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31 60

  1. 1. PROBLEM 11 .31 The velocity of a slider is defined by the relation v = v'sin(w, ,t + p). Denoting the velocity and the position of the slider at I = 0 by v0 and x0, respectively, and knowing that the maximum displacement of the slìder is 2x0, show that (a) u‘ = (v3 + xgwî)! 2x00" , (b) the maximum value of the velocity occurs when x = x0 [3 — (v0 / x0w, , )2:| / 2. SOLUTION (a) Given: v = v’ sin (wnr + p) Att=0, v= v=v’sin or sin = v—° 0 9’ 97 v, Letxbe maximum at t = ti when v = 0. Then, sin(a>, ,t, + p) = 0 and c0s(az, ,t, + p) = il Using = v or dx = vdt d! Integrating, x = C — Lcos(a), ,t + p) n v’ v’ Att= O, x= x0=C———cosp or C= x0+—cosp wn mm v! v’ Then, x = x0 + ("—cosp — —c0s(w, ,t + p) Il Il I v1 v . xmx = x0 + —cosp + — usmg coswntl + p = -1 m m» (xmax — x0)wn _1 Solving for cos p, cosp = l v With xmx = 2x0, 2 2 Using sinz p + coszp = 1, or + (n55 — l) = 1 Solving for v’ gives v’ =
  2. 2. PROBLEM 11.31 CONTINUED (b) Acceleration: a = g = 'a1,, cos (0), ; + p) Letvbemaximumatt = t2 whena=0. Then, c0s(a2,, t2 + p) = 0 From equati0n (3), the corresponding value of x is v’ v’ x00)" v' x= x0+—cosp= x0+— —l =2x0——— a) a) v’ a)
  3. 3. PROBLEM 1 1 .32 The velocity of a particle is v = v0 [l — sin(7tt/ Knowìng that the particle starts from the origin with an initial velocity v0 , determina (a) its position and its acceleration at t = 3T, (b) its average velocity during the intervalt=0tot= T. SOLUTION (a) g‘; = v = v0|:1— stria} f [x]; = | :v0t + fleosfl] 7’ o x= vt+ficos”—'—fi (1) O 7t T 7t When r = 3T, x = 3v0T + M0043”) — i = (3 — Ì)v0T T 7t 7t x = 2.36v0T 4 dv 7tv0 7t! a = — = ———c0s—— dr When t=3T, a= —%cos3n a= %4 (b) Using equation (1) with t = T, Average velocity is
  4. 4. PROBLEM 11.33 An airplane begins its take-off nm at A With zero velocity and a constant acceleration a. Knowing that it becomes airbome 30 s later at B and that the distance AB is 900 n1, deterrnine (a) the acceleration a, (b) the take- off velocity v3. SOLUTION Constant acceleration. v0 = v A = 0, x0 = x A = 0 v= v0+at= at x= x +vt+lat2 = iat2 0 0 2 2 (2) At point B, x = x0 = 900 m and t = 30 s 2 900 S0lving(2) fora, a = 2—f = @ a = 2 m/ s’ 4 r (so) Then, v0 = at = (2)(30) v0 = 60 m/ s 4
  5. 5. PROBLEM 11.34 Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a 250-m ramp at a high speed v0 and travels I80 m in 6 s at Constant deceleration before its speed is reduced to vo/ 2. Assuming the same constant deceleration, determina (a) the additional time required for the truck to stop, (b) the additional distance traveled by the truck. SOLUTION Constant acceleration. Solving (l) for a. Then, At! =6s, and I80 = î(v0 + îv0)(6) = 4.5v0 or v0 = E 4.5 = 40 m/ s v= lv0=2Omls 2 _ 20-40 10 Then, from (3), a 6 = —Î m/ sz = —3.333 m/ sz Substituting into (1) and (2), v = 40 — 3.333! x = 40 + 40: — %(3.333)F Atstopping, v = 0 or 40- 33330 = 0 r, = 12 s x = 0 + (40)(12) — %(3.333)(12)’ = 240 m (a) Additional time for stopping = 12 s — 6 s (b) Additional distance for stopping = 240 m — I80 m
  6. 6. PROBLEM 11.35 -î>v , _ A truck travels 540 fi in 8 s while being decelerated at a constant rate of ‘—"‘" ‘ “' ‘i 1.5 fi/ sz. Determine (a) its initial velocity, (b) its fmal velocity, (c) the j r” distance traveled during the first 0.6 s. SOLUTION a = —l.5 ft/ sz x0 = 0 x= x0+v0t+èalz (a) Solving for v0 using x = 540 fi when t = 8 s, 1 2 , X ‘ ‘o ‘ 50W _ 540 — 0 — (0.5)(—1.s)(s)‘ v Ì ° r 8 v = v0 + a! = 73.5 + (—l.5)(8) x = 0 + (73.s)(0.5) + à-(—l.5)(0.6)2 x = 43.8 a 4
