Hw10 - t HIT = V i sin θ 32 2 = 705 sec 2.35 Convert mph into f t sec 45 mph = 45(88 60 = 66 f t sec f N C 250 lb x y 2 To slide F rx =-f m ¨ x C

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 10 (2.15, 2.35) 2.15 ¨ x = 0 Integrating, ˙ x = C 1 @ t = 0 , C 1 = V i cos θ ˙ x = V i cos θ Integrating again, x = V i cos θ t + C 2 @ t = 0 , C 2 = 0 x = V i cos θ t ¨ y = - g Integrating, ˙ y = - gt + C 3 @ t = 0 , C 3 = V i sin θ ˙ y = - gt + V i sin θ Integrating again, y = - gt 2 2 + V i sin θ t + C 4 @ t = 0 , C 4 = 0 y = - gt 2 2 + V i sin θ t At the highest point in the trajectory, the y-component of velocity = 0. Let the time at which it reaches this point and hits the crossbar be t HIT . ˙ y HIT = 0 - gt HIT + V i sin θ = 0 t HIT = V i sin θ 32 . 2 y HIT = 8 - 32 . 2 V 2 i sin 2 θ 2(32 . 2) 2 + V i sin θ V i sin θ 32 . 2 = 8 V 2 i sin 2 θ = 16(32 . 2) ··· (1) 1
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x HIT = 60 V i cos θ V i sin θ 32 . 2 = 60 V 2 i sin θ cos θ = 60(32 . 2) ··· (2) Divide (1) by (2), tan θ = 16 60 θ = 14 . 9 V 2 i sin 2 14 . 9 = 16(32 . 2) V i = 88 . 1 f t / sec
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Unformatted text preview: t HIT = V i sin θ 32 . 2 = . 705 sec 2.35 Convert mph into f t / sec , 45 mph = 45(88 / 60) = 66 f t / sec . f N C 250 lb x y 2 To slide, F rx =-f m ¨ x C =-μ mg ¨ x C =-. 3(32 . 2) =-9 . 66 f t / sec 2 Integrating, ˙ x C =-9 . 66 t + C 1 @ t = , C 1 = 66 ˙ x C =-9 . 66 t + 66 ∴ t STOP = 66 9 . 66 = 6 . 83 sec x C , MIN =-9 . 66(6 . 83) 2 2 + 66(6 . 83) + ± ± ² C 2 = 225 f t 3...
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This note was uploaded on 05/24/2010 for the course TM 212 taught by Professor Richard during the Spring '10 term at University of Illinois, Urbana Champaign.

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Hw10 - t HIT = V i sin θ 32 2 = 705 sec 2.35 Convert mph into f t sec 45 mph = 45(88 60 = 66 f t sec f N C 250 lb x y 2 To slide F rx =-f m ¨ x C

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