PROBLEM 15.61 AA straight rack rests on a gear of radius r and is attached to a block B as shown. Denoting by wp the clockwise angular velocity of gear D and by @ the angle formed by the rack and the horizontal, derive expressions for the velocity of block B and the angular velocity of the rack in terms of r, @, and ap. SOLUTION Gear D: Rotation about D. Tooth E isin contact with rack AB. KM Vp = rey DO Rack AB. — tan 0 len Plane motion = Translation with £ + Rotation about E. : % fives : : Coe 78s = 7 ie g oe 8 Ye=VetVar — [ve—]=[%e 54 +[vaclZ 4] Draw velocity vector diagram. ee res cos@ cosO Oy = optoPROBLEM 15.62 A straight rack rests on a gear of radius r = 3 jin. and is attached to a block B as shown. Knowing that a the instant shown the angular velocity of gear D is 15 rpm counterclockwise and @ = 20°, determine (a) the velocity of block B, (b) the angular velocity of the rack. SOLUTION Gar 0: Raion thet . Teo inca wi kA ve Vp = rep = (3)(0.5n) = 4.7124 inds ? 3 Rack AB. =. = s2424i ait Plane motion = Translation with £ + Rotation about E. < fred o£ + 8 = Wate ve=VerVae [Ye —]=[%2 9] +[%e0 44] Draw velocity vector diagram. =v, 4704 = 5.02 in/s Rn oe ¥8 = 0520" ~ cos20° > Vp = 5.02 in/s— 4 ® Yas = Ve tan 20° = 4.7124 tan 20° 71517 in/s go te LSI, 49 Teg 8.2424 yy = 0.208 rad/s)a % 5G Ne “Oye PROBLEM 15.63 Bar 4B is rotating clockwise and, at the instant shown, the magnitude of the velocity of point Gis 3.6 m/s. Determine the angular velocity of each of the three bars at that instant. SOLUTION Rod AB. (Rotation about 4) Rod DE. (Rotation about £) Rod BGD. YD =e + Yop Draw velocity vector diagram, % & Plane motion = Vp =(AB)O 4p = 0.12004) S45" Vp = (ED) ope = 0.15 py 245° ‘Translation with B +Rotation about B. [vp 248°] Ys S45") +[vo0] Vpyp = V2v_ = 012020045 = 21NIew 1 By, Ip O26 2 Iyo®ap = $00 = 0.060204» ‘y+ Vo Draw vector diagram. 20) 4p sin 45° VG = vysinds® = % 36 5 ee ees = 424 O.12sinas* ou = Ards }4 — O12sinas™” 1 yp = (.2)(24), a @gp = 30.0 rad/s) 4 Wp = ¥p = 0.12045 = vp _ 5.0912 OS 0912 m/s a yg = 339 rls ) 4 ten,S PROBLEM 15.64 In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE. fr) T L SOLUTION Bar AB. Rotation about A Vp = © gy Tyg = (4K) x (-0.25j) = (1.00 mvs)i Bar ED. Rotation about E Vp = @pgk % Ky = Opgk x (0.075 — 0.15}) = 0.15@p¢ 4 ~ 0.0750 p65 Bar BD. Translation with B + Rotation about B. Vere = @ppk * Fog = @ppk x 0.21 = 0.20%—pj Y= Vat Yon 0.150%p¢4 ~ 0.0750%p¢9 = 1.001 + 0.205) ‘Components: & @pg = ~6.6667 rad/s @pg = 6.67 rad’s ) 4 ~(0.075)(~6.6667 oy = rss nw =220008)4PROBLEM 15.65 In the position shown, bar 4B has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE. SOLUTION Bar AB. Rotation about A. Inumits of in/s Va = O yg % Ty = (Ak) x (10%) = 40 Bar ED. Rotation about E. 2400 p¢1 ~ 60 pe) Vp = Ope * Foe ~ Opgk x (-6i + 2.4) Bar BD. ‘Translation with B + Rotation about B. nok x (-4j) = deagpt Yor = ap * Toe Vp = Va + Vow ~2.4epel 602p¢§ = 40} + 40h Components: bpp = 40 pg = 6.6667 rad/s 3.67 rad/s) I: -24eope = 40m (2MEE) em =4mnats 4 mp 71PROBLEM 15.66 In the position shown, bar AB has an angular velocity of 4 rad/s clockwise. Determine the angular velocity of bars BD and DE. SOLUTION Bar AB, Rotation about A. Vp = © gp % Fyig = (AK) x (164 ~ 32)) = ~(128 inst + (64 in/s)j Bar ED. Rotation about E. Vp = @pgk X tye = yg X (16K ~20}) = 2007.6 + 160d Bar BD. ‘Translation with B + Rotation about B. Vos = @pok * yp = nok x (321) = 32040} Yo = Ya + Yow 20wppi + 160 p¢§ = -1281 + 64) + 320g) Components: i: wpe = 128, @pp = -6.4 rad/s Wye = 6.40 rad/s ) 4 J: U6cpy = 64 +320, 16, 64 mp = 35008 ~ 3p 32-2 = Samadls Opp = 5.20 rad/s >PROBLEM 15.67 At the instant shown, bar AB has a constant angular velocity of 25 rad/s ‘counterclockwise. Determine at that instant (a) the angular velocity of the rectangular plate FBDH, (b) the velocity of point F, SOLUTION Bar AB. Rotation about A. Wp = 25 rad/s) "sn = (02)(25) = Sms} Bar ED. Rotation about E. YD =%— — ¥p = 0.20 pe Plate BDHF. Translation with B + Rotation about B. F % F “ Fou a h 2N/ a 30° vo= testo to—]= bel] [ro0 420] Draw velocity vector diagram. vp 5 ¥ = 5.7735 mis Mer ‘p18 *c0s30° cos30° s . Yap _ 5.7735 = TBD = STTBS | 14.4338 rads Cooney) Od “oy @ nou = 1443 rads ) 4 ‘De Veg = BFeppyp = (0.2)(14.4338) = 2.8868 mis pig = 2.8868 m/s “30° e% Vp = Va + Var = [Sivsf ] + [2.8868 mvs 30°] 135 mis 54.9" )PROBLEM 15.68 Pa At the instant shown, bar DE has a constant angular velocity of 35 rad/s 5 ‘ clockwise, Determine at that instant (a) the angular velocity of the T i /gozy Fectangular plate FBDH, (6) the point on the plate FBDH with zero tu ‘mommy TP velocty ef —t SOLUTION Rod DE. Rotation about E. Opp = 35 rad/s Vp = Moe @pe = (02)(35) = 7 m/s — Rod AB. Rotation about A. Va=Yel vp = 02049 Plate BDF. Translation with D + Rotation about D. F F F me We Nee a ‘Vow = ‘®o oe ‘> Va= Yo + Van feo b]=f0— 1+ [van £307] Draw velocity vector diagram. Mo my 7 ‘sin30°~ sin30° Yow _ 14 nour = t= og = 35 rls =14mis We, @ Once = 350 rads) 4 Point of zero velocity lies above point D. ‘7 Yow = 02m nour 35 ® 200 mm above point D.PROBLEM 15.69 ‘The 4-inradius wheel shown rolls to the left with a velocity of 45 ins. Knowing that the distance AD is 2.5 in., determine the velocity of the collar and the angular velocity of rod AB when (a) ff = 0, (b) i = 90°. SOLUTION (@ p=o Wheel AD. CA = (CD) - (DA) = 4-25 = 1.5in. ¥4 = (CA) egy = (1.5)(11.25) = 16.875 inds — Rod AB. Va = Vat Van [ve] = [16875 —}+ Deal Vp = 16.8 in/s— @yp = 11.25 rads >) = 32,005° = 4.7170 in, cosy (CA) gp = (4.7170)(11.25) = 53.066 ins v, = [53.066 in/s ~. 