  7. 7. PROBLEM 11.36 A motorist enters a freeway at 25 mi/ h and accelerates uniformly to 65 mi/ h. From the odometer in the car, the motorist knows that she traveled 0.1 mi while acoelerating. Determine (a) the acceleration of the car, (b) the time required to reach 65 mi/ h. SOLUTION Constant acceleration. v0 = 25 mi/ h = 36.667 fi/ s vf = 65 mi/ h = 95.333 fi/ s x0=0 and xf=0.lmi=528fi 2 v, = v3 + 2a(xf — x0) v} — v3 _ 95.333’ — 36.6672 ——— = 7.3333 m 2 2(x, — x0) 2(52s - 0) s (a) a= (b) vf = v0 +atf = v, — vo _ 95.333 — 36.667 t’ a 7.3333
  8. 8. PROBLEM 11.37 A group of students launches a model rocket in the vertical direction. î, I Based on tracking data, they detemtine that the altitude of the rocket was 27.5 m at the end of the powered portion of the flight and that the rocket landed 16 s later. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that g = 9.81 m/ sz, deterrnine (a) the speed v, of the rocket at the end of powered flight, (b) the maximum altitude reached 275 m by the rocket. SOLUTION Constant acceleration. Choose t = 0 at end ofpowered flight. Then, y, = 27.5 m a = -g = -9.81 m/ s’ (a) When y reaches the ground, y, = 0 and t = 16 s. Ì 2 I z yf= y'+VlÎ+-2—0Î = y,+v, !—îgt z y, — y, + tg8 = o — 27.5 + g(9.81)(16)’ v, r 16 = 76.76 m/ s v, = 76.8 m/ s 4 (b) When the rocket reaches its maximum altitude ymx, v = 0 v2 = vu’ + 2a(. v — yn) = V12 - 28(. v - m) J’ = 7| ' v2 _ vlz 2g ym = 27.5 — ° ’ 76'762 ym, = 328 m 4 (2)(9.8l)
  9. 9. PROBLEM 11.38 A sprinter in a 400-m race accelerates uniformly for the first 130 m and then runs with constant velocity. [f the sprinter’s time for the first 130 m is 25 s, determine (a) his acceleration, (b) his final velocity, (c) his time for the race. SOLUTION (a) During the acceleration phase x = x0 + v0t + àatz Using x0 = 0, and v0 = 0, and solving fora gives a = E r2 Noting thatx = 130 mwhent = 25 s, 1 a= a=0.416m/ s4 <25) (b) Final velocity is reached at t = 25 s. v, = v0 + at = 0 + (0.416)(25) v, =10.40 m/ s 4 (c) The remaining distance for the constant speed phase is Ax = 400-130 = 270m For constant velocity, At = È = E = 25.96 s 10.40 V Total time for run: t = 25 + 25.96 t = 51.0 s 4
  10. 10. PROBLEM 11.39 m. SOLUTION Constant acceleration (a, and a2) for horses I and 2. Let x = 0 and I = 0 when the horses are at pointA. Then, x = v0t + àatz = 2(x — v0t) Solving for a, a r2 Using x = 400 m and the initial velocitìes and elapsed times for each horse, 2[400 — (6.8)(61.5)] a, = x 7"‘ = 2 = 9.6239 x 10-3 m/ sz n (61.5) a, = î” ’2V2’2 = îb‘) ‘ (7'°)2(62'°)J = —17.6899 x104 m/ sz ‘z (62.0) Calculating x, — x2, x, — x2 = (v, — v2)t + %(a, — a2)? x, — x, = (6.8 — 7.0): + %[(—9.6239 x 10-3) — (—17.6899 x 104 )]r1 = —0.2r + 8.066 x 10"? AtpointB, x, — x, = 0 —0.2z, , + 4.033 x 104:}, = 0 ÎB = = S 4.033 x 10 Calculating x0 using data for either horse, Horse 1: x0 = (6.8)(49.59) + %(—9.6239 x 10‘3)(49.59)’ Horse 2: x0 = (7.0)(49.59) + %(—17.6899 x 10'3)(49.59)’ = 325 m When horse 1 crosses the finish line at t = 61.5 s, x, — x2 = —(0.2)(61.5) + (4.033 x 10'3)(6l.5)2 In a close hamess race, horse 2 passes horse I at point A, where the two velocitìes are ‘U2 = 7 m/ s and v, = 6.8 m/ s. Horse 1 later passes horse 2 C at point B and goes on to win the race at point C, 400 m fiom A. The 1 2 A E x Îî elapsed times from A to C for horse 1 and horse 2 are I, = 61.5 s and timmé I2 = 62.0 s, respectively. Assuming uniform accelerations for both horses between A and C, determine (a) the distance flom A to B, (b) the position of horse 1 relative to horse 2 when horse 1 reaches the finish line C. x0 =325m4 Ax=2.95m4
