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Dynamics Homework 4 - ENGINEERING MECHANICS Dynamics&...

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Unformatted text preview: ENGINEERING MECHANICS - Dynamics & L.D. Sturges 14-44* One end of the 2-m long bar AB shown is attached to a collar which slides on a smooth vertical shaft. The other end of the bar slides on a smooth horizontal surface. At the instant shown, 9 = 300 and collar A is moving downward with a speed of 0.4 m/s. Determine the velocity V73 of end B and the angular velocity wAB of bar A8 at this instant. Solution The collar A has no x—component of A velocity while the end of the rod B has no y-component of velocity. Therefore, the relative velocity equation ZUJAB ‘ VB = VA + vB/A 8 becomes ) A A A O A v 1 = —0.4 _] + 20A8(cos 300 1 + sin 30 J) Separating this equation .into its x— and y-components O : (A) x VB 2 AB cos 30 y. _Oo4 . 20-) 511) 30 and solving gives (I) AB 0.400 rad/s C . ....................... ........... Ans. V 0.693 m/s—) ....... .......... ..............Ans. B ENGINEERING MECHANICS - Dynamics 14-49* Movement of the solenoid plunger causes a gear to rotate. At the instant shown, the angular velocity of the gear is mo = 4 rad/s counterclockwise. Determine the angular velocity wAB of the rod AB and the velocity VA of the plunger at this instant. Solution At the instant shown, (0:0: CD 0 4 rad/s c sin ¢ = 5/13 . cos ¢ = 12/13 The gear teeth prevent slip between the gear and the control rod AB. Therefore, the velocity of the contacting points of the two rigid bodies is the same and using the relative velocity equation to write the velocity of C in terms of the motion of the two separate bodies VC= VD + vC/D = VA + vC/A gives 6 + 5(4)(-cos ¢ 3 + sin ¢ 3) W.F. Riley & L.D. Sturges IZCUAQ ¢ = VA i + 12wA8(—sin ¢ 3 - cos ¢ 3) 0%58 Separating this equation into its x— and y-components x: -20(12/13) v - 12wAB(5/13) A y: 20(5/13) —12wAB(12/13) and solving gives a.) =—. AB 0 694 rad/s = 0.694 rad/s C ...... V = -21.7 in./s 21.7 in./s 6— . ..... ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 14-75* One end of the l8-in. long bar AB is attached to a slider that moves in a vertical slot. The other end is attached to a wheel which :rolls on a horizontal surface. At the instant Eshown, the center of the wheel is moving to the gright with a speed of 5 in/s. For this instant 3. Locate C, the instantaneous center of zero velocity of the bar AB. b. Determine the angular velocity (DAB of the bar AB. c. Determine the velocity 63 of the slider 8. Solution Since slider B moves in a vertical slot, the instantaneous center of zero velocity must lie on the horizontal line through 8. Also, since the end of the rod A moves in a straight line parallel to the horizontal surface, the instantaneous center of zero ivelocity must lie on the vertical line through A. Then, & VA=VC+VA/C o 5=0+(18cos3O)AB (0118:0321 rad/Sc ....... . ...... . ........... Ans. and VB=VC+VB/C o f =0+(185in30)AB = 2.89 in./s T ............ . ....... .. ............. Ans. ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 14—90 In the four—bar linkage shown, the control link CD has a constant clockwise angular velocity of 120 rev/min. For the instant shown, in which B is directly above D a. Locate the instantaneous center of zero velocity of the bar BC. b. Determine the angular velocities wAB 0) . and BC of both bars c. Determine the velocity GB of the pin B. Solution 4n rad/s D. Bar CD is in ll At the instant shown, wCD = 120 rev/min fixed—axis rotation about point D; the velocity of point C is perpendicular to CD and V0 = 5(4H) = 20H in./s Bar AB is also in fixed-axis rotation about point A; the velocity of point B is perpendicular to AB. Therefore, the instantaneous center of zero velocity of bar BC lies at the intersection of the lines AB (which is perpendicular to the velocity 33) and CD (which is perpendicular to the velocity VG). Then -1 o 6 = tan 4/3 = 53.130 ¢ = 1800 — 120° — 9 = 6.8700 C 4’, DE _ 6 = AE sin 120o sin 6.8700 sin 53.13o° (Ar = 40.129 in. DE = 43.441 in. and = t + w Vc (DE 5’30 wBC = zen/48.441 = 1.297 rad/s D ......... ..... ...... Ans. = C + w VB (AE 12)BC° = 67.6 in./s i 30 ........................... ... Ans. = 12w AB w = 5.63 rad/s D ................................. .. Ans. AB W ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 14-96* One end of the 2-m long bar AB shown is ‘attached to a collar which slides on a smooth vertical shaft. The other end of the bar slides mia smooth horizontal surface. At the instant flwwn, 9 = 30° and collar A is moving downward with a constant speed of 0.4 m/s. Determine Um acceleration EB of end B and the angular acceleration “A8 of bar AB at this instant. Solution From the relative velocity equation between A and B, VB = VA + VB/A A A 0 A O A = - . + w + ' VB 1 O 4 J 2 AB‘COS 30 1 Sin 30 J) the x- and y-components /\ u) 0 A8 x: VB = ZQAB cos 30 30F OKAG O y: 0 = -0.4 + 2w sin 30 AB B luAB give V wAB = 0.4 rad/s C 1 am“ VB = 0.69282 m/s -+ Then, from the relative acceleration equation between A and B, ‘=§+§ 38 A B/A a ’1‘ _. o A o A + + ‘ B O ZaAB(cos 30 1 Sin 30 J) 2 o A A + 2(o.4) (-sin 30 1 + cos 30° J) The x— and y— components x: a3 = 2aAB cos 30 — 0.1600 0 : = ' + . y o 2aAB Sln 30 0 27713 give 2 2 AB = —o.27713 rad/s = 0.277 rad/s D ............... Ans. 2 2 a = —0.640 m/s = 0.640 m/s 6— . ........... ... Ans. :ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 14—101* Movement of the solenoid plunger causes a gear to rotate. At the instant shown, the angular velocity of the gear is w = 4 rad/s counterclockwise and the 0 speed of the plunger is decreasing at a 2 rate of 0.25 ft/s . Determine the angular acceleration (to of the gear and the angular acceleration “A3 of the rod AB at this instant . Solution At the instant shown, C (o — w — 4 d/ C . CD — o — m S 5 [2. sin 43 = 5/13 cos ¢ = 12/13 A and the relative velocity equation gives 'D ‘3 fl = —. + —9 = _. + _. Vc VD VC/D VA vC/A 6 + 5(4)(—cos 4) I + sin 4) 3) 1 _ A _ A A 5w. - VA 1 + 12wAB(-Sin ¢ 1 — cos d) J) 50" I ,l : - = - w ‘F\\<::; x 20(12/13) VA 12 ”(s/la) C y: 20(5/13) = -12w (12/13) AB ¢ wo = — _ = . a (DAB 0 6944 rad/s 0 69444 rad/s C D 0 VA = -21.667 in./s = 21.667 in./s <— Then, if the speed of A is decreasing, 8A = 0.25 ft/s —) = 0.25 1 ft/s and the relative acceleration equation gives -. = —o —o = * " 2. ac an + ac/D aA + aC/A lawAB 0 + (5/12)ao(—-c:s <13 ’1‘ + sin 4’ 3) B C was + (5/12)(4) (-sin 43 ’1‘ — cos (b j) c"A8 = 0.25 ’1‘ + (12/12)a (-sin dz ’1‘ — cos d: 3) AB 2 A A mum A + (12/12)(0.6944) (cos 4) 1 — sin 42 J) ¢ X: -O.38462(Io - 2.56410 = 0.25 — 0.384621'1AB + 0.44510 : . a - . = - . a - . y 0 16026 0 6 15365 0 92308 AB 0 18546 2 “AB = 6.76 rad/s C .............................. Ans. 2 2 or = -1.711 rad/s = 1.711 rad/s D ............... Ans. O ...
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