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# homework9 - \Problem 17.128 The angular velocity mm = 5°...

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Unformatted text preview: \Problem 17.128 The angular velocity mm = 5° per second. Determine the angular velocity of the hydraulic actuator BC and the rate at which the actuator is extend— mg 1 l l \ \ Solution: The point C effectively slides in a slot in the arm BC. The angular velocity of “ME = 5 = 0.0873 rad/s. The velocity of point C with respect to arm AC is i k V(' = cuAc X rc/A = 0 0 wAC 2.6 2.4 0 = —0.2094i + 0.2269j (In/s), The unit vector parallel to the actuator BC is 1.2i + 2.4j J 1.22 + 2.42 The velocity of point C in terms of the velocity of the actuator is e = = 0.4472i + 0.8944j. vc = vcme+ wgc X l‘C/B. l J k VC 2 qu1(0.4472i+ 0.8944j) + 0 0 (age 1.2 2.4 0 VC = vcm1(0.4472i + 0.8944j) + ch(—2.4i + l.2j). Equate like terms in the two expressions: ~0.2094 = 0.4472vcm — 2.40)“, 0.2269 = 0.8944vcrel + 1.2w3c. (use = 0.1076 rad/s = 6J7 deg/s , vac] = 0.109 (m/s) . which is also the velocity of extension of the actuator. EM“ Problem 17.132 Block A slides up the inclined surface 3’ at 2‘ft/s. Determine the angular velocity of bar AC and the velocity of point C. 1 \ Solution: The velocity at A is given to be VA 2 2(——icos 20° +jsin 20°) = —1.879i + 0.6840j (ft/s). From geometry, the coordinates of point C are 7 (7, 2.5 = (7, 3.89) (ft). The unit vector parallel to the bar (toward A) is e = (72 + 3.892)“/2(—7i — 3.89j) = —0.8742i — 0.4856j. The velocity at A in terms of the motion of the bar is l j k VA = UArele + 0MB >< mm = UAmle‘l' 0 0 wAC , —4.5 —2.5 0 VA 2 —0.87420Ami — 0.4856vA.31j+ 2.5wAci — 4.5a)ch (ft/s). Equate the two expressions for VA and separate components: —1.879 = —0.8742UAiel + 2,5wAc, 0.6840 2 —0.4856vAm1 —- 4.5wA_c. Solve: mm] = 1.311 ft/s, («MC 2 —0.293 rad/s (clockwise). Noting that VA = 2 ft/s, the velocity at point C is i j k vC : vA(—0.8742i ~ 0.4856j) + [ 0 o 4.293], 2.5 3.89 — 2.5 0 vc = —0.738i —- 1.37j (ft/s) . Problem.n\17.133 In Problem 17.132, the block A slides up the inclined surface at a constant velocity of 2 NS. Determine the angular acceleration of bar AC and the accelerat on of point C. Solution: The‘velocities: The velocity at A is given to be 3 i VA : 2(—icos 20° Ajsin 20°) = —l.879i + 0.6840j (ft/s). \ From geometry, the coordinates of point C are 7 (.7, 2.5 = (7, 3.89) (ft). The unit vector parallel to the bar (toward A) is ' —-7i — 3.89j e = _______ = —0.8742i — 0.4856 . #72 + 3.892 j The velocity at A in terms of the motion of the bar is l J k VA = vArele + «Mr: X mm = vAreie + 0 0 wAC , —4.5 ——2.5 0 VA = —0.8742vme1i -— 048561),“va + 2.5wAci — 4.50)ch (ft/S). Equate the two expressions and separate components: —l.879 = —0.8742u,4m| + 2.5wAc, 0.6842 2 -O.4856UBm| — 45(qu Solve: vAml = 1.311 ft/s, wAc = —0.293 rad/s (clockwise). The accelerations: The acceleration of block A is given to be zero. In