Spring Problems Part 2-Engineering Mechanics-Assignment Solution - Document's extract

Exercises, Dynamics

Description: This is assignment solution for Dynamics course. It was submitted to Deendayal Chandar at Aligarh Muslim University. It includes: Spring, Armature, Electric, Drill, Displacement, Velocity, Angular, Motor, Point, Contact, Speed

Engineering Mechanics - Dynamics Chapter 15 e( −vB1y cos ( θ ) − vB1x sin ( θ ) ) = vB2n ⎛ vB1x ⎞ ⎜ ⎟ ⎜ vB1y ⎟ ⎜ vB2n ⎟ ⎜ ⎟ = Find ( vB1x , vB1y , vB2n , vB2t , t , R) ⎜ vB2t ⎟ ⎜t⎟ ⎜ ⎟ ⎝R⎠ ⎛ vB2n ⎞ ⎛ 1.70 ⎞ ft ⎜ ⎟=⎜ ⎟ vB2t ⎠ ⎝ 6.36 ⎠ s ⎝ *Problem 15-72 ⎛ vB1x ⎞ ⎛ 3.00 ⎞ ft ⎜ ⎟=⎜ ⎟ ⎝ vB1y ⎠ ⎝ −6.00 ⎠ s t = 0.19 s R = 0.79 ft ⎛ vB2n ⎞ ft ⎜ ⎟ = 6.59 s ⎝ vB2t ⎠ The drop hammer H has a weight WH and falls from rest h onto a forged anvil plate P that has a weight WP. The plate is mounted on a set of springs that have a combined stiffness kT . Determine (a) the velocity of P and H just after collision and (b) the maximum compression in the springs caused by the impact. The coefficient of restitution between the hammer and the plate is e. Neglect friction along the vertical guideposts A and B. Given: WH = 900 lb WP = 500 lb h = 3 ft Solution: kT = 500 g = 32.2 e = 0.6 lb ft ft s 2 δ st = Guesses WP kT vH1 = 2g h vH2 = 1 Given ft s vP2 = 1 ft s δ = 2 ft ⎛ WH ⎞ ⎜ ⎟ vH1 = ⎝g⎠ ⎛ WH ⎞ ⎛ WP ⎞ ⎜ ⎟ vH2 + ⎜ ⎟ vP2 ⎝g⎠ ⎝g⎠ e vH1 = vP2 − vH2 349 docsity.com Engineering Mechanics - Dynamics Chapter 15 1 1 2 2 1 ⎛ WP ⎞ 2 kTδ st + ⎜ ⎟ vP2 = kTδ − WP( δ − δ st) 2 2⎝ g ⎠ 2 ⎛ vH2 ⎞ ⎜ ⎟ ⎜ vP2 ⎟ = Find ( vH2 , vP2 , δ ) ⎜ ⎟ ⎝δ⎠ ⎛ vH2 ⎞ ⎛ 5.96 ⎞ ft ⎜ ⎟=⎜ ⎟ ⎝ vP2 ⎠ ⎝ 14.30 ⎠ s δ = 3.52 ft Problem 15-73 It was observed that a tennis ball when served horizontally a distance h above the ground strikes the smooth ground at B a distance d away. Determine the initial velocity vA of the ball and the velocity vB (and θ) of the ball just after it strikes the court at B. The coefficient of restitution is e. Given: h = 7.5 ft d = 20 ft e = 0.7 g = 32.2 Solution: Guesses vA = 1 ft s ft s vB2 = 1 ft s t =1s ft s 2 vBy1 = 1 θ = 10 deg Given h= 12 gt 2 d = vA t vBy1 =