  11. 11. PROBLEM 11 .4O explosion. SOLUTION Choose x positive upward. Constant acceleration a = —g Rocket launch data: ROCkCÎ A: X = 0, V = Vo, Î = 0 R0cketB: x=0,v= v0,t= t,=4s Velocities: Rocket/ î: v4 = v0 — gt RocketB: v, = v0 - g(! — r5) Positions: Rocket A: x A = v0! — Ègrz 1 kocketBzx, =v0(t—r, ,)-%g(r—r, )‘, 121,, For simultaneous explosions at x A = x3 = 80 m when! = 1E, l 2 l 2 1 3 l 2 VME '58’: = V0('5 "n)"îg('e "'B) = "da " V001 "E875 + 875% ‘îgîs Solving for v0, v0 = gtE — % 2 Themwhent = r5, x14 = [gtE — %)! E — àgtè, or 1,2; — tBIE — i = 0 t i hl + 4 1) 34 4- ’(4)2+(4)(1)(2}{80) Solving rom» ÎF - B B ( X ( g ) - L43‘; = 6.5075 2 2 Two rockets are launched at a fireworks performance. Rocket A is launched with an initial velocity v0 and rocket B is launched 4 seconds later with the same initial velocity. The two rockets are timed to explode simultaneously at a height of 80 m, as .4 is falling and B is rising. Assuming a constant acceleration g = 9.81 m/ sz, determine (a) the initial velocity v0, (b) the velocity of B relative to A at the time of the (l)
  12. 12. PROBLEM 11.40 CONTINUED (a) From equation (1), q, = (9.s1)(6.507) — w vo = 442 m/ s t 4 At time r5, VA = v0 — gtE v3 = vo — g(tE —— t”) v, — v, = gr, = (9.81)(4) v3” = 39.2 m/ s H
  13. 13. PROBLEM 1 1 .41 A police ofiìcer in a patrol car parked in a 65 mi/ h speed zone observes a passing automobile traveling ata slow, constant speed. Believing that the driver of the automobile might be intoxicated, the officer starts his car, aecelerates uniformly to 85 mi/ h in 8 s, and, maintaining a constant velocity of 85 mi/ h, overtakes the motorist 42 s afier the automobile passed him. Knowing that 18 s elapsed before the officer began pursuing the motorist, detemiine (a) the distance the officer uaveled before overtaking the motorist, (b) the motorist’s speed. SOLUTION Motion ofmotorist: v, ,, = v, " = constant, x, " = v, ,,t Motion of patrol car: Starts at r = 18 s with constant acceleration At r=18+8=26s, vp=85mi/ h=l24.67fi/ s V 124.67 Vp = a(t—l8) d= j=èT8=li583 fi/ Sz 1 2 xp = îafi — 18) At t = 26 s, x, = %(15.583)(26 — 18)’ = 498.67 n For t 2 26 s, vp = 124.67 fi/ s xl, = 498.67 +124.67(t — 26) (a) At t = 42 s, x, = 498.67 +124.67(42 — 26) = 2493.4 a x, = 0.472 mi 4 (b) At r = 42 s, x", = v, ,,(42) = xp = 2493.4 a _ 2493.4 v", 42 = 59.37 fi/ s v, ,, = 40.5 mi/ h 4
  14. 14. PROBLEM 11.42 v‘. n lilla As relay runner A enters the GS-ft-long exchange zone with a speed of ‘n -v " 42 lì/ s, he begins to slow down. He hands the baton to mnner B 1.82 s later as they leave the exchange zone with the same velocity. Detennine (a) the uniform acceleration of each of the mnners, (b) when runner B should begin to run. SOLUTION Let x = O at the start of the exchange zone, and t = 0 as runner A enters the exchange zone. Then, (v A )° = 42 fi/ s. Motion ofrunnerA: v, = (vno + a‘! 1 2 x, = (vd)ot+îa, ,t (a) Solving for a A, and noting that x, = 65 fi when t = 1.82 s 2 x4 —(v, )or 2| 65 —(42)(l.82)| “A = 4.91a“: 4 a‘ ' z’ 1.822 m: = 1.8 s, v = 42+ -6.91 l.82 = 29.43 fi/ s ( A); ( X l Motion of runner B: (v, )o = 0 and x, = 0 at the starting time t, of runner B. Then, v}, — (v, É = 20,24;, Solving for 0,, and noting that v, = (VA )f = 29.43 fl/ s = 3% 29'432 ‘ ° a, = 6.66 n/ s’ 4 ha _ (2)(65) When x, = 65 fi, a, v, = (v, )o + a, (l — tu) = a, ,(t -t, ) r—r, ,=v—” "n (b) Solving for 1,, and noting that v, = 29.43 fl/ s when t = 1.82 s E = —2.60 s 6.66 Runner B should begin 2.60 s before mnner A reaches the exchange zone. 4
  15. 15. PROBLEM 11.43 i ‘In a boat race, boat A is leading boat B by 38 m and both boats are traveling at a constant speed of 168 km/ h. At t = 0 , the boats accelerate at constant rates. Knowing that when B passes A, t = 8 s and v A = 228 km/ h, determine (a) the acceleration of A, (b) the acceleration of B. SOLUTION (a) Acceleration of A. vA. = (vA)0 + aAt, (m0 =168 km/ h = 46.67 m/ s Att=8s, vA=228krn/ h=63.33m/ s aA = 11,4 —t(v, ), = 63.33 g 46.67 “A = M8 m, A 1 1 (b) xA = (xA)0 + (vA)0t + îaAtz x, = (x, )0 +(vB)0t + 50,12 "A ‘ x8 = ("Ala ‘ (‘alo + [Mlo ’ W001‘ * ÈÙA ‘ da)’ When t = 0, (xA)0 — (xB)0 = 38 m and (vB)o — (vA)0 = 0 Whent=8s, xA—x, ,=0 Hence, 0 = 38 + %(aA — a, ,)(8)’, or aA — a, = —l.1875 a, = a, + 1.1875 = 2.08 + 1.1875 a, = 3.27 m/ s’ 4
