A must accelerate...

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