DynamicsH.W.15-10-11

DynamicsH.W.15-10-11 - PROBLEM 15.153 7 wk€ . D Pin P is...

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Unformatted text preview: PROBLEM 15.153 7 wk€ . D Pin P is attached to the collar shown; the motion of the pin is guided by a a” slot cut in rod BD and by the collar that slides on rod AE. Knowing that at :5 the instant considered the rods rotate clockwise with constant angular velocities, determine for the given data the velocity of pin P, G;VM‘- wAE : 7 rad/S, WED : rad/s 500 mm SOLUTION 05 AB = 500 mm = 0.5 m, AP :: 0.5tan30°, BF 2 cos30° QJAE = 7 rad/s (08D : 4.8 rad/s) Let P’ be the coinciding point on AE and H1 be the outward velocity of the collar along the] rod AE. 5 M4» F“ v“ VP = VP: + VPME :- [(AP)a)AE 1+,[ul a] Let P" be the coinciding point on BD and u2 be the outward speed along the slot l1 J A A W VP : VP, + VFW) : [(513)5080 V; 30°] + [u7 ,1: 60”] Equate the two expressions for VP and resolve into components. $1.: u1 =[ 0'5 )(4.8)(cos30°) + u2 cos60° 313’ leb (1) or ul 2 2.4 + 0.5142 / + i: —(0.5tan30°)(7) = ~L C0:350!))(4.8)sin30" + u2 sin 600 * 1 “(Jim chow “2 : w[2.4 tan30° ~ 3.5tan30°] : 4173333 m/s .wwuufln sin 60° 6 From (1), u, = 2.4 + (0.5)(-0.73333) : 2.0333 m/s hau— VP = [(0.5tan30°)(7) iJ—l— [2,0333 ma] 3 [20207 m/S l]+ [20333 m/S M, ] M v,, = «2.03332 + (2.0207)2 = 2.87 m/s tan/3 = ——i-m, /3 a 444.80 l V]; 2 2.87 m/s QC 448° { WA. “A “.4, A’s/c m. PROBLEM Two rotating rods are connected by a slider block P. The velocity V0 of the slider block relative to the rod on which it slides has a constant velocity of 30 in./s and is directed outward. Determine the angular velocity of each rod for the position shown. ,6 = 50°—20° = 30° AB _ 20 _ BP _ AP sin fl sin30° sin 20° sin 1 30° BP = 13.6808 in., AP = 30.642 in. vP = (BP)a)BP {50° V... = (AP)(0AE 3%. 20° 4%”) u VF”: : \A 200 Resolve into components. 1.: (31°)pr cos 50° 2 (AP)wAE cos 20° + v0 sin 20° (0 +3: (BP)cuBP sin 50° = (AP)mAE sin20° -— v0 cosZO° (7' 7 Rearranging, (13.6808c0550°)a),31D — 30.642005 20°60“ = 305in20° (13.68088in50°)a)BP — 30.6423in 20°60” = —30cosZO° Solving Eqs. (1) and (2), QJBP : rad/S, 1.6958 rad/s wAE = 1.696 rad/s )4 wAE 6(in = 4.39 rad/s I} 4 Note that instead of resolving into components, the triangle of vectors VP, VP, and VP”: can be constructed. Then, V0 S, W _ 30 in /s, v = 3:9— : 51.962 in./s tan VP = .30 = 60 in./s sin/6 VP. 51.962 , a) = : =1.696 rad/s, ‘5 (AP) 30.642 amp 2 VP 60 = 4.39 rad/s (BP) 2 13.6808 PROBLEM 15.169 A chain is looped around two gears of radius 2 in. that can rotate freely with respect to the 16-in. arm AB. The chain moves about arm AB in a clockwise direction at the constant rate of 4 in./s relative to the arm. Knowing that in the position shown arm AB rotates clockwise about A at the constant rate a) = 0.75 rad/s, determine the acceleration of each of the chain links indicated. Links 3 and 4. i— SOLUTION Let arm AB be a rotating frame ofreference. Q = 0.75 rad/s )2 —(0.75 rad/s)k Link 3: r3 = (181n.)1 vm : u l = —(4 in../s)j a3. = ~er3 = ~(0.75)2(18i) = —(10.125in./32)i a3/AB = = —2—— = 8in./szi «— = —(8 in./sz)i 29 x v3,” = (2)(—0.75k) x (—41) : —(61n./52)1 a3 : a3. + am + 29 x v.48 = —24.1251n./s2 a3 : 241111152 «- 4 Link 4: r4 = (8 1n.)1 — (2111.) j v4,” : u «A : —(4 1n./s)1 a4. = 42% = —(0.75)2(8i — 2j) = —(4.5 in./s2)i + (1.1251n./s2)j aw = 0 29 x VW = (2)(—0.75k) x (—4i) = (61n./s2)j a4 = a4, + 34W, + 29 X VW = —(4.5 in./s2)i + (7.125 in./sz)j j haw a4 = \/(4.5)2 + (7.125) = 8.43 in./s2 7.125 tan = ——————, = 57.7° fl 4.5 fl a4 : 8.43 1n./s2 :2, 57.7° 4 0 1 H‘V‘k (Jain. “5; «NM :rmsIZw Cums :flS’mfl/X A), Rchgafi’Z/H, 4v 2:: T2: M I \CM 2 fax. n L (t. W - 1 ~ ' “"’ s I Z I“ r ) “Led-Act % ti ‘z/K’V‘A 2. {W { , a g t [I ~w r kw M 2, Liana a C :25 w m :( '2>((~.75/2) x W) K‘i'\~ a: J“ WAIT} ‘V , _>’ k L’ <4 =3) q<5“- 613' 743/43 ~+ ZA M's/A3, 0% Qs ,3 ~10.IL§'1 #8 o .. 