32.005] @ = 18.663 with 4-+ Rotation aboutPROBLEM 15.69 CONTINUED A i a> Y : 'p Ve=Va+ Van [ve —}=[a207] +E veh] Draw velocity vector diagram, 5 = 180° - y (90° + 9) = 90° — 32,008° - 18,663° = 39.332° Law of snes. s % 32° aos0 r % vp, = Yasin _ (63.066)sin32.005° m4 ~ sin(90° + 9) ssin108.663° = 29.686 ins Yara «, 29.686 «9 37 sadly a = yay = 2.37 rads Oy = ES Oy = 2.37 rads)PROBLEM 15.70 ‘An automobile travels to the right at a constant speed of 80 km/h. If the diameter of the wheel is 560 mm, determine the velocities of | ts B,C, ‘D, and E on the rim of the wheel. SOLUTION v4 = 80 kmh = 22.222 m/| vo =04 ve = 560 mm 280 mm = 0.28 m 2222 _ 79.364 rad's ) 0.28 2 Yau = You = You = 70 = (0.28)(79.364) = 22.222 mvs Vy = Vg + Vg = [22.222 ind —- ] + [22.222 in/s—- ] vy = 444m's— 4 Vo = V+ Voy = [22222 inds— ] + [22.222 in/s «2 30°] Vp = 42.9 mis 215.0° 4 Ve = V4 + ¥p = [22.222 ins —~ ] + [22.222 inJs| ] vp = 314 mis S450PROBLEM 15.71 ‘A Sem beam AE is being lowered by means of two overhead cranes. At > the instant shown it is known that the velocity of point D is 1 m/s downward and the velocity of point £ is 1.5 m/s downward. Determine ve smo] (@) the instantaneous center of rotation of the beam, (6) the velocity of 2150-20 basa point 4 ‘SOLUTION LS als @ Lye =15+2~3=05m 0.500 mto the right of 4 o ee (03) = 0.1667 ms vq = 0.1667 ms | 4PROBLEM 15.72 ‘A 5m beam AE is being lowered by means of two overhead cranes. At the instant shown it is known that the velocity of point A is $42 mm/s downward and the velocity of point £ is 292 mm/s upward. Determine 7 , neat (2) the instantaneous center of rotation of the beam, (b) the velocity of ee SOLUTION @ C lies 3.25 m to the right of 4. io ~ lac = 3.5 ~ 3.2494 = 0.2506 m 1.0418 mvs} ® Icoe = (0.2506)(0.1668) Vp = 41.8 mmis | 4PROBLEM 15.73 ‘At the instant shown during deceleration, the velocity of an automobile is 40 fs to the right. Knowing that the velocity ofthe contact point 4 of the wheel with the ground is 5 fs to the right, determine (a) the instantaneous center of rotation of the wheel, (b) the velocity of point B, (6) the velocity of point D. SOLUTION C lies 1.714 in. below 4. Vp = 75.0 fis—- Vp = 53.2 is 241.2° €PROBLEM 15.74 At the instant shown during acceleration, the velocity of an automobile is 40 f/s to the right. Knowing that the velocity of the contact point A of the ‘wheel with the ground is 5 fl/s to the left, determine (a) the instantaneous center of rotation of the wheel, () the velocity of point B, (c) the velocity of point E. SOLUTION @ O) Vp = 85.0 fis—~ o 1.0 = (1.5186)(45) = 60.2 fs tang = 08, = 484° leo Vp = 60.2 fis S48.4°PROBLEM 15.75 ‘A 60-mm-radius drum is rigidly attached to a 100-mm-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum. Knowing that end E of the cord is Fe oh 0 pulled to the let with a velocity of 120 mv, determine (a) the angular A velocity of the drums, (b) the velocity of the center of the drums, (c) the ° length of cord wound or unwound per second. 2 SOLUTION Since the drum rolls without sliding, its instantaneous center lies at D. @ © vq = (100)(3) = 300 mas v4 = 300 mms — Since v, is greater than vp, cord is being wound. 