This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 = aAiele + ZwAchieI(—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 ...
View
Full
Document
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.
 Spring '08
 N/A
 Dynamics

Click to edit the document details