6T“ :‘~2%.ILS;V‘/g2’<—"‘ 3"qu am"? *‘“lfi)1 a“ :2 c5 ‘( 54¢“ ’— ( c3 b’rcleia4}\pa «by C; U U fK/‘m (Ft t? [L .51 v.“ 5) xi) (:2; 11;“ l/Cfié g ' "’m'zfl TZ'M‘VTI‘MY via «a wpfg. c ma /5 M Mr A 9092 Q): M. l :1: PROBLEM 1 5.1 58 Pin P is attached to the collar shown; the motion of the pin is guided by a slot cut in bar 130 and by the collar that slides on rod AE. Rod AE rotates with a constant angular velocity of 5 rad/s clockwise and the distance from A to P increases at a constant rate of 2 m/s. Determine at the instant shown (a) the angular acceleration of bar BD, ([7) the relative acceleration of pin P with respect to bar BD. SOLUTION M AP = 400 mm = 0.4 m, BP = 400d? mm : 0.4V?va - . __ x _ _ Gwen. (DAE m Srad/sj, (1A5 — O, VP/AE v 2m/s (, ans/M 7 O. F1nd: (1,81) and aP/BD. Velocity of coinciding point P’ on rod AE. VP: = (AP)(()AE = (0.4)(5) = 2 m/s w... or (2 m/s)i Velocz'zy ofP relative to rod AE. vp/AE : (2 rn/s)j Velocity ofpoint P. VP 2 VP. + VP,” : (2 m/s)i + (2 m/s)j Velocity of coinciding point P" on rod BD. v,” = (BP)a)BD 5, 45° = 0.4J2wBD :52 45° = —0‘4a),,fi,i + 0.4mm Velocity ofP relative to rod 190. sip/ED = (cos 45°)ui + (sin 45°)uj Velocity ofpointP. VP = VP» + vP/BD VP = ——O.4a)BDi + 0.4wBDj + (cos45°)ui + (sin 450)uj Equating the two expressions for VP and resolving into components. i: 2 = —0.4a)BD + (cos45°)u (1) j: 2 = 0.44231) + (sin 45°)u (2) Solving(l) and (2), wBD = 0, u = 5J2 m/s, VFW) = (2 m/s)i + (2 m/s)j Acceleration of coinciding point P' on rod/IE. apr = (AP)aAEi —(AP)w§Ej = o — (0.4)(5)2j : —(10 ml);- y/ .M ’———-l PROBLEM 15.160 At the instant shown the length of the boom AB is being decreased at the constant rate of 0.6 W3 and the boom is being lowered at the constant rate of 0.08 rad/s. Determine (a) the velocity of point B, (b) the acceleration ofpoint B. 47% > r, /57// (£50.09 .5 c2 vp/bomw .c/% 7 go GOLUTION ‘1 v3, = r0) = (18)(0.08) = 1.44 fi/s ‘5': 60° Velocity of coinciding point B’ on boom. Velocity of point B relative to the boom. VB/boom = 0.6 ft/S 7 30° (a) Velocity of point B. VB : v8. + vwboom .3; : (v8)x = 1.44cos60° - 0.600330° = 0.20038 ft/S +1; (v3) = ~1.44sin60° — 0.6sin30° = ~1.54708 ft/s y v3 = 00.200382 +1.547082 21.560 ft/s 1.54708 , = ~82.6° 0.20038 fl tanfl=— v8 = 1.560 ft/s V: 82.6° 4 “W Acceleration of coinciding point B' on boom. a3, = 7032 = (18)(0.08)2 = 0.1152 1052 7 30° Acceleration of B relative to the boom. a “mom 2 0 Coriolis acceleration. 25”“ = (2)10‘08XO-6l = 0'096 fi/SZ 5“ 600 (b) Acceleration of point B. 33 = 331+ aB/boom ‘1' 250“ 2:. : (a3)x = —0.1152cos30° + 0 — 0.096cos60° = —0.14777 fi/s2 +1 : (a3)y = “0.1 152s1n 30° + 0 + 0.096sin 60° = 0.025538 £st a3 = (0.14777)2 + (0.025538)2 = 0.1500 ft/sz 0.025538 , = 9.8° 0.14777 fl tanfl = 33 = 0.1500 ft/s2 be 9.8° 1 PROBLEM 15.158 CONTINUED Acceleration of P relative to rod AE. am”, = 0 Coriolis acceleration. Acceleration of point P. 21,; : 3,: + zip/AB + 20345 x VFW; = (20 m/s2)i * (10 I‘D/32>] Acceleration of coinciding point P” on rod BD. 20)” x vP/AE z (2)(—5k) >< 2j = (20 m/sz)i 7 o l ap~ = aBDk x rP/B — (030er = —0.4aBD1 + 0.40530; + 0 Acceleration of P relative to rod BD. Coriolis acceleration. 20030 X vP/BD : 0 Acceleration of point P. 3P I 3P" + aP/AE + 20331) X VP/b’z.) = —0.4aBDi + 0.4a80j + (cos 45°)a,,i + ( sin 45%,; Equating the two expressions for a P and resolving imo components. i: 20 = —O.4aBD + (C0545°)ar j: —10 = 0.40:”) + (sin 45°)a, Solving (3) and (4), aw : —37.5 rad/32, exp/ED = (cos45‘”)a,.i + (sin45°)a,.j (3) (4) (1,. = 5V6 [Tl/82 aim = 37.5 rad/s2 )4 aP/BD 2 7.07 m/s.2 A: 45° 4 ._____J ...
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DynamicsH.W.15-10-11 - PROBLEM 15.153 7 wk€ . D Pin P is...

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