4 Yp = 300 ~ 120 = 180 mms ) ‘Cord wound per second= 180.0 mmPROBLEM 15.76 A 60-mm-radius drum is rigidly attached to a 100-mm-radius drum as shown. One of the drums rolls without sliding on the surface shown, and ‘ cord is wound around the other drum. Knowing that end E of the cord is pulled to the left with a velocity of 120 mm/s, determine (a) the angular velocity of the drums, (b) the velocity of the center of the drums, (c) the length of cord wound or unwound per second. SOLUTION Since the drum rolls without sliding, its instantaneous center lie at B. Ve = Vp = 120 mms — @ © v4 = 180 mms— Since v4 isto the right and vp isto the left, cord is being unwound. v4 —¥g = 180 +120 = 300 mms © Cord unwound per second = 300 mmPROBLEM 15.77 A double pulley is attached to a slider block by a pin at A. The 1.5-in.- : radius inner pulley is rigidly attached to the 3-in.-radius outer pulley. Knowing that each of the two cords is pulled at a constant speed as shown, determine (a) the instantaneous center of rotation of the double pulley, (6) the velocity of the slider block, (c) the number of inches of cord wrapped or unwrapped on each pulley per second, SOLUTION vo= Winds}, vp =8inds| fe ptme eae BD 45 =a Losin fo) 0 = 23 CA = 30-25 = 05 in, C lies 0.500 in. tothe right of 4. F o = 0.50 = (0.5)(4) = 2 inds v4 = 2.00 ins | © Yp-¥4= inst Cord DE is unwrapped at 12.00 in/s. vp - V4 = 6inJs| Cord BF is unwrapped at 6.00 inJ/s.PROBLEM 15.78 Solve Prob. 15.77, assuming that cord £ is pulled upward at a speed of 8 inJs and cord F is pulled downward at a speed of 10 inJ/s. SOLUTION we a ) F © vo=8ins}, vp =10ins| Yp tv, _ 8+10 BD 45 c=" =8 =2in a4 C4 =3-2=1in = 4rads Cllies 1,000 in, to the right of 4. € ¥4 = (1.000)q = (1.000)(4) = 4 ins v, = 4.00 inis | Vp - V4 = 12inJ/s{ Cord DE is unwrapped at 12.00 in./s. Vy -V4 = 6 ins} Cord BF is unwrapped at 6.00 in/s.PROBLEM 15.79 BS —~ Knowing that at the instant shown the angular velocity of bar DC is a RE — m7, 18 rd counerclockwis, determine (a) he angular velo of bee AB, (6) the angular velocity of bar BC, (c) the velocity of the midpoint of bar Dome 78 2 Y> BC. SOLUTION Bar DC. (rotation about D) Ye = @ep(CD) = (18)(0.25) =45 ms Yo = 4.5 mis “30° Bar AB. (rotation about A) vp = vp 430° Locate the instantaneous center (point /) of bar BC by noting that velocity directions at two points are known, Extend lines AB and CD to intersect at J. For the given configuration, point / coincides with D. IC = 0.25 m, 1B = 0.25¥3 m oe = = 45. = 18 rads} IC 0.25 Vg = (1B) gc = (0.25V3)(18) = 7.7942 mis? yy = 31.2 rads ) 4 (b) gc = 18.00 rad's) (©) Locate point M, the midpoint of bar BC. gle ICM is an equilateral triangle. IM = 0.25 m vy = (IM) age = (0.25)(18) = 4.5 ms Vy = 450mbs 730°PROBLEM 15.80 “ Knowing that at the instant shown bar AB is rotating countercloc ER tm and that the magnitude of the velocity of the midpoint of bar BC is Some ot a7. 