MECH Statics & Dynamics

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1 Chapter D Problem 3 Solution 1/7/8 1:8 PM MECH 11 - Static & Dynamic Chapter D Problem 3 Solution Page 7, Engineering Mechanic - Dynamic, 4th Edition, Meriam and Kraige Given: Particle moving along a traight line (-axi) with velocity, v, given in term of time, t, by: v = A - B t + C t 3/ A = m/, B = 4 m/ C = 5 m/ 5/ The poition of the particle at time t = i given by equal to 3 m. Find: The poition,, velocity, v, and acceleration, a, when the time, t, i equal to 3.. Obervation: 1. Intereted in motion only without regard to the force cauing the motion, free body diagram i not of interet.. The motion i along a ingle traight line. The motion diagram i imple enough that it can be omitted. 1. Mechanical Sytem - Particle during the interval from t = to t = Equation v = A - B t + C t 3/ Relationhip between velocity, acceleration and time: a = dv/dt = -B + 3/ C t 1/ Relationhip between velocity, poition, and time: d/dt = v = A - B t + C t 3/ Separating variable and integrating: ( ) dx = ( ) t v dt = ( ) t { A - B t + C t 3/ } dt - = { A t - B t / + /5 C t 5/ } t = + A t - B t / + /5 C t 5/ 4. Solve Evaluating each of the three expreion at t equal to 3 : v = A - B t + C t 3/ v(t=3) = m/ - 4 m/ m/ 5/ (3 ) 3/ = m/ a = -B + 3/ C t 1/ a(t=3) = -4 m/ + 3/ 5 m/ 5/ (3 ) 1/ = 8.99 m/ = + A t - B t / + /5 C t 5/ (t = 3) = 3 m + m/ 3-4 m/ 1/ (3 ) + /5 5 m/ 5/ (3 ) 5/ file:///uer//document/nel_web/nels_mech11/p_d_3.html Page 1 of

2 Chapter D Problem 3 Solution 1/7/8 1:8 PM =. m Reult Poition = (t=3) =. m Velocity = v(t=3) = m/ Acceleration = a(t=3) = 8.99 m/ file:///uer//document/nel_web/nels_mech11/p_d_3.html Page of

3 8 Chapter Kinematic of Particle Sample Problem / A particle move along the x-axi with an initial velocity v x 5 ft/ec at the origin when t. For the firt 4 econd it ha no acceleration, and thereafter it i acted on by a retarding force which give it a contant acceleration a x 1 ft/ec. Calculate the velocity and the x-coordinate of the particle for the condition of t 8 ec and t 1 ec and find the maximum poitive x-coordinate reached by the particle. Helpful Hint Learn to be flexible with ymbol. The poition coordinate x i jut a valid a. Solution. dv a dt The velocity of the particle after t 4 ec i computed from v x 5 dv x 1 t dt and i plotted a hown. At the pecified time, the velocitie are 4 v x 9 1t ft/ec Note that we integrate to a general time t and then ubtitute pecific value. t 8 ec, v x 9 1(8) 1 ft/ec The x-coordinate of the particle at any time greater than 4 econd i the ditance traveled during the firt 4 econd plu the ditance traveled after the dicontinuity in acceleration occurred. Thu, d v dt t 1 ec, v x 9 1(1) 3 ft/ec x 5(4) t (9 1t) dt 5t 9t 8 ft 4 v x, ft/ec t, ec For the two pecified time, 3 t 8 ec, t 1 ec, x 5(8 ) 9(8) 8 3 ft x 5(1 ) 9(1) 8 8 ft The x-coordinate for t 1 ec i le than that for t 8 ec ince the motion i in the negative x-direction after t 9 ec. The maximum poitive x-coordinate i, then, the value of x for t 9 ec which i x max 5(9 ) 9(9) 8 35 ft Thee diplacement are een to be the net poitive area under the v-t graph up to the value of t in quetion. Show that the total ditance traveled by the particle in the 1 ec i 37 ft.

4 Chapter D Problem 9 Solution 1/7/8 1:9 PM MECH 11 - Static & Dynamic Chapter D Problem 9 Solution Page 31, Engineering Mechanic - Dynamic, 4th Edition, Meriam and Kraige Given: The acceleration of an arrow decreae linearly with ditance,, from a maximum of a equal to 16, ft/ upon releae of the arrow to zero after a ditance of travel L equal to ft. Find: The maximum velocity of the arrow.. Obervation: 1. Intereted excluively in the motion of the arrow independent of the force producing that motion, thu no free body diagram i of interet.. The motion i along a ingle traight line. The motion diagram i imple enough that it can be omitted. 3. The arrow will travel nearly in a traight line during that brief interval between releae of the aroow and the launch point. 4. A the arrow continue accelerating until it reache the ditance L, the maximum velocity will occur at that point. 1. Mechanical Sytem - Arrow from releae until it ha traveled a ditance L. 3. Equation Acceleration, a, i linear with ditance, : a = m + b The acceleration i known at two point: a(=) = -a /L a(=l) = The "intercept", b, i the value of the acceleration at =, that i a. The "lope", m, i the change in acceleration, a, divided by the change in ditance,, between two point where both of thoe quantitie are known: m = ( - a ) / ( L - ) = -a /L The dependence of the acceleration on poition can be expreed a: a = -a /L + a = a { 1 - /L } The relationhip between acceleration, velocity, and poition i: a = v dv/d v dv/d = a { 1 - /L } Separating variable and integrating: ( ) vmax v dv = ( ) L a { 1 - /L }d 1/ v vmax = a { - 1/ /L } L 1/ v max = a { L - 1/ L /L } v max = a L file:///uer//document/nel_web/nels_mech11/p_d_9.html Page 1 of

