Dynamics hw #5 - 14-45. An automobile having a mass of 2 Mg...

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Unformatted text preview: 14-45. An automobile having a mass of 2 Mg travels up 3 7° slope at a constant speed of v = 100 km/h. If mechanical friction and wind resistance are neglected, determine the power developed by the engine if the auto— mobile has an efficiency 6 = 0.65. Equation of Motion: The force F which is required to maintain the car’s constant speed up the slope must be determined first. ' + Z Er.- = max: F — 2(103)(9.81)sin 7° 2 2(103)(0) F :2 2391.08 N ' . l00(10") m ( l h P : ‘ e d 1h .‘ *: ————- = ower Here,l.he spec of eczema, [ h I x 3 5 27.78 m/s. The power output can be obtained using Eq. l4-l0. P = F - v :. 239] .08(27.78) : 66.418(103) W = 66.4]8 kW Using Eq. 14-l l, the required power input from the engine to provide the above power output is power output power input: a w=102kw Ans 0.65 14-50. Acar .__has a mass m and acceleratcs' along 'a horizontal _s__t'raight_road from rest Suchfthat'the 'pQWer is always 'a 'cgnstaht'tamount P. Determine how far itm-USt travel to’ reaChh speed-ofv. Power : Sine: the: power ouput is constant. than the traction forceF-vatics. with 1.). Applying Eq'. 14- 10. w: have; P = F W P P 2 F1) F= ~— 1: Equation (pf-Marion : + P P -.-+ 2F = ; -- = = ......_ _ , 11‘anr U ma a mu . . . _ va'u K Humane: : Applymg cquatmn d5 = —. we. have a . s " mt)2 mu’ d: = d = ID In P v _: 7 3P Ans 14-77. The S-Ib collar is released from rest at A and travels along the smooth guide. Dotormjhe itsirspeed when its center reaches point C and the normal force: it exerts 15% + VA = Tc + Vc on the ‘rod at this point. The spring has-an :gnstrotohed length of 12 in.. and point C is locatedjust before the and 0 + 0 +}[2(12m£’)2 + 5(2) : 1(_5_)v2 + lumm REV + (E9): 12 2 2 12 12 12 " ‘1 of the curvod portion of the rod. 12 2 32.2 2 12 v = 12.556 ft/s =12.6ft/s Ans $15. * _ . 5 (12.556)! nZFu — m ; N + Fsm50.19¢4° = _ a" C ‘ 32.2( 1 ) c ) F5 _ . 12 10 12 “a E—ks. Ii: 12 *2 _2__= 5°” 2( )[ (12) + {12) 121 7.241011: Thus, NC = Ans 14-86. The roller-coaster car has a speed of 15 ft/s when it is at the crest of a vertical parabolic track. Determine the car’s velocity and the normal force it exerts on the track when it reaches point B. Neglect friction and the T mass of the wheels. The total weight of the car and the "film, passengers is 350 lb. y=_2;3(mooo—2) MMLVMMKVA s dy—_l __2 9_tan-l(2_ o —- — — . — - )—-63.43 dx 100 x=2oo 200“ d_zy___1_ dxz ' 100 DammatA T4+VA =Tb+ls 3501b 1 350 15, 0 1 350 1 30200 (333° aim)‘ ’+ 71%)“) ‘ 5‘ ’ v,=114.48=n4ru3 Ans Q dy I? , N [1%)] 1+(—2)2 ’ a P=T= l =1118.0fl 5 I‘m 350 + 2F. =ma..; 350 63.43°—N =(_._) / c°s B 32.2 (114.48)2 1118.0 N, = 29.1 lb Ans *14-92. The 2-lb collar has a speed of 5 ft/s at A. The attached spring has an unstretched length of 2 ft and a J' stiffness of k = 10 lb/fl. If the collar moves over the smooth rod, determine its speed when it reaches point B, the normal force of the rod on the collar, and the rate of decrease in its speed. Damm aLB : H+M=E+fi ‘2 11 -2 :13. 