terms of the bar AC , the acceleration of A is 3A = 0 = aAieie + ZwAc >< vAieie + «Ac X l‘A/B - wiCI‘A/B- I J k 0 = (Mme + 2(4),“: 0,4le 0 0 l —0.8742 -0.4856 0 I J k + 0 0 «AC —w§C(—4.5i—2.5j). —4.5 —2.5 0 0 = aAi-ele + ZwAchi-eI(—eyi + exj) + aAc(2.5i - 4.5j) — wf‘C(——4.5i — 2.5j). Separate components to obtain: 0 = —0.8742(I,m| — 043736 + 2.5aAC + 0.3875, 0 = —-0.4856aAm1 + 0.6742 — 4.5mm + 0.2153. Solve: am] = 0.4433 (ft/\$2) (toward A). 0mg 2 0.1494 rad/52 (counterclockwise) The acceleration of point C is ac = tlAreie + ZwAc >< VAlel + “AC X rC/B — wﬁch/a -i . j k ac = “Arcle + 2wAC “Ale! 0 0 1 ex ey 0 l j k + 0 0 «Ac — wf‘caji + (3.89 — 2.5)j). 2.5 3.89 -— 2.5 0 Substitute numerical values: ac = —|.184i + 0.711j (ft/32) ., Problem 17.146 Suppose that at the instant shown in \Problem 17.145, xA = 1 m, dxA/dt = —3 m/s, d?“ /dt2 = 4 m/82, and the plate’s angular velocity and ahgular acceleration are a) = —4j + 2k (rad/s), and at = 3i "— 6j (rad/82). What are the x, y, z components of the velocity and acceleration of A relative to a non rotating reference frame that is stationary with respect to 0? Soluir'on: The velocity is VA = vo + VAM + w x rA/a. The rela— tive velocity is clx _ dy . dz VArel —- (217)14' (dt_).]+(dt)k, dx dy d dx dz ‘—=—3 ,—=—0.252=. -——- ——15 ,——-= where W m/s dt dt x 05x dt m/s d, 0, and mm 2 xi + yj + zj = i+ 0.25j + 0, from which vA : —3i — 1.5j + a) x (i + 0.25j). i j k vA=—3i—l.5j+ 0 —4 2 =—3i—l.5j—0.5i+2j+4k 1 0.25 0 = —3.5i + 0.5j + 4k (m/s) The acceleration is aA = ao + aAM + 21» x VAml + at x mm + w x (w x mm). Noting d2): 42y dzz = (2175)” (it?) + "~ where 2 2 2 2 d—y = 5—025):2 = 0.5 + 0.5x = 6.5 (m/sz), (122 . . 2 . . -—— = 0,01 2 3| — 6J (rad/s ), vAml 2 ~31 — 1.53, (It2 and from above: a) X rA/o = ~0.5i + 2j + 4k. Substitute: i j k i j k aA:4i+6.5j+2 0 ——4 2 + 3 -6 0 —3 —l.5 0 l 0.25 0 i j k + 0 —4 2 . —0.5 2 4 rm 2 4i + 6.5j + 2(3i + ()j -— 12k) + (6.75k) + (—20i —j — 2k) Problem 17.150 In Problem 17.149, determine the acceleration of passenger A that passenger B observes relative to the coordinate system ﬁxed to the car in which B is riding. Solution: Rearrange: Use the solution to Problem 17.149: amt = 3A — as — 20) X VAiel — 0) >< (‘0 >< l‘A/B)- vAre] : _120j (fl/S)~ i j k i j k _50_ aRcl=—2j+5i—2 0 0 a) —w2 kx 0 o l . w‘soo‘o'lmd/S‘ 0 —120 o 500 0 0 rA/B = 500i (ft). 1 J k am=—2j+5i——2(l20w)i—w2|:0 0 I], 0 500 0 and = —l4i — 2j (ft/32) . The acceleration of A is 8,4 = —2j (ft/s), The acceleration of B is as = —500(m2)i = —5i ft/sz, and or = 0, from which 8A = 83 + aAnzl + 20) X VAIcl + (0 X (w X l'A/B)- \ r ...
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## This note was uploaded on 03/20/2008 for the course EM 311 M taught by Professor N/a during the Spring '08 term at University of Texas.

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homework9 - \Problem 17.128 The angular velocity mm = 5°...

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