  16. 16. PROBLEM 11 .44 Car A is parked along the northbound lane of a_ highway, and car B is traveling in the southbound lane at a constant speed of 96 km/ h. At t = 0, A starts and accelerates at a constant rate a A, while at I = 5 s, B begins to slow down with a constant deceleration of magnitude a A /6. Knowing that when the cars pass each other x = 90m and 11A = '12,, determine (a) the acceleration a A, (b) when the vehicles pass each other, (c) the distance between the vehicles at t = 0. SOLUTION (a) Acceleration of A. 1 vA = (vA)o+aAr and xA = (xA)0+(vA)0t= îaAt2 (vA)o = 0 and (xA)o = 0 gives 1 2 vA = aAt and xA = îaAt When carspass at r = t], xA = 90m 2 = a - = L82 and vA = aAt, A t, _ “A a “A For 0 s 1 s 5 s, v, = (v, )o = -96 km/ h = —26.667 m/ s For! > 5 s, v3 = (v5), + aB(t — 5): —26.667 + Î15—aA(t — 5) When vehicles pass, VA = ‘V8 aAt, = 26.667 — è-aAfil — 5) ÌaAr, — ÎaA = 26.667 or 7:, — 5 = —- 6 6 al‘ t] = flgives 7180 —5=É2 “A E “A 7J180u — 5 = 160u’ 160142 — 7J180u + 5 = 0
  17. 17. PROBLEM 11.44 CONTINUED Solving the quadratic equation, A = 74/180 i , )(49)(180)-(4)(l60)(5) = 93315 i 74357 u (2)(160) 320 = 0.0592125 and 0.52776 aA = L, = 285.2 m/ s and 3.590 m/ s Il The corresponding values for t, are 1, = (E =0.794s, and 1, - i = 7.08s 285.2 3.590 Reject 0.794 s since it is less than 5 s. Thus, (b) Time ofpassing. (c) Distance d. 0 S t S 5 s, x, = (163), ) — (v, )ot = d - 26.6671 At 1 = 5 s, x, = d — (22.667)(5) = d —133.33 F01‘! > s s, x, = a — 133.33 + (v, )o(t — 5) + %a, ,(1 — s)’ x, = d -133.33 — 26.667(t — 5) + %(3‘T59)(1 - 5)’ Whent= t,=7.08s, xB= xA=90 90 = d - 133.33 - (26.667)(2.08) + èîgfèysy d = 90 +133.33 + 55.47 + 1.29 aA = 3.59 m/ s 4 1=1,=7.08s4 d =278m4
  18. 18. PROBLEM 11.45 Two automobiles A and B traveling in the same direction in adjacent lanes are stopped at a traffic signal. As the signal tums green, automobile A accelerates at a constant rate of 6.5 fi/ s’. Two seconds later, automobile B starts and accelerates at a constant rate of 11.7 fi/ s’. Detennine (a) when and where B will overtake A, (b) the speed of each automobile at that time. SOLUTION For 1> 0, x, = (11)), +(v, )01 fio)’ = 0 + 0 +%(6.5)t’ or x, = 3.251’ For 1 > 2s, x, 41,), +(vB)0(t—2)+%aB(t—5)2 = 0 +0+%(11.7)(1- 2)’ or x), = 5.85(1 — 2)’ = 5.851’ - 23.41 + 23.4 For x A = x, ,, 3.251’ = 5.851’ — 23.41 + 23.4, or 2.601’ - 23.41 + 23.4 = 0 Solving the quadratic equation, t = 1.1459 and t = 7.8541 s Reject the smaller value since it is less than 5 s. (a) 1 = 7.85 s 4 r, = x, = (3.25)(7.8541)’ x = 200 a 4 v, = (u), + aAt = o + (6.5)(7.854l) v, = 51.1 fi/ s 4 v, = (1), ), + a, ,(1 — 2) = 0 + (lI.7)(7.8S4l — 2) v, = 68.5 fi/ s 4
  19. 19. PROBLEM 11 .46 Two automobiles A and B are approaching each other in adjacent highway lanes. At t = 0, A and B are 0.62 mi apart, their speeds are 04:68 mi/ h and v, =39 mi/ h, and they are at points P and Q, respectively. Knowing that A passes point Q 40 s afcer B was there and that B passes point P 42 s afier A was there, determine (a) the unifoxm accelerations of A and B, (b) when the vehicles pass each other, (c) the speed of B at that time. SOLUTION Let x be the position relative to point P. Then, (xA)° = 0 and (xB)0 = 0.62 mi = 3273.6 fi Also, (vA)0 = 6.8 mi/ h = 99.73 fi/ s and (v, )0 à —39 mi/ h = -57.2 fi/ s (a) Uniform accelerations. 