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Dynamics_CH13_HW

# Dynamics_CH13_HW - *13—4 The van is traveling at 20 km/h...

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Unformatted text preview: *13—4. The van is traveling at 20 km/h when the 20 km/h coupling of the trailer at A fails. 1f the trailer has a mass of 250 kg and coasts 45 m before coming to rest, determine the constant horizontal force F created by rolling friction which causes the tiailer to stop. 20(10’) km/h = = .55 . 20 3600 5 6 m/9 25009.81) N 25(l(0.3429) (*1) 02 = “3 + 211.15 — No) i Q 0 = 5.5562 + 2(1))(45 — 0) F I a = —0.3429 m/s2 = 0.3429 m/sZ ~> N i, g F. = mar; F = 250(03429) = 85.7 N Ans 13-5. A block having a mass of 2 kg is placed on a spring scale located in an elevator that is moving down— ward. If the scale reading, which measures the force in the spring, is 20 N, determine the acceleration of the elevator. Neglect the mass of the scale. t2 Ft = my: 20 — 2(9.81) = 2a 11 = 0.19 m/s2 T The elevator is slowing down. The baggage truck A has a mass of 800 kg and is LEO pull the two cars, each with mass 300 kg. If the 3V6 force F on the truck is F = 480 N, determine the all acceleration of the truck. What is the acceleration E truck if the coupling at C suddenly fails? The car ‘are free to roll. Neglect the mass of the wheels. r = ""0: 480 = {800 + 2(300)lu n = 0.3429 = 0.343 mm? = mm; 480 : (800 + 300)“ a = 0.436 m/s2 13-18. The man pushes on the 60-lb crate with a force F. The force is always directed down at 30° from the h0rizontal as shown, and its magnitude is increased until the crate begins to slide. Determine the crate’s initial acceleration if the static coefficient of friction is p.“ = 0.6 and the kinetic coefficient of friction is M = 03. Force to produce motion : .3 2F. =0; Fc0530°-O.6N=0 +TZF,=0; N—60—Fsin30°=0 N=9L80lb F=63.60lb SinceN=9l.80 lb. .3 U, = M: 6160 cos30° —0.3(91.30) = (ﬁgﬂa a = l4. 8 fI./s2 Ans 13-19. A 40-lb suitcase slides from rest 20 ft down the smooth ramp. Determine the point where it strikes the ground at C. How long does it take to go from A to C? +\.):F;=ma,; 405in30°=§42254 a = 151 M’ (A) v2 = v: + 2041-10): v; = o + 2(16.l)(20) v, = 15.38 fl/s (v\.) v = v0 + a, I: 3.38 0 + 16.11“ I“ = 1.576 s (10:. = (3.)9 + (V.)o I R = o + 25.38 cos30° (Inc) I z (+ i) .r, = (1,)0 + (we H gaz’ 4 = o + 25.38 amen” + 2132.30.02 I": = 0.2413 s R = 5.30 n An: Tom m = I" + I" = 1.82 8 Ann t 2. *13—24. At a given instant the lO—lb block A is moving downward with a speed of 6 ft/s. Determine its speed 2 5 later. Block B has a weight of 4 lb, and the coefficient of kinetic friction between it and the horizontal plane is M = 0.2. Neglect the mass of the pulleys and cord. 10 +lﬂ;=ma,: 10~2T=mﬂn 24. = 4- Solving: T = 3.38 lb ,1, = 10.403 {W a, = 401mm2 (+l) VA = (v.00 + “A‘ v4 = a + 10.4039) = 26.8 ftls fblock A so that the 5—kg block B e in I: 2 S. 13-25. Determine the required mass 0 when it is released from rest it moves 0.75 m up along the smooth inclined plan Neglect the mass of the pulleys and cords. 1 Kinematic: Applying equation 5 = so + 110 !+ Eartz, we have I ( +) 0.75=0+O+Ea,,(22) a, =0.375 mls2 Establish the position - coordinate equation, we have 2sA+(:A ~sB)=l 35A —:B=l T Taking time derivative twice yields 30A —a9 =0 Front qul], 312A —o.375 = o a, = 0.125 m/s2 Equation of Motion : The tension T developed in the cord is the same throughout the entire cord since the cord passes over the smooth pulleys. 