Home_work_set_04

# Home_work_set_04 - Problem 14—2 The crate of weight W has...

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Unformatted text preview: Problem 14—2 The crate of weight W has a velocity VA when it is at A. Determine its velocity after it slides down the plane to s = s'. The coefﬁcient of kinetic friction between the crate and the plane is ,uk. Given: W=201b a=3 ft : VA 2 12 — b 4 s s’ = 6 ft #k = .2 ‘7 ‘8 I XVII-r, ,1} Solution: \ Akit‘ a 0 = atan(;) NC = Wcos(6) F = ,ukNC m Guess v’ = 1— s 1 W 1 W ft Given — — vAz + Wsin(6)s'—Fs' = — — v’2 v’ = Find(v’) v' = 17.72— 2 g 2 g s _——————_—I——_n_———-———___—_ *Pr0blem 14-12 The force F, acting in a constant direction on the block of mass M, has a magnitude which varies with position x of the block. Determine the speed of the block after it slides a distance d ,. When x = 0, the block is moving to the right at vo‘The coefﬁcient of kinetic friction between the block and surface is ,uk Given: M=20kg c=3 d1 : 3 m d = 4 m N v0 = 2 — k = 50 — s 2 m ,uk 2 0.3 g = 9.81 — Solution: 62 “,2 Guess v1 = 2r—n- 5 Given 1 d I 1 c —Mv02+ —— kxzdx—ykMng —,uk kxzde—MVIZ 2 2 d2 0 2 d2 2 c + c + V] = Find(v1) v1 = 3.774 31- S *Problem 14-16 The block A of weight WA rests on a surface for which the coefﬁcient of kinetic friction is yk. Determine the distance the cylinder B of weight W3 must descend so that A has a speed VA starting from rest. ft R 1 I? v = — ___ g A S *‘x f \“ mgm AN ' Solution: W1“. L = SA + 233 Guesses d = 1 ft Given 2 1 2 VA WBd—ykWA2d= 2— WAvA + W3 7 g d = Find(d) d: 0.313 ft Problem 14-21 The crash cushion for a highway barrier consists of a nest of barrels ﬁlled with an impact—absorbing material. The barrier stopping force is measured versus the vehicle penetration into the barrier. Determine the distance a car having weight W will penetrate the barrier if it is originally traveling at speed v0 when it strikes the ﬁrst barrel. Units Used: kip = 103 1b Given: W = 4000 lb ft v0 = 55 — s - 32 2 ﬂ g — - 2 5 Solution: 2 g 1(W) 2 Area=— — v0 2 g — — v0 —Area=0 Harrier stopping fume [kiwi :- ﬁ 111% 35 In 25 Vehicle mnumtinn 1ft] Area = 187.888 kip- ft We must produce this much work with the barrels. Assume that 5 ft < x < 15 ft Area = (2 ft)(9 kip) + (3 ft)(18 kip) + (x — 5 ft)(27 kip) x _ Area -— 72 kip-ft — 27 kip + 5 ft x = 9.292 ft Check that the assumption is corrrect! Problem 14-23 The block of weight W is released from rest at A and slides down the smooth circular surface AB. It then continues to slide along the horizontal rough surface until it strikes the spring. Determine how far it compresses the spring before stopping. Given: W = 5 lb yk = 0.2 a = 3 ft :9 = 90 deg l b = 2 ft k = 40 —b ft Solution: Guess d : 1 ft Given 1 Wa—ykW(b+d)—§kd2=0 d = Find(d) d: 0.782 ft Problem 14—26 Cy1inder A has weight WA and block B has weight W8. Determine the distance A must descend from rest before it obtains speed VA. Also, what is the tension in the cord supporting block A? Neglect the mass of the cord and pulleys. Given: ft WA 2601b VA =8— 5 ft T“ WB=101b g=32.2—2 s Solution: L=2sA+sB 0=2vA+vB W System ’1 1 WA 1 W3 0 + WAd— WBZd = -[—j VA2 + —(—j (2VA)2 2 g 2 g WA + 4W3 2 _ VA 2g r d = d = 2.484 ft WA — 2W3 Block A alone 2 1 WA 2 WAVA 0+WAd—TdZE—VA T=WA— T=361b g *Problem 14—36 A block of weight W rests on the smooth semicylindrical surface. An elastic cord having a stiffness k is attached to the block at B and to the base of the semicylinder at point C. If the block is released from rest at A( 0: 0°), determine the unstretched length of the cord so the block begins to leave the semicylinder at the instant 6 = 6,. Neglect the size of the block. Given W = 2 lb lb k = 2 — ft 61 2 45 deg 4 a = 1.5 ft G g = £2 txxx a; s x 55:”. - . F. i f" Solution. 