Đang tải... (xem toàn văn)
A car starts from rest and reaches a speed v after traveling a distance d along a straight road. Determine its constant acceleration and the time of travel.A car starts from rest and reaches a speed v after traveling a distance d along a straight road. Determine its constant acceleration and the time of travel.A car starts from rest and reaches a speed v after traveling a distance d along a straight road. Determine its constant acceleration and the time of travel.A car starts from rest and reaches a speed v after traveling a distance d along a straight road. Determine its constant acceleration and the time of travel.
Table of Contents Chapter 12 Chapter 13 145 Chapter 14 242 Chapter 15 302 Chapter 16 396 Chapter 17 504 Chapter 18 591 Chapter 19 632 Chapter 20 666 Chapter 21 714 Chapter 22 786 Engineering Mechanics - Dynamics Chapter 12 Problem 12-1 A truck traveling along a straight road at speed v1, increases its speed to v2 in time t If its acceleration is constant, determine the distance traveled Given: km hr v1 = 20 v2 = 120 km hr t = 15 s Solution: a = v2 − v1 t d = v1 t + m a = 1.852 s at d = 291.67 m Problem 12-2 A car starts from rest and reaches a speed v after traveling a distance d along a straight road Determine its constant acceleration and the time of travel v = 80 Given: ft s d = 500 ft Solution: v a = 2d v = 2a d v = at t = a = 6.4 ft s v a t = 12.5 s Problem 12-3 A baseball is thrown downward from a tower of height h with an initial speed v0 Determine the speed at which it hits the ground and the time of travel Given: h = 50 ft g = 32.2 ft v0 = 18 s ft s Solution: v = v0 + 2g h v = 59.5 ft s Engineering Mechanics - Dynamics t = v − v0 Chapter 12 t = 1.29 s g *Problem 12–4 Starting from rest, a particle moving in a straight line has an acceleration of a = (bt + c) What is the particle’s velocity at t1 and what is its position at t2? b = Given: m m c = −6 t1 = s s t2 = 11 s s Solution: t a ( t) = b t + c ⌠ v ( t ) = ⎮ a ( t ) dt ⌡0 v ( t1 ) = d ( t2 ) = 80.7 m m s t ⌠ d ( t ) = ⎮ v ( t ) dt ⌡0 Problem 12-5 Traveling with an initial speed v0 a car accelerates at rate a along a straight road How long will it take to reach a speed vf ? Also, through what distance does the car travel during this time? Given: v0 = 70 km hr km a = 6000 hr vf = 120 km hr Solution: vf = v0 + a t t = vf − v0 2 vf = v0 + 2a s t = 30 s a vf − v0 s = 2a s = 792 m Problem 12-6 ( −bt ) where t is the elapsed time Determine the distance A freight train travels at v = v0 − e traveled in time t1, and the acceleration at this time Engineering Mechanics - Dynamics Chapter 12 Given: ft s v0 = 60 b = s t1 = s Solution: ( −bt v ( t) = v0 − e ) d ( t1 ) = 123.0 ft t ⌠ d ( t ) = ⎮ v ( t ) dt ⌡0 d v ( t) dt a ( t) = a ( t1 ) = 2.99 ft s Problem 12-7 The position of a particle along a straight line is given by sp = at3 + bt2 + ct Determine its maximum acceleration and maximum velocity during the time interval t0 ≤ t ≤ tf Given: a = ft b = −9 s ft c = 15 s ft s t0 = s tf = 10 s Solution: sp = a t + b t + c t vp = d sp = 3a t + 2b t + c dt ap = d d vp = s = 6a t + 2b p dt dt Since the acceleration is linear in time then the maximum will occur at the start or at the end We check both possibilities amax = max ( 6a t0 + b , 6a tf + 2b) amax = 42 ft s The maximum velocity can occur at the beginning, at the end, or where the acceleration is zero We will check all three locations tcr = −b 3a tcr = s Engineering Mechanics - Dynamics ( Chapter 12 2 ) vmax = max 3a t0 + 2b t0 + c , 3a tf + 2b tf + c , 3a tcr + 2b tcr + c vmax = 135 ft s *Problem 12-8 From approximately what floor of a building must a car be dropped from an at-rest position so that it reaches a speed vf when it hits the ground? Each floor is a distance h higher than the one below it (Note: You may want to remember this when traveling at speed vf ) vf = 55 mph Given: h = 12 ft g = 32.2 ft s Solution: ac = g vf = + 2ac s Number of floors N Height of one floor h = 12 ft N = H = H vf H = 101.124 ft 2ac N = 8.427 h N = ceil ( N) The car must be dropped from floor number N = Problem 12–9 A particle moves along a straight line such that its position is defined by sp = at3 + bt2 + c Determine the average velocity, the average speed, and the acceleration of the particle at time t1 Given: a = m s b = −3 m c = 2m t0 = s s Solution: sp ( t) = a t + b t + c vp ( t) = d dt sp ( t) Find the critical velocity where vp = ap ( t ) = d dt vp ( t) t1 = s Engineering Mechanics - Dynamics t2 = 1.5 s vave = Chapter 12 vp ( t2 ) = Given t2 = Find ( t2 ) t2 = s sp ( t1 ) − sp ( t0 ) vave = t1 vavespeed = sp ( t2 ) − sp ( t0 ) + sp ( t1 ) − sp ( t2 ) m s vavespeed = t1 a1 = ap ( t1 ) a1 = 18 m s m s Problem 12–10 A particle is moving along a straight line such that its acceleration is defined as a = −kv If v = v0 when d = and t = 0, determine the particle’s velocity as a function of position and the distance the particle moves before it stops Given: k = s ap ( v) = −k v Solution: m v0 = 20 s v d v ⌠ ⎮ dv = −k sp ⌡v v = −k v ds Velocity as a function of position v = v0 − k sp Distance it travels before it stops = v0 − k sp sp = v0 sp = 10 m k Problem 12-11 The acceleration of a particle as it moves along a straight line is given by a = bt + c If s = s0 and v = v0 when t = 0, determine the particle’s velocity and position when t = t1 Also, determine the total distance the particle travels during this time period Given: b = m s c = −1 m s0 = m s v0 = m s t1 = s Engineering Mechanics - Dynamics Chapter 12 Solution: v ⌠t ⌠ ⎮ dv = ⎮ ( b t + c) dt ⌡0 ⌡v v = v0 + bt 2 + ct t ⌠ s ⎞ ⌠ bt ⎮ ⎛ ⎜ = + c t⎟ dt ⎮ ds ⎮ v0 + ⌡s ⎠ ⌡0 ⎝ v1 = v0 + When t = t1 b t1 s = s0 + v0 t + b c t + t 2 + c t1 v1 = 32 b c t1 + t1 s1 = s0 + v0 t1 + m s s1 = 67 m The total distance traveled depends on whether the particle turned around or not To tell we will plot the velocity and see if it is zero at any point in the interval t = , 0.01t1 t1 v ( t) = v0 + bt 2 + ct If v never goes to zero then d = s1 − s0 40 d = 66 m v( t ) 20 0 t *Problem 12–12 A particle, initially at the origin, moves along a straight line through a fluid medium such that its velocity is defined as v = b(1 − e−ct) Determine the displacement of the particle during the time < t < t1 Given: b = 1.8 m s c = 0.3 s t1 = s Engineering Mechanics - Dynamics Chapter 12 Solution: v ( t) − c t) = b( − e ⌠t sp ( t) = ⎮ v ( t) dt ⌡0 sp ( t1 ) = 1.839 m Problem 12–13 The velocity of a particle traveling in a straight line is given v = bt + ct2 If s = when t = 0, determine the particle’s deceleration and position when t = t1 How far has the particle traveled during the time t1, and what is its average speed? b = Given: m c = −3 s t0 = s d a ( t) = ⌠t sp ( t) = ⎮ v ( t) dt ⌡0 v ( t) dt a1 = a ( t ) Deceleration t1 = s s v ( t) = b t + c t Solution: m a1 = −12 m s Find the turning time t2 t2 = 1.5 s v ( t2 ) = Given t2 = s d = sp ( t1 ) − sp ( t2 ) + sp ( t2 ) − sp ( t0 ) Total distance traveled vavespeed = Average speed t2 = Find ( t2 ) d vavespeed = 2.667 t1 − t0 d=8m m s Problem 12–14 A particle moves along a straight line such that its position is defined by s = bt2 + ct + d Determine the average velocity, the average speed, and the acceleration of the particle when t = t1 m m Given: b = c = −6 d = 5m t0 = s t1 = s s s Solution: sp ( t) = b t + c t + d Find the critical time vavevel = v ( t) = d dt t2 = 2s sp ( t) a ( t) = sp ( t1 ) − sp ( t0 ) v ( t) dt v ( t2 ) = Given d t2 = Find ( t2 ) vavevel = t1 t2 = s m s Engineering Mechanics - Dynamics vavespeed = Chapter 12 sp ( t1 ) − sp ( t2 ) + sp ( t2 ) − sp ( t0 ) t1 vavespeed = a1 = a ( t ) a1 = m s m s Problem 12–15 A particle is moving along a straight line such that when it is at the origin it has a velocity v0 If it begins to decelerate at the rate a = bv1/2 determine the particle’s position and velocity when t = t1 Given: v0 = m s m b = −1.5 t1 = s a ( v) = b v s Solution: v ⌠ ⎮ ⎮ ⌡v d a ( v) = b v = v dt v ( t) = v dv = ( v− ) v0 = b t ⎛ v + b t⎞ ⎜ ⎟ ⎠ ⎝ v ( t1 ) = 0.25 t ⌠ sp ( t) = ⎮ v ( t) dt ⌡0 m s sp ( t1 ) = 3.5 m *Problem 12-16 A particle travels to the right along a straight line with a velocity vp = a / (b + sp) Determine