08/ rm From FBD(b) , +25. =may; T—5(9.81)sir160°=5(0.375) T Ewe/M T=44.35 N From FBD(a), 3(4435) —9.SlmA = "1‘ (—0.125) mA =13.7kg Ans +T2ﬁ=ma,; 128 13-37. The conveyor belt is moving at 4 m/s. If the coefﬁcient of static friction between the conveyor and the 10—kg package B is p3 = 0.2, determine the shortest time the belt can stop so that the package does not slide on the belt. 1.1:): = mg; 0.2(934) = 10a ‘8“ 1 a = 1.962 / “‘5 Thu-mm) #981 (1v)v=vo+nkl 4 = 0 +1362: 1 = 204 s Ans 13-38. The 2-lb collar C fits loosely on the smooth shaft. [f the spring is unstretched when 5 = 0 and the collar is given a velocity of 15 ft/s, determine the velocity of the collar when 5 = l fL Ans 133:;m32tejzﬂzﬁyézr belt delivers each 12-kg crate to Ed down do it the crate 5 speed is A = 2.5 m/s, who“ between 2g th e ramp. If the coeffrcrent of kinetic inc th ac crate and the ramp is pk = 0.3, e speed at which each crate slides off the am a’ - tB. Assume that no tipping Occurs. lake 8 = 30°. Ag}; =ma’; Nc — 12(9.81)m530° = NC = 101.95 N 11mm Bc' mm; = n.1,; 12(9.81)stn30° _ 0.3(10135) = 12.,c tum \ ac = 2.356 mzs‘ M/ \°= 6,» v, = v: + 2&0" — J") (2.5)1 + 2(2.356)(3—0) v, = 4.5152 = 452 111/: An: 135 13-42. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not move relative to B. All surfaces are smooth. Require “A = a5 = a BlockA f 4.ng =0; NcosB-mg=0 (.25:ma,; NsinB=ma I N a=gmn9 Block B 1 LEE =ma,; P-Nsin9=ma P-mg tan9=mg m9 P : 2mg tans Ans 13-43. Blocks A and B each have a mass m. Determine the largest horizontal force P which can be applied to B so that A will not slip up B. The coefﬁcient of static friction between A and B is ,u.,.. Neglect any friction between B and C. “m J A. 4—— ﬁt] NcosG—p,Nsin6—mg=0 \ 9 N N5m9+p,NcosB=ma N: "'3 , cosB—p,sm9 (sin9+p,cose) a =5 cosB-#,sir19 P‘#,NC059~Nsin9=ma P _mg(si.n9 +u,c039) = mg£sin€+mcosej cosB—u,sin9 cosB-uﬁme P = 2mg(sin9+y,coso A c059 -;1,sin8 m 137 13-54. Using the data in Prob. 13-53, determine the minimum speed at which the car can travel around the track without sliding down the slope. + T 2F, = 0; Ncos20° +0i ZNSin 20° —l700(9.81)= 0 N=16543.1N {— 2F, = man: 16543715in20° —0,2(16543.1)c0520° =1700(‘1’3"03) um... = [2.2 m/s Am 13-55. A girl, having a mass of 15 kg, sits motionless relative to the surface of a horizontal platform at a distance of r = 5 m from the platform‘s center. If the angular motion of the platform is slowly increased so that the girl’s tangential component of acceleration can be neglected, determine the maximum speed which the girl will have before she begins to slip off the platform. The coefﬁcient of static friction between the girl and the platform is u = 0.2. t: Equation of Motion : Since the girl is on the verge ofslipping, 1-} = AN =0.2N. Applying E2113 - 8, we have 25:0; N—15(9i81)=0 N= l47.15N . 01 LE, =ma,.; 0.2(l47.15) = l’S(—5—) v=3.l3 mls ' Ans lAA —I—————~ 13-58. Prove that if the block is released from rest at point B of a smooth path of arbitrary shape, the speed it attains when it reaches point A is equal to the speed it attains_when it falls freely through a distance [1; Le, v = VZgh. riZF. =Iman mgsinﬂ =ma, a, =gsin6 M udu=a. dx=gsin8¢tr However dyadrsino 9 " J:udu=t~gdy N \ g- ,. t 2 —s u: 23;. (2.5.0. 0 13-59. At the instant 6 = 60°, the boy‘s center of mass I: G has a downward speed 115 = 15 ft/s. Determine the rate of increase in his speed and the tension in each of lo ﬂ 5' the two supporting cords of the swing at this instant. The _~ boy has a weight of 60 lb. Neglect his size and the mass .~ of the seat and cords. 60 .\t.ﬁ=ma,; 60ms60°=ﬁ4 “i=‘5-l N5: 1"“ 60 151 _ - _ ' ° s -— —— T: 46.9 lb A /'+}:ﬁ—ma,,. 21' 60511160 322(10) ” *13-60. At the instant 6 = 60°, the boy‘s center of mass G is momentarily at rest. Determine his speed and the tension in each of the two supporting cords of the swing when 6 = 90°. The boy has a weight of 60 lb. Neglect his size and the mass of the seat and cords, 1 "DE "-4.: 27—603in9=ﬂ 1’. m 322 m vdv=ads howevcrds=|odg 10"v dv = :g’fszzmada v = 9.289 ﬁ/s From .[1 2T— ‘ :32 93392 E“ I 60"“9‘” 32.2 10 T=38.01b Ans #13410. The block has a weight of 2 lb and it is free to move along the smooth slot in the rotating disk. The spring has a stiffness of 2.5 lb /ft and an unstretched length of 1.25 ft. Determine the force of the spring on the block and the tangential component of force which the slot exerts on the side of the block, when the block is at rest with respect to the disk and is traveling with a constant speed of 12 ft/s. I: = ks: I; = 2.5(p—l.25) 2.5(322)(p1 - 1.25;» = 288 p2 — 1.25;: —3.58 = 0 Choosing the positive root. p=162tt F; = 2.5(2.62- 1.25) = 3.421!) 