21103—220 flmjrs) +§(10)(4.5 2) +2045) 2(32_2](va) +2( )( ) + vs = 34.060 ftis = 34.1 ftfs Am 1 F45»? §=ma=1i =4, 2“: dx x=3 2 /g -2 N 7/.5_7° 9=—71.57° Err {3 J (3’13" N WT ’ /V 1-“ __ _3 1 I ,9a df; =Q‘LI—iilllgsmzsn ,1 El . a n__ 2 (34'060)2 yzr; =ma,., —N+10c0518.43 +2cos7l.57 31.623 ) N=7.84lb Ans . D . D 2 +39»: mm; 2sm71.57 —105m18.43 =[3—2—2Ja. a: = —20.4 Ms2 Ann 15-6. A man hits the SO—g golf ball such that it leaves the tee at an angle of 40° with the horizontal and strikes the ground at the same elevation a distance of 20 m away. Determine the impulse of the club C on the ball. Neglect the impulse caused by the ball's weight while the club is striking the ball. (in (+ T) (V) S); = (so); + (VOL! 20 = 0 + v cos40°(t) 1 s = so +vut+§act2 1 0 = o + v sin40°(t) ~ E(9131)? t:1.85 s v = 14.115 m/s rrtvl + Xi’th : mv2 o + J'th : (0.05)(14.115) dez = 0,706 N‘s 4:840“ Ans *15-28. Block A weighs 10 lb and block 8 weighs 3 lb. If B is moving downward with a vetocity (#3)] =3 ft/s at ! = 0, determine the velocity of A when I = 1 s; The coefficient of kinetic friction between the horizontal plane and block A is ,u,‘ = 0.15. :A -+ 255 = I vA = -2v, ((1) mvl + SJ.th = mu: —fl2)3 T1 151.0 ' m 32_2< u— 0+0. ( i-g—fiivm (+l) mv1+EIth= va T T i I. 3 (vnh 1011) 322(3) + 3(1) — 2T(1)— 572(- 2 ) ET 1.5!)» T —32.2T— 10002 = 11.70 ion 3a, —64.4T+ 1.5m)2 = —105.6 T = 1.50 lb (VA): = —6.00 ft/s = 6.00 ft/s —) Ans 15-37. A 40-lb box slides from rest down the smooth ramp onto the surface of a ZO-lb cart. Determine the speed of the box at the instant it stops sliding on the cart; If someone ties the cart to the ramp at B, determine the horizontal impulse the box will exert at C in order to stop its motion. Neglect friction and the size of the box. Datum atB : n+m=u+e (£%wa+o 0+40<15):% v3 = 31.08 ft/s Box and cart: 4. (H) va1 =Emv2 0+(%)(31.08):(42;:9)V1 v2 = 20.7 ft/s Ans '6) mv1,+£Jth=mvz (§%J(31.0‘8)~1F 41:0 #3 dez=38.61b- s Ans 15-42. The man M weighs 150 lb and jumps onto the boat 3 which has a weight of 200 lb. If he has a horizontal component of velocity relative to the boat of 3 ft/s, just before he enters the boat, and the boat is traveling v5 = 2 ft/s away from the pier when he makes the jump. determine the resulting velocity of the man and boat. + (a) VM = VB + VM/B vM = 2+3 vM = 5ft/s (:9 val = Em v2 150 200 322(5) + 32.2(2 (V3); = 3.29 ft/S ) 350 3 3720392 Ans 15-54. Blocks A and B have masses of40 kg and 60 kg, respectively.They are placed on a smooth surface and the spring cannected between them is stretched 2 m. If they are released from rest, determine the speeds of both blocks the instant the spring becomes unstretched. (L) 21an = 2va 0+0=40vA “60% n+w=n+% 1 1 l 0 + 508mm: = 5010va + imam? VA 2 3.29 mls Ans v5 = 2.19 m/s Ans ...
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This note was uploaded on 02/13/2010 for the course ME 324 taught by Professor Neptune during the Spring '08 term at University of Texas.

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Dynamics hw #5 - 14-45. An automobile having a mass of 2 Mg...

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