1 2 x —(x ) —(v ) t x4 = (xA)o +(vA)0t+îa, t2 or aA = A ‘t: A ° a, = 2 32736 ' o " (99'73)(4°) = —0.895 m’ a, = 0.895 n/ s’ —— Q1 <40)’ 2 ‘a “("90 ‘(Vfiofl l x, = (x, )0 + (v, )0 + îafltz or a, = r2 a = 2[0 - 3273.6 — (—57.2)(42) ” <42)’ (b) When vehicles pass each other x A = x, . = —0.988 01s’ a, = 0.988 ius’ -— 4 . 1 1 (x, )o + (mo: + —a, :2 = (x, ,)0 + (v, )0t + —a, :’ 2 2 0 + 99.73: + -%(—0.895)t2 = 3273.6 — 57.2: + à-(—0.988)t2 0.046512 - 156.93: + 3273.6 = 0 Solving the quadratic equation, t = 20.7 s and —3390 s Reject the negative value. Then, (c) Speed ofB. v, = (ma + a5! = -57.2 + (-0.988)(20.7) = —77.7 fi/ s
  20. 20. PROBLEM 11.47 Block A moves down with a constant velocity of 2 fi/ s. Detennine (a) the velocity of block C, (b) the velocity of collar B relative to block A, (c) the relative velocity of portion D of the cable with respect to block A. SOLUTION Let x be positive downward for all blocks and for point D. v A = 2 fi/ s Constraint ofcable supponingA: x À + (x A — x8) = constant 2v,4—v, =0 or v, ,=2vA= (2)(2)=4fi/ s Constraint of cable supporting B: 2x, + xc = constant vc + Zv, = 0 or vc = —2v, = —(2)(4) = -8 fi/ s (a) vc = 8 fl/ s f «t (b) v, ,,, =v, ,-v, =4—z v, ,,, =2fi/ sl< (c) x0 + xc = constant, vD + vc = 0 v0 = —8vc = 8 fi/ s vDM= vD—vA=8——2 vDM=6fi/ s1‘(
  21. 21. PROBLEM 11.48 Block C starts fiom rest and moves down with a constant acceleration. Knowing that afier block A has moved 1.5 fi its velocity is 0.6 tì/ s, determine (a) the accelerations of A and C, (b) the velocity and the change in position of block B alter 2 s. SOLUTION Let x be positive downward for all blocks. Constraint of cable supportingA: x A + (x A — x3) = constant ZVA-VB =0 or v, =2VA and a, =2aA Constraint of cable supporting B: 2x, + xc = constant 2V, + Vc '= »0, or vc = —2vB, and ac = 42a, = —-4aA Since vc and ac are down, VA and a, areup, i. e. negative. V3 -(VA): = 2aA[xA —(xA)0:l 2 _ 2 2 _ (a) a, = V‘ ("k = (oflîo = —0.12fi/ s2 a, = 0.12 a/ s’ i4 2| ‘A ‘ (xÙo 0X45) ac = 4a, ac = 0.48 a/ s’ 1 4 (b) a, = 2a, =(2)(—0.12)= —0.24 tvsî 71v, = a, ,: = (—0.24)(2) = -0.48 fi/ s Av, = 0.48 a/ s î 4 1 1 Ax, = îafi = î(—o.24)(2)2 = -o.4s a Ax, = 0.48 a I 4
  22. 22. PROBLEM 11.49 Block C starts from rest and moves downward with a constant acceleration. Knowing that afier 12 s the velocity of block A is 456 mm/ s, determine (a) the accelerations of A, B, and C, (b) the velocity and the change in position of block B afler 8 s. SOLUTION Let x be position relative to the fixed supports, taken positive if downward. Constraint of cable on lefi: 2x , + 3x, = constant 2 2 2V, + 3V, = 0, or v, = —3V, , and a, = -Îa, Constraint ofcable on right: x, + Zxc = constant l l l v, + Zvc = 0, or vc = ——V, , = 51', , and ac = 3a, Block C moves downward; hence, block A also moves downward. (a) Accelerations. v; = (VAL, + 414T or a4 = m” x04)‘, = “Îz- o = 38.0 mm/ sz a, = 38.0 mm/ sz l 4 a, = —%a, = —[%)(3s.0) = —25.3 mai/ s’ a, = 25.3 mm/ sz 1 4 ac = àa, = G-)(38.0) = 12.67 mm/ s’ nc = 12.67 mm/ sz 1 4 (b) Velocity and change in position of B afier 8 s. V, = (V, )° + 0,! = 0 + (—25.3)(8) = —203 mm/ s v, = 203 mm/ s I 4 1 1 x, — (x, )° = (V, )0t + 30,12 = o + î(—25.3)(8)2 = -811 mm Ax, =811mmî4
  23. 23. PROBLEM 11.50 the cable with respect to block B. SOLUTION Let x be position relative to the fixed supports, taken positive if downward. Constraint ofcable on Ieft: 2x, + 3x, = constant 3 V, = ——V, 2V, + 3V, = 0, or 2 Constraint of cable on right: x, + 2xc = constant V, + 2Vc = 0 or vc = ——v, Constraint of point on D on cable: 2x , + xD = constant 2V, + VD = 0 or VD = —2V, (a) Velocity of block A. 3 3 v, = 7V, = —î(6l0) = —915 mm/ s (b) Velocity of block C. Vc = ——V, = —%(6l0) = —305 mm/ s (c) Velocity ofpoint D. VD = -(2)(—9l5) (a) Velocity relative to B. vu, = VD — V, = 1830 — 610 Block B moves downward with a constant velocity of 610 mm/ s. Detennine (a) the velocity of block A, (b) the velocity of block C, (c) the velocity of portion D of the cable, (d) the relative velocity of portion D of v, =915mnvsî4 vc=305mm/ sT4 v, , =1830mnvs14 vu, = 1220 mm/ s 14