13-81. If the bicycle and rider have a total weight of 180 lb. determine the resultant normal force acting on the bicycle when it is at point A while it is freely coasting at 124 = 6ft/s. Also. compute the increase in the bicyclist's speed at this point. Neglect the resistance due to the wind . and the size of the bicycle and rider. ’ g ZONE" 9 = u"(—222|) = ~6S.76° d‘y = 1‘“ t )I 0m __ __ _. a — 'kz 20 m‘ "a n ' (1<l'(~2.221)7]"1 p ——.—————03‘89 — 41.43 f1 [80 +£F=m ; 130'65.76'=-— v \. I “I "" 32.2“ a, = 29.4ft131 All: 1 [4,”; = "10.; ISOMGSJQS" — N = ELF-6") 32.2 41.43 N = 69.0 lb All! -—‘) l e- ) —- ~0 cos 20.: All: All AF; 3“ Fr 18011» 64776" 13-87. The Z—kg rod AB moves up and down as its end slides on the smooth contoured surface of the cam, where r = 0.1 m and z = (0.02 sin 26) m. lfthe cam is rotating at a constant rate of 5 rad/s, determine the maximum and minimum force the cam exerts on the rod. Kinematic: Taking the required time derivatives. we have 2 = 0.02 sin 20 t9 = 5 rad/s (i). = 0 .: = 0.02 sin 29 i = 0.04 cos 2919 = 0.04(Cos 2967 — Zsin 26512) Thus, a; : = 0.04IC0529(O) - 2 sin 2961)] = —2 sin 29 Al 9 = 45°. a. = —2 sin 90° = —2 m/s2 At 9 = 45“. a. = —2sin(—90c) = 2 m/s2 Equation of Motion: AI 9 = 45°. applying Eq. [39, we have E F: : mug; (szmin —- 2(9.8l) : 2(~2) (F:)min : N At 6 = —45°, we have 2 F; 2 17111:; (FJnm —» 20181) = 2(2) (12)"m = 23.6 N *13—88. The boy of mass 40 kg is sliding down the spiral slide at a constant speed such that his position, measured from the top of the chute, has components r =2 1.5 m, 6 = (0.7!) rad, and z = (—0.5!) m, where t is in seconds. Determine the components of force F,, F9, and F; which the slide exerts on him at the instant I = 2 s. Neglect the size of the boy. r=15 20.7/ 22—05! ::r=() 9:0.7 12—05 6:0 2:0 40(9.81)N (1,. 2 F — as)? = 0 - i.5(0.7)2 = —0.735 .18 <1¢,:ré‘l+2H)=O r F3 9 a; = 'z' = O E F,. = mu,.‘, F, = vim—0.735; : —29.4 N Ans E F” = mm); ‘,, : 0 Ans 2F; =maz'. I"; -40(9.8l) :0 F. = 392 N Ans 1£n 13-109. The collar, which has a weight of 3 lb, slides along the smooth rod lying in the horizontal plane and having the shape of a parabola r = 4/(1 — cos 0), where 0 is in radians and r is in feet. If the collar's angular rate r; 4 is constant and equals 0 = 4 rad/s, determine the 1-”9 tangential retarding force P needed to cause the motion . 4mg 5 and the normal force that the collar exerts on the rod at ’ ' (l—cmﬂ)’ > __ 0 the 1nstant0 — 90 . 4mg 5 + 400.8(8): 13:4, r=4 k: —16 'r'=1zs a, - r'- as)“ = [ZS-4(4):- 54 .,=ré+2'n‘i-0+2(-16)(4)= —128 4 l—coss r- dr —4|in8 n ' (hm-ma)2 4 I--u a." (I—u-I)I I - In- 4 s—n—l r “ya—- (ﬁ) y - —45- - 135° 3 P Iin45°—Ncol45' = —-(64) +112; sum: 32.2 . . 3 9—D; : mg; -Pcos‘5' — N "1145‘ = 3—u(‘17-3) Solving, P = 12.6 lb N: 4.22 lb w‘___________________.___——— 13-110. The pilot of an airplane executes a vertical loop A which in part follows the path of a cardioid, r = 600(1 + M cos 0) ft, where 0 is in radians. If his speed at A (0 = 0°) _, is a constant up = 80 ft/s, determine the vertical force the V, ' If belt of his seat must exert on him to hold him to his seat 3 / when.the plane is upside down at A. He weighs 150 lb. 4/ I H605“ “2059) n See hint related to Prob. 13—108. _/ / 7:600“ +c059)ls=o= =1200fl r r = —6005m99i9=o~ = 0 ,_ z 1 r = -6005m88-600c0588 9:0: = -6009 it; = r2 +(r9)2 9/ 0 5/ Z , (30)z = 0 {12009) 9 = 0.06667 h \, f 2vP VP 2 2r? + 2(r8)(r9 + r9) 6 0=0+0+2rleb 9:0 2 a, = r - r6 = —600(0.06667)2 —1200(0t06667)2 = —8 {1/52 A WW gum“... 4.“... ag=r9+2r9=0+0=0 +T):F,=ma,; N—150=(3125—(:)(—8) N=113lb Ans ...
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