5 Chapter D Problem 9 Solution 1/7/8 1:9 PM 4. Solve v max = a L v max = (a L) 1/ = (16, ft/ ft) 1/ = ft/ Reult Maximum velocity = v max = ft/ file:///uer//document/nel_web/nels_mech11/p_d_9.html Page of

6 Article / Rectilinear Motion 9 Sample Problem /3 The pring-mounted lider move in the horizontal guide with negligible friction and ha a velocity v in the -direction a it croe the mid-poition where and t. The two pring together exert a retarding force to the motion of the lider, which give it an acceleration proportional to the diplacement but oppoitely directed and equal to a k, where k i contant. (The contant i arbitrarily quared for later convenience in the form of the expreion.) Determine the expreion for the diplacement and velocity v a function of the time t. Solution I. Since the acceleration i pecified in term of the diplacement, the differential relation vdv admay be integrated. Thu, v dv k d C 1 a contant, or When, v v, o that C 1 v /, and the velocity become v v k v k C 1 Helpful Hint We have ued an indefinite integral here and evaluated the contant of integration. For practice, obtain the ame reult by uing the definite integral with the appropriate limit. The plu ign of the radical i taken when v i poitive (in the plu -direction). Thi lat expreion may be integrated by ubtituting v d/dt. Thu, d v k dt C a contant, or 1 k in1 k v t C With the requirement of t when, the contant of integration become C, and we may olve the equation for o that Again try the definite integral here a above. v in kt k The velocity i v ṡ, which give v v co kt Solution II. Since a, the given relation may be written at once a k Thi i an ordinary linear differential equation of econd order for which the olution i well known and i A in Kt B co Kt where A, B, and K are contant. Subtitution of thi expreion into the differential equation how that it atifie the equation, provided that K k. The velocity i v ṡ, which become v Ak co kt Bk in kt The initial condition v v when t require that A v /k, and the condition when t give B. Thu, the olution i v k in kt and v v co kt Thi motion i called imple harmonic motion and i characteritic of all ocillation where the retoring force, and hence the acceleration, i proportional to the diplacement but oppoite in ign.

7 3 Chapter Kinematic of Particle Sample Problem /4 A freighter i moving at a peed of 8 knot when it engine are uddenly topped. If it take 1 minute for the freighter to reduce it peed to 4 knot, determine and plot the ditance in nautical mile moved by the hip and it peed v in knot a function of the time t during thi interval. The deceleration of the hip i proportional to the quare of it peed, o that a kv. Helpful Hint Recall that one knot i the peed of one nautical mile (676 ft) per hour. Work directly in the unit of nautical mile and hour. Solution. The peed and the time are given, o we may ubtitute the expreion for acceleration directly into the baic definition a dv/dt and integrate. Thu, kv dv dt 1 1 Now we ubtitute the end limit of v 4 knot and t 6 6 hour and get 4 The peed i plotted againt the time a hown. The ditance i obtained by ubtituting the expreion for v into the definition v d/dt and integrating. Thu, 8 1 6t d dt 1 v 1 8 kt v 8 1 8kt 8 1 8k(1/6) t dv k dt v k 3 4 mi1 8 dt 1 6t d 4 ln (1 6t) 3 v dv 8 t v k dt v 8 1 6t We chooe to integrate to a general value of v and it correponding time t o that we may obtain the variation of v with t. v, knot The ditance i alo plotted againt the time a hown, and we ee that the hip 4 4 ha moved through a ditance ln (1 6 ln.94 mi (nautical) during the ) 3 minute t, min 1., mi (nautical) t, min