its deceleration when sp = sp1 Given: Solution: a = m s b = 4m sp1 = m −a −a a ap = vp = = dsp b + sp (b + sp) (b + sp)3 dvp a vp = b + sp ap1 = −a (b + sp1)3 ap1 = −0.116 m s Engineering Mechanics - Dynamics k = 75 lb ft Chapter 22 Cratio = 0.8 δ = 0.15 ft ft g = 32.2 s Solution: ωn = kg W ωn = 18.57 rad s MF = 2⎤ MF = 0.997 ⎡ ⎛ω⎞ ω ⎢1 − ⎜ ⎟ ⎥ + ⎢⎡2( Cratio) ⎛⎜ ⎞⎟⎤⎥ ⎣ ⎝ ωn ⎠ ⎦ ⎣ ⎝ ωn ⎠⎦ *Problem 22-60 The bar has a weight W If the stiffness of the spring is k and the dashpot has a damping coefficient c, determine the differential equation which describes the motion in terms of the angle θ of the bar’s rotation Also, what should be the damping coefficient of the dashpot if the bar is to be critically damped? Given: W = lb b = ft lb ft a = ft k = c = 60 lb⋅ s ft Solution: ⎛ W ⎞ ( a + b) θ'' + c b2 θ' + k ( a + b) θ = ⎜ ⎟ ⎝g⎠ M = ⎛ W ⎞ ( a + b) ⎜ ⎟ ⎝g⎠ C = cb K = k ( a + b) Mθ'' + Cθ' + Kθ = where M = 1.55 lb⋅ ft⋅ s C = 540.00 lb⋅ ft ⋅ s K = 200.00 lb⋅ ft 828 Engineering Mechanics - Dynamics Chapter 22 To find critical damping K M ωn = c = C = 2Mωn C c = 3.92 b lb⋅ s ft Problem 22-61 A block having mass M is suspended from a spring that has stiffness k If the block is given an upward velocity v from its equilibrium position at t = 0, determine its position as a function of time Assume that positive displacement of the block is downward and that motion takes place in a medium which furnishes a damping force F = C |v| Given: M = kg k = 600 N m C = 50 N⋅ s m v = 0.6 Solution: ωn = 9.258 rad s Cc = 2Mωn Cc = 129.6 N⋅ s m s < Cc = 129.61 N⋅ k M ωn = If C = 50.00 N⋅ m −C b = ωn Cc y = 0m Given ωd = ωn y' = −v s m the system is underdamped ⎛ C⎞ 1−⎜ ⎟ ⎝ Cc ⎠ A = 1m φ = rad t = 0s sin ( ωd t + φ ) bt y = Ae bt y' = A b e sin ( ωd t + φ ) + A ωd e bt ⎛A⎞ ⎜ ⎟ = Find ( A , φ ) ⎝φ⎠ 829 cos ( ωd t + φ ) m s Engineering Mechanics - Dynamics bt y = Ae Chapter 22 sin ( ωd t + φ ) A = −0.0702 m b = −3.57 rad s ωd = 8.54 rad s φ = 0.00 rad Problem 22-62 The damping factor C/C c, may be determined experimentally by measuring the successive amplitudes of vibrating motion of a system If two of these maximum displacements can be approximated by x1 and x2, as shown in Fig 22-17, show that the ratio ln (x1 / x2) = 2π (C / C c ) / ( − ( C /Ce )2 )1/2 The quantity ln(x1 / x2) is called the logarithmic decrement Solution: ⎛− C t ⎞ ⎜ 2m ⎟ x = D⎝ e sin ( ωd t + φ )⎠ − xmax = D e − x1 x2 = De − De C 2m C 2m C 2m − t x1 = D e t1 C =e 2m C 2m − t1 x2 = D e C 2m (t2− t1) t2 Since ω d t − ω d t = 2π so that ln ⎜ t2 − t1 = then 2π ωd ⎛ x1 ⎞ Cπ ⎟= ⎝ x2 ⎠ mωd Cc C⎞ = ⎟ 2m ⎝ Cc ⎠ ωd = ωn − ⎛⎜ Cc = 2mωn So that, ⎛ x1 ⎞ ⎟= ⎝ x2 ⎠ ln ⎜ ⎛ C⎞ ⎟ ⎝ Cc ⎠ 2π ⎜ ⎛ C⎞ 1−⎜ ⎟ ⎝ Cc ⎠ Q.E.D 830 1− ⎛ C⎞ ⎜C ⎟ ⎝ c⎠ t2 Engineering Mechanics - Dynamics Chapter 22 Problem 22-63 Determine the differential equation of motion for the damped vibratory system shown.What type of motion occurs? Given: M = 25 kg N m k = 100 N⋅ s m c = 200 Solution: M g − k( y + yst) − 2c y' = M y'' M y'' + k y + 2c y' + k yst − M g = Equilibrium k yst − M g = M y'' + 2c y' + k y = y'' + (1) 2c k y' + y=0 M M By comparing Eq.