  24. 24. PROBLEM 11.51 ln the position shown, collar B moves to the left with a constant velocity of 300 mm/ s. Detennine (a) the velocity of collar A, (b) the velocity of portion C of the cable, (c) the relative velocity of portion C of the cable with respect to collar B. SOLUTION Let x be position relative to the right supports, increasing to the lefi. Constraint of entire cable: 2x A + x, + (x5 — x A) = constant 2V, , + v, ‘ = 0 VA = —2v, Constraint ofpoint C of cable: 2x A + xc = constant 2vfl+vc =0 vf = —2vA (a) Velocity of collar A. v, = —2v, = —(2)(300) = —600 mm/ s v, = 600 mm/ s —- ‘C (b) Velocity of point C of cable. vc = —2v, = —(2)(—600) = l200 mm/ s vC = 1200 mm/ s % 4 (c) Velocity of point C relative to collar B. v0, = vc — v, = 1200 - 300 = 900 mm/ s v0, = 90o mm/ s -— 4
  25. 25. PROBLEM 11.52 Collar A starts fiom rest and moves to the right with a constant acceleration. Knowing that afier 8 s the relative velocity of collar B with respect to collar A is 610 mm/ s, determine (a) the accelerations of A and B, (b) the velocity and the change in position of B after 6 s. SOLUTION Let x be position relative to the right supports, increasing to the lefi. Constraint of entire cable: 2x A + x1, + (x, — x A) = constant, 2V, , + VA = 0, or v3 = —-; -vA, and a, — —îaA (a) Accelerations of A and B. 1 2 vB/ A = V13 ‘ VA = ‘EVA ‘ VA VA = ‘ÎVa/ A v, = ——: -(610) = —406.67 mm/ s v, - m)‘, = —406.67 — o t = —50.s mm/ s’ VA —(vA)o = aAt, aA — a, = 50.8mm/ s2 —— 4 1 1 a, = 7 A = —î(—114.4) a, = 25.4 mm/ sz -— 4- (b) Velocity and change in position of B afier 6 s. v1, = (v, ,)o + a”! = 0 + (50.8)(6) v3 = 152.5 mm/ s ——- 4 1 1 x, — (x, )0 = (vB)Ot + —2—a, t2 = î(25.4)(6)2 Ax, = 458 mm -— 4
  26. 26. PROBLEM 11.53 At the instant shown, slider block B is moving to the right with a constant acceleration, and its speed is 6 in. /s. Knowing that after slider block A has moved 10 in. to the right its velocity is 2.4 in. /s, determine (a) the accelerations of A and B, (b) the acceleration of portion D of the cable, (c) the velocity and the change in position of slider block B afier 4 s. SOLUTION Let x be position relative to lefi anchor. At the right anchor, x = d. Constraint ofcable: x, + (x, — xA) + 2(d — xA) = constant 2V3 — 3vA = 0 or vA = àv, and aA = àa, Constraint ofpointD of cable: (d — xA + d — x0) = constant vA +vD = 0 or vD = —vA and a0 = —aA (a) Accelerations of A and B. . 2 . (v, )0 = 6 m. /s (vA)o = = 4 mJs V124—(VA)à = 2aA[xA ‘(XA)OJ vi - (mi _ 2.4’ — 4’ = _î= -0.512' . /’ = 0.512‘ 120-4 "‘ (2)00) '“ “‘ "” a, = àa, = %(0.s12) = —0.768 ia/ s’ a, = 0.768 art/ s‘ -— «t (b) Acceleration ofpoint D. a0 = —aA = —(—0.512) a0 = 0.512 in. /s2 —- <4 (c) Velocity of block B afier 4 s. v, = (v, ,)° + a, ,r = 6 + (—0.768)(4) v, = 2.93 m/ s’ —— 4 Change in position of block B. x, — (x, )o = (mo: + èagtz = (s)(4) + %(—0.768)(4)2 Ax, = 17.86 in. —— 4
  27. 27. PROBLEM 11 .54 Slider block B moves to the right with a constant velocity of l'2 in. /s. I Detennine (a) the velocity of slider block A, (b) the velocity ofportion C D of the cable, (c) the velocity of portion D of the cable, (d) the relative velocity of portion C of the cable with respect to slider block A. SOLUTION Let x be position relative to lefi anchor. At right anchor x = d. Constraint ofentire cable: x5 + (x3 — xA) + 2(d — xA) = constant 2115 — 3vA = 0 (a) Velocity ofA: vA = %v, = %(12) vA = 8.00 in. /s —— 4 Constraint ofpoint C of cable: x3 + x3 — xc = constant 2V, , — vc = 0 (b) Velocity of C: vc = 2V, , = 2(12) vC = 24 in. /s ——— 4 Constraint ofpoint D of cable: d — x A + d — xc = constant v A + vD = 0, (c) Velocity of D: vD = —v A = -8.00 in. /s vD = 8.00 in. /s «——- 4 (d) Relative velocity. vc, A = vC — v A = 24 — 8 vc, A = 16.00 in. /s ——— 4