8 46 Chapter Kinematic of Particle Sample Problem /5 The curvilinear motion of a particle i defined by v x 5 16t and y 1 4t, where v x i in meter per econd, y i in meter, and t i in econd. It i alo known that x when t. Plot the path of the particle and determine it velocity and acceleration when the poition y i reached. Solution. The x-coordinate i obtained by integrating the expreion for v x, and the x-component of the acceleration i obtained by differentiating v x. Thu, dx v x dt x dx t (5 16t) dt x 5t 8t m y, m t = [a x v x] a x d (5 16t) dt a x 16 m/ The y-component of velocity and acceleration are [v v y d dt (1 y ẏ] 4t ) v y 8t m/ t = 5 4 A x, m 6 8 [a y v y] a y d dt (8t) a y 8 m/ Path Path We now calculate correponding value of x and y for variou value of t and plot x againt y to obtain the path a hown. When y, 1 4t, o t 5. For thi value of the time, we have v x 5 16(5) 3 m/ v y 8(5) 4 m/ v (3) (4) 5 m/ a (16) (8) m/ The velocity and acceleration component and their reultant are hown on the eparate diagram for point A, where y. Thu, for thi condition we may write v 3i 4j m/ a 16i 8j m/ v x = 3 m/ a x = 16 m/ A A θ = 53.1 v = 5 m/ Helpful Hint a = m/ a y = 8 m/ v y = 4 m/ We oberve that the velocity vector lie along the tangent to the path a it hould, but that the acceleration vector i not tangent to the path. Note epecially that the acceleration vector ha a component that point toward the inide of the curved path. We concluded from our diagram in Fig. /5 that it i impoible for the acceleration to have a component that point toward the outide of the curve.

9 Problem 1-11 The acceleration of a particle a it move along a traight line i given by a = b t + c. If = and v = v when t =, determine the particle' velocity and poition when t = t 1. Alo, determine the total ditance the particle travel during thi time period. Given: b := m 3 c := 1 m := 1m v := m t 1 := 6 Solution: v t bt 1 dv = ( bt + c) dt v = v v + + ct t bt 1 d = v + + ct t d = + v t + b 6 t3 bt 1 When t = t 1 v 1 := v + + ct 1 v 1 = 3. m + c t b 1 + v t 1 6 t 3 c + 1 t := = 67. m The total ditance traveled depend on whether the particle turned around or not. To tell we will plot the velocity and ee if it i zero at any point in the interval bt t :=,.1t 1.. t 1 vt ():= v + + ct If v never goe to zero then 4 d := 1 d = 66. m vt () 4 6 t

10 Problem 1-15 A particle travel to the right along a traight line with a velocity v p = a / (b + p ). Determine it poition when t = t 1 if p = p when t =. Given: a := 5 m b := 4 m p := 5m t 1 := 6 Solution: d p dt = a b + p p p ( b + p ) dp = t a dt o b p p + b p p = at Gue p1 := 1m Given b p1 p1 p + b p = at 1 p1 := Find( p1 ) p1 = 7.87 m

11 Problem 1-39 A freight train tart from ret and travel with a contant acceleration a. After a time t 1 it maintain a contant peed o that when t = t it ha traveled a ditance d. Determine the time t 1 and draw the v-t graph for the motion. Given : a :=.5 ft t := 16 d := ft Solution : Guee t 1 := 8 v max := 3 ft Given v max = at 1 d v max t 1 ( ) ( ) 1 a t = 1 + v max t t 1 := Find v max, t 1 v max = ft t 1 = 7.34 The equation of motion t a :=,.1 t 1.. t 1 t c := t 1, 1.1 t 1.. t ( ) := at a v a t a ft ( ) := v max v c t c ft The plot Velocity in ft/ ( ) v a t a 1 v c ( t c ) t a, t c Time in econd

12 Problem 1-44 A motorcycle tart from ret at = and travel along a traight road with the peed hown by the v-t graph. Determine the motorcycle' acceleration and poition when t = t 4 and t = t 5. = 1. Given: v := 5 m t 1 := t := t 3 := t 4 := t 5 := dv Solution: At t := t 4 Becaue t 1 < t 4 < t then a 4 = = dt ( ) v 1 4 := v t 1 + t 4 t 1 4 = 3. m At t := t 5 Becaue t < t 5 < t 3 then v m a 5 := a 5 = 1. t 3 t ( ) 1 5 := t 1 v + v t t 1 + ( ) 1 v t 3 t 1 t 3 t 5 v t 3 t 5 t 3 t ( ) 5 = 48. m

13 Problem 1-48 The velocity of a car i plotted a hown. Determine the total ditance the car move until it top at time t = t. Contruct the a-t graph. Given : v := 1 m t 1 := t := 4 8 Solution : ( ) 1 d := v t 1 + v t t 1 d = 6. m The graph τ 1 :=,.1 t 1.. t 1 a 1 τ 1 ( ) := m ( ) τ := t 1, 1.1 t 1.. t a τ := v t t 1 m Acceleration in m/^. a 1 ( τ 1 ) a ( τ ) τ 1, τ Time in econd

14 Problem 1-75 The path of a particle i defined by y = 4kx, and the component of velocity along the y axi i v y = ct, where both k and c are contant. Determine the x and y component of acceleration. Solution : y = 4 k x y v y = 4 k v x v y + y a y = 4 k a x v y = a y = ct c ( ct ) + y c = 4 k a x a x = c k ( y + c t )