(1) to Eq 22-27 p = k M p = 2.00 cc = 2M p rad s cc = 100.00 N⋅ Since c = 200.00 N⋅ s s m > cc = 100.00 N⋅ s , the system is overdamped m m and will not oscillate The motion is an exponential decay *Problem 22-64 The block of mass M is subjected to the action of the harmonic force F = F cosω t Write the equation which describes the steady-state motion Given: M = 20 kg k = 400 N m F = 90 N C = 125 N⋅ s m ω = rad s 831 Engineering Mechanics - Dynamics Chapter 22 Solution: ωn = 2k M ωn = 6.32 Cc = 2Mωn rad s Cc = 253.0 N⋅ s m Cω ⎤ ⎡ ⎢ ⎥ 2k φ = atan ⎢ ⎥ ⎢ ⎛ω⎞ ⎥ − ⎜ω ⎟ ⎥ ⎢ ⎣ ⎝ n⎠ ⎦ F0 2k A = ⎡ ⎛ ω ⎞ 2⎤ ⎡ ⎛ C ⎞ ⎛ ω ⎞⎤ ⎢1 − ⎜ ⎟ ⎥ + ⎢2⎜ ⎟ ⎜ ⎟⎥ ⎣ ⎝ ωn ⎠ ⎦ ⎣ ⎝ Cc ⎠ ⎝ ωn ⎠⎦ x = A cos ( ωt − φ ) A = 0.119 m ω = 6.00 rad s φ = 83.9 deg Problem 22-65 Draw the electrical circuit that is equivalent to the mechanical system shown What is the differential equation which describes the charge q in the circuit? Solution: For the block, m x'' + c x' + 2k x = Let m=L c=R L q'' + R q' + x=q k= C ⎛ ⎞q = ⎜ ⎟ ⎝ C⎠ Problem 22-66 The block of mass M is continually damped If the block is displaced x = x1 and released from rest, determine the time required for it to return to the position x = x2 832 Engineering Mechanics - Dynamics Chapter 22 Given: M = 10 kg k = 60 N m x1 = 50 mm C = 80 N⋅ s m x2 = mm Solution: ωn = k M ωn = 2.45 Since C = 80.00 N⋅ −C λ1 = + 2M t = 0s Given s m Cc = 2Mωn > Cc = 48.99 N⋅ ⎛ C⎞ − k ⎜ ⎟ M ⎝ 2M ⎠ x = x1 x' = s m m s Cc = 48.99 N⋅ A1 = m ⎛ C⎞ − k ⎜ ⎟ M ⎝ 2M ⎠ A2 = m λ1 t λ2 t x' = A1 λ e + A2λ e ⎛ λ ⎞ ⎛ −0.84 ⎞ rad ⎜ ⎟=⎜ ⎟ ⎝ λ ⎠ ⎝ −7.16 ⎠ s λ1 t λ2 t x2 = A e + A2 e s m the system is overdamped −C λ2 = − 2M λ1 t λ2 t x = A1 e + A2 e ⎛ A1 ⎞ ⎜ ⎟ = Find ( A1 , A2 ) ⎝ A2 ⎠ Given rad s t = Find ( t) 833 ⎛ A1 ⎞ ⎛ 0.06 ⎞ ⎜ ⎟=⎜ ⎟m ⎝ A2 ⎠ ⎝ −0.01 ⎠ t = 3.99 s ... 0.01tf tf t2 = , 1.01 tf tf s1 ( t1 ) = tf s2 ( t2 ) a t1 m ⎡ ⎛ tf ⎞ tf ⎛ = ⎢ a ⎜ ⎟ + a ⎜ t2 − 2 2 ⎝ ⎠ tf ⎞ ⎟ 2 tf ⎞ ⎛ − a ⎜ t2 − ⎟ 2 ⎝ 2 ⎥1 ⎦m Distance in m 0 .2 s1( t1) 0.1 s2( t2) 0 0.05... ( t1 ) = s2 ( t2 ) a1 ( t ) = a a t1 m ⎡ ⎛ tf ⎞ tf ⎛ = ⎢ a ⎜ ⎟ + a ⎜ t2 − 2 2 ⎝ ⎠ tf ⎞ ⎟ 2 s m tf ⎞ ⎛ − a ⎜ t2 − ⎟ 2 ⎝ 2 ⎥1 ⎦m s a2 ( t ) = − a m Distance in m s1( t1) 0.5 s2( t2) 0 0.05... time t2 Given: d = 80 ft t1 = s t2 = s g = 32. 2 ft s Solution: aA = g vA = g t aB = g vB = g( t − t1 ) g t g sB = ( t − t1 ) sA = At time t2 sA2 = g t2 sA2 = 64.4 ft sB2 = g t2 − t1 ) ( sB2 = 16.1