2.6 m/s, determine (a) the angular velocity of bar AB, (b) the angular Vi velocity of bar BC, (c) the angular velocity of bar DC. SOLUTION Bar AB. (rotation about A) Bar CD. (rotation about D) Bar BC. Locate its instantaneous center (point 1) by noting that velocity directions at two points are known. Extend lines AB and CD to intersect at J. For the given configuration, point / coincides with D. Locate point M, the midpoint of bar BC. From geometry, triangle ICM is an equilateral triangle. IM = AB=CD=025m, 1B = 0.253 m Eeeaere > ne = THe = 73g 7 OAT) (@ ¥g = (JB)@gc = (0.25V3)(10.4) = 4.5033 mis yy = 2B. = 43033 ~ 18.0133 rads 4g = 18.01 rads )4 AB (0.25 a (b) @pc = 10.40 radis ) < © ¥e = (IC) og = (0.25)(10.4) = 2.6 ms cp = MC = © 2 10.4 rads ep = 10.40 rads) Dc 0.25PROBLEM 15.81 Knowing that at the instant shown the velocity of collar 4 is 45 inJs to the let, determine (a) the angular velocity of rod ADB, (6) the velocity of point B. SOLUTION a Collar A. (Rectilinear motion) v4 = 45 ins — Rod DE. (Rotation about E) Vp = Yoh, Locate the instantaneous center (point C) of bar ADB by noting that velocity directions at points 4 and D are known. Draw AC perpendicular to v,and DC perpendicular to ¥p, (a) © gpg = 6.00 rads ) ® BF = 4+3tanf = 5.6in. CF =3in. = 6.353 in Vp = (CB)dgpg = (6.353)(6.00) = 38.1 nis 90° — 9 = 61.8° 8.1 inJs S61.8° VaPROBLEM 15.82 Knowing that at the instant shown the angular velocity of rod DE is 2.4 rad/s clockwise, determine (a) the velocity of collar 4, (6) the velocity of point B. Rod DE. (Rotation about E) Ope = 2.4 rads) Vp = (ED) apg = (6)(2.4) = 14.4 inst Collar A. (Rectilinear motion) Wye Locate the instantaneous center (point C) of bar ADB by noting that ‘velocity directions at points 4 and D are known. Draw AC perpendicular to v, and DC perpendicular to vp @ ¥%4 = (CA) apy = (7.8)(36) = 27 inds v4 =210inis— 4 (6) nea) BF = 4+3tanf = 5.6'in. CF =3in, F3 BF” 36 (cry + (Bry = V3 +56 = 6.353 in, tang p= BP (CB) app = (6.383)(3.6) = 22.9 ins 90° - p = 61.8° Vp = 22.9 ins 061.84PROBLEM 15.83 An overhead door is guided by wheels at 4 and B that roll in horizontal and vertical tracks. Knowing that when @ = 40° the velocity of wheel B is 0.6 m/s upward, determine (a) the angular velocity of the door, (b) the velocity of end D of the door. O6msf vy =vy — Locate the instantaneous center (point C) by noting that velocity directions at points 4 and B are known. Draw AC perpendicular to v,, and BC perpendicular to BC = (AB) sin = 2sin40* = 1.28557 m @ yg) = 28 06 _ 9.46672 rads “BC ~ 128557 4pp = 0.467 radi) 4 Toc = ac + Foe = [1.28557 m— ] +[2mN4o%] = 2.9930 m "530.79" = 30.79° ® Vp = Foc ano = (2.9930)(0.46672) = 1397 ms vp =1397 V8 90° ~ B = 59.2°PROBLEM 15.84 Rod ABD is guided by wheels at 4 and B that roll in horizontal and vertical tracks. Knowing that at the instant shown = 60° and the velocity of wheel B is 800 mm/s downward, determine (a) the angular velocity of the rod, (6) the velocity of point D. SOLUTION