EGR245_CHAPTER13b

# EGR245_CHAPTER13b - EGR 245 EGR ENGINEERING MECHANICS...

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EGR 245 EGR 245 ENGINEERING ENGINEERING MECHANICS DYNAMICS DYNAMICS CHAPTER 13 – KINEMATICS OF PARTICLES: ENERGY AND MOMENTUM METHODS PROBLEM 13.119 t A 3-lb particle is acted upon by a force where F is expressed in pounds and t in seconds. Determine the magnitude and direction of the velocity of the particle at knowing that its velocity is zero at ( ) ( ) 20sin 2 24cos2 , tt =+ Fij 6s t = 0. t =

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EGR 245 EGR 245 ENGINEERING ENGINEERING MECHANICS DYNAMICS DYNAMICS CHAPTER 13 – KINEMATICS OF PARTICLES: ENERGY AND MOMENTUM METHODS PROBLEM 13.119 6 06 0 md t m += vF v () ( ) 6 6 0 3 02 0 s i n 2 2 4 c o s 2 32.2 tt d t ⎡⎤ ++ = ⎣⎦ ij v [] 6 6 0 3 01 0 c o s 2 1 2 s i n 2 32.2 +− + = v [ ] 6 10 cos12 cos0 12 sin12 sin0 0.09317 −− + = v 6 1.5615 6.4389 0.09317 −= v [ ] 6 16.759 69.109 ft/s =− vi j 6 71.1 ft/s v = 76.4 ° W W
EGR 245 EGR 245 ENGINEERING ENGINEERING MECHANICS DYNAMICS DYNAMICS CHAPTER 13 – KINEMATICS OF PARTICLES: ENERGY AND MOMENTUM METHODS PROBLEM 13.124 t A 10-ton truck enters a safety ramp at a high speed v 0 and travels 540 ft in 5.5 s before its speed is reduced to . Assuming constant deceleration determine ( a ) the initial speed v 0 , ( b ) the magnitude of the braking force. Neglect air resistance and rolling resistance. 15 ° 0 /2 v

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EGR 245 EGR 245 ENGINEERING ENGINEERING MECHANICS DYNAMICS DYNAMICS CHAPTER 13 – KINEMATICS OF PARTICLES: ENERGY AND MOMENTUM METHODS PROBLEM 13.124 t Momentum, 20,000 lb W = 2 20,000 621.118 lb s ft 32.2 m == ( ) 01 :s i n 1 5 x mv F mg t mv −+ ° = () ( ) 0 0 621.118 sin15 5.5 621.118 2 v vF m g ⎛⎞ ° = ⎜⎟ ⎝⎠ ( )( ) 0 310.559 sin15 5.5 m g =+ ° (1) Conservation of energy: 22 11 sin15 mv F mg x mv ° =
EGR 245 EGR 245 ENGINEERING ENGINEERING MECHANICS DYNAMICS DYNAMICS CHAPTER 13 – KINEMATICS OF PARTICLES: ENERGY AND MOMENTUM METHODS PROBLEM 13.124 t () ( ) 2 2 0 0 310.559 310.559 sin15 540 4 v vF m g −+ ° = ( ) ( ) 2 0 3310 .559 sin15 540 4 v Fm g =+ ° Using (1) eliminate ( ) sin15 : g + ° ( ) 2 0 0 3 310.559 5.5 310.559 41 8 0 v v ⎛⎞ = ⎜⎟ ⎝⎠ ( a ) ( )( ) 0 4 540 130.909 ft s 89.3 mi h 35 .5 v == = W

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EGR 245 EGR 245 ENGINEERING ENGINEERING MECHANICS DYNAMICS DYNAMICS CHAPTER 13 – KINEMATICS OF PARTICLES: ENERGY AND MOMENTUM METHODS PROBLEM 13.124 t F = 2220 lb () ( ) 0 310.559 130.909 310.559 + sin15 7391.8 5.5 5.5 v Fm g °= = = ( ) 7391.8 20000 sin15 2215 lb F =− ° = ( b ) W
EGR 245 EGR 245 ENGINEERING ENGINEERING MECHANICS DYNAMICS DYNAMICS CHAPTER 13 – KINEMATICS OF PARTICLES: ENERGY AND MOMENTUM METHODS PROBLEM 13.129 t A tractor-trailer rig with a 2000-kg tractor, a 4500-kg trailer, and a 3600-kg trailer is traveling on a level road at 90 km/h. The brakes on the rear trailer fail and the antiskid system of the tractor and front trailer provide the largest possible force which will not cause the wheels to slide. Knowing that the coefficient of static friction is 0.75, determine ( a ) the shortest time for the rig to a come to a stop, ( b ) the force in the coupling between the two trailers during that time. Assume that the force exerted by the coupling on each of the two trailers is horizontal.

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EGR 245 EGR 245 ENGINEERING ENGINEERING MECHANICS DYNAMICS DYNAMICS CHAPTER 13 – KINEMATICS OF PARTICLES: ENERGY AND MOMENTUM METHODS PROBLEM 13.129 t ( a ) Combined 90km h 25 m s v == () ( )( ) ( ) 12 6500 9.81 63765 N; 3600 9.81 35316 N WW 11 2 2 1 ;0 . 7 5 NW N W F N = Impulse = 0 – mv 0 + ←⎯⎯ ( ) 1 0.75 10,100 kg 25 m s Nt −= ( ) ( ) 10,100 25 5.2798 s 0.75 63765 t 5.28 s t = W
EGR 245 EGR 245

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## This note was uploaded on 04/14/2010 for the course EGR 120 taught by Professor Rassai during the Spring '10 term at Algoma University.

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EGR245_CHAPTER13b - EGR 245 EGR ENGINEERING MECHANICS...

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