  28. 28. PROBLEM 11.55 Collars A and B start from rest, and collar A moves upward with an acceleration of 312 mm/ sz. Knowing that collar B moves downward with a constant acceleration and that its velocity is 200 mm/ s afier moving 800 mm, determine (a) the acceleration of block C, (b) the distance through which block C will have moved afier 3 s. SOLUTION Let x be position relative to upper support, positive downward. Let d = value of x at Iower support. Constraint ofcable: (d — xA) + (xc — xA) + 2xc + (xc — x, ) = constant 4vc—v, —2vA =0 and 4ac—a, ,—2aA =0 Accelerations: vÎ, — (v, L2) = Zagh, — (x, )0:| a = v}, —(va)É = 2003-0 flsmmz ” 2(s00—0) 1 1 (a) ac = Z(a, , + 20A) = Z[25 + (2)(—3:’)] ac = (6.25 — 1.5:’) mm/ sz 1 4 (b) Velocity and position: vc = (ma + Eacdl = 0 + 6.25: — 0.5:‘ xc — (xc)0 = gvca: = 3125:’ — 0125:‘ xc — (xc)o = (3.125)(3)2 — (0.125)(3)‘ Axc = l8.00mm 14
  29. 29. PROBLEM 11.56 Collar A starts fiom rest a1 t = 0 and moves downward with a constant acceleration of l80 mm/ sz. Collar B moves upward with a constant acceleration, and its initial velocity is 200 mm/ s. Knowing that collar B moves through 500 mm between I = 0 and t = 2 s, determine (a) the accelerations of collar B and block C, (b) the time at which the velocity of block C is zero, (c) the distance through which block C will have moved at that time. SOLUTION Let x be position relative to upper support, positive downward. Let d = value of x at Iower support. Constraint ofcable: (d — xA) + (xc — xA) + 2xc + (xc — x, ) = constant 4vc—v, ,-—2vA=0, and 4ac—a, —2aA=0 (a) Accelerations of B and C. a, = I80 mm/ sz, (ma = o l v, = (v, )o + agì, x, — (x, )o = (v, )or + EaBIZ a = 2 x’ _(x”)° —(v”)°‘ = 2 ‘sin-Hooxz) = -50 mm/ s’ a = 50 mm/ s’ l 4 B I2 (2): B ac = %(a, , + 20A) = %(—s0 + (2)(1a0)) ac = 77.5 mm/ sz 1 4 (b) When vc = 0. )o = [(0,1) + 2(vA)o] = %(—200 + 0) = -50 mm/ s t: v. ‘(Velo = o-(—so) ac 77.5 xc — (xc )0 = (vc)o! +%a,12 = (—50)(o.64s) + %(77.5)(0.645)2 = —l6.l3mm Axc = l6.l3mm14
  30. 30. PROBLEM 11.57 Slider block B moves to the lefi with a constant velocity of 2 in. /s. At t = 0, slider block A is moving to the right with a constant acceleration and a velocity 014 in. /s. Knowing that at : = 2 s slider 111061: c has moved i 1.5 in. to the right, detennine (a) the velocity of slider block C at t = 0, (b) the velocity of portion D of the cable at t = 0, (c) the accelerations of A and C. SOLUTION Let x be position relative to the anchor, positive to the right. Constraint ofcable: -—x3 + (xc — x3) + 3(xc — x A) = constant 4vc—2v3—3vA=0 4ac—2a3-3aA=0 When t = 0, v3 = -2 in. /s and (v, )0 = 4 in. /s (a) (vc), = H21), + 303), ] = %[(2)(—2) + (3)(4)] (wc), = 2 in. /s _. 4 Constraint ofpointD: (x3 — xA) + (xc — xA) + (xc — x3) — x3 = constant vD+2vc—2vA—2v3=0 (vo), = 202). , + 2V» - 2(vC)o = (’)(4)+(2)(-2) - (2)(2) 1 xc — (xc)0 = (vc)°t + îact’ (c) dc = lw xc Ù 0C)“ — 0C)", = 142 l'5 _ mm] = —1.250 1117s’ ‘z (’)2 ac =1.250 1117s’ -— 4 a, = %(4ac — 2a3) _= %[(4)(—1.250) —«(2)(0)] = —1.667 1117s’ a, = 1.667 1117s’ -— 4