vp=800mms| vy =¥y— Locate the instantaneous center of rod ABD by noting that velocity directions at points A and B are known, Draw AC perpendicular to v, and BC perpendicular to Vp vp _ 800 BC ~ 300c0530° .0792 rads} @ 480 ay = 3.08 rails 4 ((600c0s30°)° + (300sin30")> = $40.83 mm 300sin30° 600c0s30° 7 =1610° 9-7 = 3. o ¥p = leo®any = (540.83)(3.0792) = 1.665 x 10°mm/s Vp = 1.665 ms73.9°PROBLEM 15.85 Small wheels have been attached to the ends of bar AB and roll freely along the surfaces shown. Knowing that the velocity of wheel B is 2.5 m/s tothe right at the instant shown, determine (a) the velocity of end A of the bar, (6) the angular velocity of the bar, (c) the velocity of the ‘midpoint of the bar. Vga vy A045, vy = 2.5ms— Locate the instantaneous center (point C) of rod AB by noting that velocity directions at points A and B are known. Draw AC perpendicular to v, and BC perpendicular to vp. Let 1 = AB = 600 mm = 0.6m Law of sines for triangle ABC. 6 a 1 0.84853 m sin75° sin60® sin45® a= 0.73485 m, b= 0.81962 m on Be = 3.0502 rads 962 (@ v4 = aay = (0.73485)(3.0502) = 2.24 mis V4 = 2.24 mis £45.09 .05 radis ® (© Let Mbe the midpoint of 4B. Law of cosines for triangle CMB. 1Y _ opt o£} - hese (0.81962)* + (0.3)° - (2)(0.81962)(0.3) cos 60° 0.3sin 60° 0.71825 * Baur My = map = (0:71825)(3.0502) = 2.19 mss, Vy = 219 mis 621.294PROBLEM 15.86 At the instant shown, the angular velocity of bar DE is 8 rad/s ‘counterclockwise. Determine (a) the angular velocity of bar BD, (b) the angular velocity of bar AB, (c) the velocity of the midpoint of bar BD. mo A (60 mn ee | SOLUTION Bar DE. vp = ew,, = (0.6)(8) = 48 ms Vp =4.8m/s — Bar AB. vp = a0,, = 0.2 Oyp Vp = 0.204g > 30° Locate the instantaneous center (point C) of bar BD by noting that velocity directions at points B and D are known. Draw BC perpendicular to vj, and DC perpendicular to vy. seat bet 1 = BD =06m. Law of sines for triangle CBD. 6 a 1 06 sin120° ~ 5in30® $in30° ~ $in30° 5 =103923m, d= 06m @ =— =8nds Opp = 8.00 radis ) 4 1.03923)(8) = 8.3138 m/s yy = 41.6 rad’s ) 4 (6) Law of cosines fr triangle CMD. mad + (3) = 24 eos120° Ble? = 06 +(03 0.793725 m — (2)(0.6)(0.3) ¢0s120°PROBLEM 15.86 CONTINUED Law of sines. sin B 120" ing = (03)sin120° m™ © 0.19375 » B=19.1° Velocity of M. vy = magn = (0.793728)(8) = 6.35 mis Vy = 6.35 mis%19.1°-¢PROBLEM 15.87 Arm ABD is connected by pins to a collar at B and to crank DE. Knowing that the velocity of collar B is 16 in/s upward, determine (a) the angular velocity of arm ABD, (6) the velocity of point 4 36% 64 SOLUTION vpei6inst tny-H 5 yang Locate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D are known. Draw BC perpendicular tov», and DC perpendicular to vp. CI =(Dd)uany ~ (64/5) = 2.6667 in, CB = JB ~ CJ = 12.8 ~ 2.6667 = 10.1333 in. 579 rads) ve 16 7 @ ann = 285 = Fpaagag 7 157895 rs "a0 CK = CB + BK = 10.1333 + 7.2 = 17.3333 in, KA 3.6 === = 11.733°, 90° - 6 = 78.3" CK 