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g t e vBy1 = vB2 sin ( θ ) vA = vB2 cos ( θ ) 350 docsity.com Engineering Mechanics - Dynamics Chapter 15 ⎛ vA ⎞ ⎜ ⎟ t⎟ ⎜ ⎜ vBy1 ⎟ = Find ( v , t , v , v , θ ) A By1 B2 ⎜ ⎟ ⎜ vB2 ⎟ ⎜ ⎟ ⎝θ⎠ vA = 29.30 ft s vB2 = 33.10 ft s θ = 27.70 deg Problem 15-74 The tennis ball is struck with a horizontal velocity vA, strikes the smooth ground at B, and bounces upward at θ = θ1. Determine the initial velocity vA, the final velocity vB, and the coefficient of restitution between the ball and the ground. Given: h = 7.5 ft d = 20 ft θ 1 = 30 deg g = 32.2 Solution: Guesses ft s 2 θ = θ1 vA = 1 h= ft s t =1s d = vA t vBy1 = 1 ft s vB2 = 1 ft s e = 0.5 Given 12 gt 2 vBy1 = g t vA = vB2 cos ( θ ) e vBy1 = vB2 sin ( θ ) ⎛ vA ⎞ ⎜ ⎟ ⎜t⎟ ⎜ vBy1 ⎟ = Find ( v , t , v , v , e) A By1 B2 ⎜ ⎟ ⎜ vB2 ⎟ ⎜e⎟ ⎝ ⎠ Problem 15-75 vA = 29.30 ft s vB2 = 33.84 ft s e = 0.77 The ping-pong ball has mass M. If it is struck with the velocity shown, determine how high h it rises above the end of the smooth table after the rebound. The coefficient of restitution is e. 351 docsity.com Engineering Mechanics - Dynamics Chapter 15 Given: M = 2 gm e = 0.8 a = 2.25 m b = 0.75 m g = 9.81 m s 2 θ = 30 deg m v = 18 s Solution: Guesses v1x = 1 t1 = 1 s m s v1y = 1 t2 = 2 s m s v2x = 1 m s v2y = 1 m s h =1m v1y = g t1 + v sin ( θ ) b = v2x t2 h = v2y t2 − Given v1x = v cos ( θ ) v2x = v1x a = v cos ( θ ) t1 e v1y = v2y ⎛ g⎞t 2 ⎜ ⎟2 ⎝ 2⎠ ⎛ v1x ⎞ ⎜⎟ ⎜ v1y ⎟ ⎜ v2x ⎟ ⎜⎟ ⎜ v2y ⎟ = Find ( v1x , v1y , v2x , v2y , t1 , t2 , h) ⎜t ⎟ ⎜ 1⎟ ⎜ t2 ⎟ ⎜⎟ ⎝h⎠ ⎛ v1x ⎞ ⎛ 15.59 ⎞ ⎜⎟⎜ ⎟ ⎜ v1y ⎟ = ⎜ 10.42 ⎟ m ⎜ v2x ⎟ ⎜ 15.59 ⎟ s ⎜⎟⎜ ⎟ ⎝ v2y ⎠ ⎝ 8.33 ⎠ ⎛ t1 ⎞ ⎛ 0.14 ⎞ ⎜ ⎟=⎜ ⎟s ⎝ t2 ⎠ ⎝ 0.05 ⎠ h = 390 mm *Problem 15-76 The box B of weight WB is dropped from rest a distance d from the top of

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the plate P of weight WP, which is supported by the spring having a stiffness k. Determine the maximum compression imparted to the spring. Neglect the mass of the spring. 352 docsity.com Engineering Mechanics - Dynamics Chapter 15 Given: WB = 5 lb lb k = 30 ft WP = 10 lb d = 5 ft g = 32.2 e = 0.6 ft s 2 Solution: δ st = Guesses WP k vB1 = ft s 2g d vP2 = 1 ft s vB2 = 1 δ = 2 ft Given ⎛ WB ⎞ ⎛ WB ⎞ ⎛ WP ⎞ ⎜ ⎟ vB1 = ⎜ ⎟ vB2 + ⎜ ⎟ vP2 ⎝g⎠ ⎝g⎠ ⎝g⎠ e vB1 = vP2 − vB2 1 12 2 1 ⎛ WP ⎞ 2 kδ st + ⎜ ⎟ vP2 = kδ − WP( δ − δ st) 2 2⎝ g ⎠ 2 ⎛ vB2 ⎞ ⎜ ⎟ ⎜ vP2 ⎟ = Find ( vB2 , vP2 , δ ) ⎜ ⎟ ⎝δ⎠ ⎛ vB2 ⎞ ⎛ −1.20 ⎞ ft ⎜ ⎟=⎜ ⎟ ⎝ vP2 ⎠ ⎝ 9.57 ⎠ s δ = 1.31 ft Problem 15-77 A pitching machine throws the ball of weight M towards the wall with an initial velocity vA as shown. Determine (a) the velocity at which it strikes the wall at B , (b) the velocity at which it rebounds from the wall and (c) the distance d from the wall to where it strikes the ground at C. Given: M = 0.5 kg vA = 10 m s a =3m b = 1.5 m e = 0.5 θ = 30 deg g = 9.81 m s 2 353 docsity.com Engineering Mechanics - Dynamics Chapter 15 Solution: Guesses m s m s vBx2 = 1 vBy2 = 1 d =1m t2 = 1 s m s m s vBx1 = 1 vBy1 = 1 h =1m t1 = 1 s Given vA cos ( θ ) t1 = a vBy2 = vBy1 d = vBx2 t2 vA cos ( θ ) = vBx1 b + vA sin ( θ ) t1 − 1 2 g t1 = h 2 vA sin ( θ ) − g t1 = vBy1 h + vBy2 t2 − e vBx1 = vBx2 1 2 g t2 = 0 2 ⎛ vBx1 ⎞ ⎜ ⎟ ⎜ vBy1 ⎟ ⎜ vBx2 ⎟ ⎜ ⎟ ⎜ vBy2 ⎟ = Find ( v , v , v , v , h , t , t , d) Bx1 By1 Bx2 By2 12 ⎜h⎟ ⎜ ⎟ ⎜ t1 ⎟ ⎜ t2 ⎟ ⎟ ⎜ ⎝d⎠ ⎛ vBx1 ⎞ m ⎜ ⎟ = 8.81 s ⎝ vBy1 ⎠ ⎛ vBx2 ⎞ m ⎜ ⎟ = 4.62 s ⎝ vBy2 ⎠ d = 3.96 m Problem 15-78 The box of weight Wb slides on the surface for which the coefficient of friction is μk. The box has velocity v when it is a distance d from the plate. If it strikes the pla

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