  31. 31. PROBLEM 11.58 Slider block A starts with an initial velocity at t = O and a constant acceleration of 9 in. /s2 to the right. Slider block C starts from rest at t = 0 and moves to the right with constant acceleration. Knowing that at t = 2 s, the velocitìes of A and B are 14 in. /s to the right and l in. /s to the lefi, respectively, determine (a) the accelerations of B and C, (b) the initial velocitìes of A and B, (c) the initial velocity of portion E of the cable. SOLUTION Let x be position relative to the anchor, positive to the right. Constraint ofcable: -x, + (xc — x, ) + 3(xc — x, ) = constant 4vC —2v, -3vA =0 and 4aC—2a, ,—3aA =0 (a) Accelerations of B and C. Al t = 2 s, v’, = 14 in. /s and v, = —l in. /s vc = a2», + su) = %[(2)(—| ) + (3)(14)] = 10 in. /s (vC)o = 0 vc = (vdo + act ac = e = —- ac = 5 in. /s2 —-— 4 a, = —(4ac — 3a, ) = %[(4)(s) — (3)(9)] = —3.5 in. /s2 a, = 3.5 ìn. /s2 -— 4 (b) lnitial velocitìes of A and B. v4 = (vA)0—a4t (vno= vA-aAt= l4—(9)(2)= —4in. /s (v, )0 = —4 in. /s-— 4 v, = (v, )0 — 42,: (v, )o = v, — a, ,: = —1—(—3.5)(2) (v, )0 = 6 in. /s —-— < Constraint ofpoint E: 2(xc — x, ) + (x5 — xA) = constant vE-3vA-4-2vc- =0 (c) (vE)o = 3(v, )o — 2(vC)0 = (3)(—4) — (2)(0) = —12 in. /s (vE)0 = 12 in. /s o- 4
  32. 32. PROBLEM 11 .59 The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and afier l s the relative change in position of block C with respect to block A is 140 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 40 rrun/ s downward, the displacements of A and B are 80 mm downward and 160 mm downward, respectively, determine (a) the accelerations of A and B if a, > 5 mm/ sz , (b) the change in position of block D when the velocity of block C is 300 mm/ s upward. SOLUTION Let x be position relative to the support taken positive if downward. Constraint of cable connecting blocks A, B, and C: 2xA+2xB+xC = constant, 2vA+2vB+vC=0, 2aA+2a, +aC =0 (u), = m), = ("Ho = o. (x11, 4x51, = (xc), (xx/ A), = 0» (van), = 0 (v, ,,, )2 — (vm = 2a, ” [xm - (x, ,,)o] vi, ” — 0 = 2a, ,,, (x, , — x, — 0) "i" = î-‘Oz = 10 mm/ s’ a“ = 2(x, — x, ) 2(l60 -s0) 1 2 1 2 ‘su = (‘a/ A)“ + (VE/ Ah! + ""01! = o ‘l’ o + ‘ala/ A’ 2 2 r2 = 2x, “ or r: 2(x, , —xA) = ‘2060-80) = 45 "su “su lo l x, — (x, )o = (v, )0r + 342412 2 x4 -(x4), , -(v4), ,t = 2(s0- 0) ‘z (4)2 a, =a, +a, ,,, =1o+1o 8B=20mfll/ S21"( (a) a, = a, = 10 mm/ sz 1 «t a‘. = -(2a, , + 211,) = —[(2)(20) + (2)(10)] = -60 mm/ s Vc -(vc). , = -300—0 = aC -60 vc = (vc)o + 0c! t: 5 s Constraint of cable supporting block D: (xD—xA)+(xD—xB)= c0nstant, 2vD—vA—v, , =0 l l 2aD—aA—a, ,=0, a0=î(a,4+a, )=î(l0+20)= l5nun/ s (b) x0 — (xD)0 = (vD)ol + Èabtz = 0 + %(15)(5)’ AxD = 187.5 mm 14
  33. 33. PROBLEM 1 1.60 10 s. SOLUTION Let x be position relative to the support taken positive if downward. Constraint of cable connecting blocks A, B, and C: 2x, + 2x0 + xc = constant 2V, , + 2v0 + vc = 0 2a, + 2a, + ac = 0 Constraint of cable supporting block D: (xD—x, )+(x, _,—x, )=constant, ZVD-vfl-v, =0 20D — a, — a, = 0 Given: a0, = ac — a, = -120 or ac = a, -120 Given: 0D, , = aD — a, = 220 or aD = a, + 220 Substituting (3) and (4) into (1) and (2), 2aA+2aB+(a, —l20)=0 or 2al+3ag = l20 2(aA+220)—aA—a, =0 or a, ,—a, =—440 Solving (S) and (6) simultaneously, a, = —240 mm/ sz and a0 = 200 mm/ s’ From (3) and (4), a0 = 80 mm/ sz and a0 = —2o mm/ sz (a) Velocity of C afier 6 s. v0 = (v0), + a0: = 0 + (80)(6) (b) Change in position of D afier I0 s. l x0 — (x0), = (v0), + €1.012 = o + î(—20)(1o)’ = -1000 mm The system shown starts from rest, and each component moves with a constant acceleration. If the relative acceleration of block C with respect to collar B is 120 mm/ sz upward and the relative acceleration of block D with respect to block A is 220 mm/ sz downward, determine (a) the velocity of block C afier 6 s, (b) the change in position of block D afier (l) (2) (3) (4) (5) (6) vc =480mm/ s1‘( m0=1.o00mî-t

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