17.3333) B= M733 6 . c= SE 173333177032 in cos cos .733° tan p ® ¥4 = (AC) sap = (17.7032)(1.57895) = 28.0 ins, v4 = 28.0 in/s.78.3°PROBLEM 15.88 ‘Arm ABD is connected by pins to a collar at B and to crank DE. Knowing that the angular velocity of crank DE is 1.2 rad/s counterclockwise, determine (a) the angular velocity of arm ABD, (6) the velocity of point 4. SOLUTION tny = F223, y= 226007, Ep = 72-2 1310, DF 12 087” e087 (ED) eopy = (13)(1.2) = 15.6 inds ve= al Locate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D are known. Draw BC perpendicular to vp and DC perpendicular to v., DI _ 6A Vp = 156 inis ay 5 ; Cy = (Di)tany = (a5) = 2.6667 ing CD = LE = 2 = 69333 in to. = 156 725 rads) gy = 2.25 ad's) . ano ~ Cy ~ 69333 CK = KJ ~ CJ = 20 ~2.6667 = 17.3333 in. AK 36 — = 11733" 90°- B= 78.9 mee Tia? a ac = E1733 «177932 in cosh cosh ¥4 = (AC) yap = (17.7032)(2.25) = 39.8 ins ) v4 = 398 in/s 778.3"PROBLEM 15.89 ‘The pin at B is attached to member ABD and can slide freely along the slot cut in the fixed plate. Knowing that at the instant shown the angular velocity of arm DE is 3 rad/s clockwise, determine (a) the angular velocity of member ABD, (b) the velocity of point 4. SOLUTION y Opp = 3 rads ¥p = (DE)ene = (160)3) 480 mms Vp is perpendicular to DE. v= ve} Locate the instantaneous center (point C) of bar ABD by noting that velocity directions at points B and D are known, Draw BC perpendicular to vp and DC perpendicular to Vp. BD = 120mm, DK = (BD)cos30* = 120c0s30° DK _ 120cos30° cos p = DK _ 1200830" =D 160 , B= 49.495", 80° - 30° - f = 100.505°PROBLEM 15.89 CONTINUED Law of sines for triangle BCD. (BD)sin30° _ 120sin 30° = 78911 mm sin sin ac = (BD)sing _ 120806 155.177 mm sing ~~ sin Law of cosines for triangle ABC. (Ac) = (BC) + (4B) - 2(4B)(BC) cos 150° (4c)? = 155.177? + 200? - (2)(155.177)(200)cos150°, AC = 343.27 mm. Law of sines. sing = y= 169" pene 5 @ gp = 22, = EO = 60828 nds) @4gp = 6.08 rad/s >) 4 Oy ¥4 = (AC) ean = (343.27)( 6.0828) = 2088 mms v4 = 2.09 ms 773.19PROBLEM 15.90 ‘Two identical rods ABF and DBE are connected by a that at the instant shown the velocity of point D determine the velocity of (a) point E, (6) point F. at B. Knowing 100 mm/s upward, SOLUTION Locate the instantaneous center (point C) of bar DBE by noting thatthe velocity directions at points B and D are known. Draw BC perpendicular to v, and DC perpendicular to Vp. Law of sines for triangle BCD. BC = BD = 180mm vp _ 200 oat = Cy ~ 347.73 0.57515 rad/s Vp = (BC)epme = (180)( 0.57515) = 103.528 mmisPROBLEM 15.90 CONTINUED = 0.57515 rad/s) BF = vp _ 103.528 80 ve = (AF) gap = (300)(0.57515) = 172.546 mm's Law of cosines for triangle DCE. (cey = (coy + (DEY -2(€D)(DE)eos15° (cEY = 347.73? + 300° - (2)(347.73)(300)cos15°, CE = 96.889 mm EH = DE sin15° = 300 sin15° _ BH _ 300sin15° =36.7° ce” 9689 = cos B @ ve = (CE) sco = (96.889)(0.57515) = (55.7) mms, vp = 55.7 mms 36.7° 4 ® vp = 172.5 mms 475.0°