5 H «ex ft {\“x ? Guess 6 : 1 ft v1 = l — x j S “"ef Given N5 : U W v12 ‘/ Wsin(61) = / g a In! 1 1 W V]2 2 2 —k7za—6 ——k 7r—6 a—6 —Wasinl9 = —— 2< >2[( 1)] (ngz V1 . ft [ j = Find(v1,5) v1 = 5.843— a = 2.77 ft 5 s ——-—-——————-—_—_—_ Problem 14-75 The bob of the pendulum has a mass M and is released from rest when it is in the horizontal position shown. Determine its speed and the tension in the cord at the instant the bob passes through its lowest position. Given: M = 0.2 kg m r = 0.75 m g = 9.81 —2 S is Solution: Datum at initial position: T1+V1=T2+V2 1 0 + 0 = EMvg2 —Mgr v2=\/Er 2F" = Ma" T: 5.89 N Problem 14-77 The collar of weight W is released from rest at A and travels along the smooth guide. Determine its speed when its center reaches point C and the normal force it exerts on the rod at this point. The spring has an unstretched length L, and point C is located just before the end of the curved portion of the rod. Given: W = 5 lb L = 12 in W h = 10in lb k = 2 .— 1n — 32 2 ft g — - 2 5 Solution: TA+VA=Tc+VC 2 k k ft vC = 2gL + (—gjhz — (—th/Lz + h2 — L) vC = 12.556 — W W Narmtlﬁ] ital] 2 W vc kL NC = —[—] — L2 + h2 — L) NC = 18.919lb *Problem 14-80 The roller-coaster car has mass M, including its passenger, and starts from the top of the hill A with a speed VA. Determine the minimum height h of the hill crest so that the car travels around both inside loops without leaving the track. Neglect friction, the mass of the wheels, and the size of the car. What is the normal reaction on the car when the car is at B and when it is at C? Units Used: kN = 103 N Given: m M=800kg vA=0-— r3 = 10 m ft rC = 7 m g = 32.2 —2 5 Solution: Check the loop at B ﬁrst We require that N B = 0 V32 in _N _ M = _ — v : r V = — B g VB B \[g B B S l 2 1 2 TA+VA=TB+VB EMvA +Mgh=§MvB +Mg2rB V32 — VA2 h = —— + ZrB h = 25 m 2g Now check the loop at C 1 2 l 2 TA+VA=Tc+VC EMVA +Mgh=EMvC +Mg2rC vC = J vA2 + 2g(h — 2rC) vC = 14.694 3 S -NC - Mg VC "C NC: 2 VC rC —Mg NC: 16.825kN Since NC > 0 then the coaster successfully passes through 100p C. Problem 14-85 The bob of mass M of a pendulum is ﬁred from rest at position A. If the spring is compressed to a distance 5 and released, determine (a) its stiffness k so that the speed of the bob is zero when it reaches point B, where the radius of curvature is still r, and (b) the stiffness k so that when the bob reaches point C the tension in the cord is zero. Units Used: kN = 103 N Given: “ ﬂ f, m M = 0.75 kg g = 9.81 —2 s y, (3: 50mm r=0.6m i 3,, Solution: A AtB: ,V M 1 52 p33 4444 4 H —k = Mgr 3 2 ‘E 2Mgr kN k = k = 3.53 —— E 52 m z At C ‘ vcz —Mg=- 2— VC=V2g1 r l l M kN E1:52 = Mg3r+ —MvC2 k = —(6gr+ VCZ) k = 14.13— 2 62 m Problem 14-91 The ride at an amusement park consists of a gondola which is lifted to a height h at A. If it is released from rest and falls along the parabolic track, determine the speed at the instant y = 0'. Also determine the normal reaction of the tracks on the gondola at this instant. The gondola and passenger have a total weight W. Neglect the effects of friction. Given: W=5001b d=20ft h=120ft a=260ft Solution: 2 X ,v(x) = —- p(x) = a X y'(x) = 2- a (I + y'(x)2)3 y"(x) 6(x) = atan(y'(x)) Guesses x2 = 1 ft Given 1 Wk 2 — 2 x2 V2 g ft v2 =10: (W) 2 — v2 +Wd = Find(x2 , v2 , FN) FN H 2 y (x) = - a FN =11b d=y(x2) x2 = 72.1ft I," {f \- 1 FN — Wcos(6(x2)) 2 ft V2 = —80.2 : K g p(x2) FN = 952 lb *Problem 14-96 The double-spring bumper is used to stop the steel billet of weight W in the rolling mill. Determine the maximum deﬂection of the plate A caused by the billet if it strikes the plate with a speed v. Neglect the mass of the springs, rollers and the plates A and B. Given: lb W = 1500 lb k1 = 3000 — ft ft lb v = 8 — k2 = 4500 — 3 ft Solution: ka1 = k2x2 1 W 1 1 -— — v2 2 —k1x12 + —k2x22 2 g 2 2 2 1 W 2 1 2 1 km -- — v = -k1x1 + —k2 2 g 2 2 kg 2 2 (W) 2 k1 x1 2 — v = k1+ x1 g k2 x1 = 0.235 m ...
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Home_work_set_04 - Problem 14—2 The crate of weight W has...

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