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been E :32? ESP? 9: $2200 3%: E9395 in [he xy are Problem 17.170 Points B and C city vectors of arms AB and BC are (0,13 = —0.5k (rad/s) and mgr Determine the velocity of point C. plane. The angular vclo 2.0k (rad/s). The vector 0
o Solution jsin 15") = 734.” — 196.7j (mm). 760(icos [5° rB/A The vector rc/B = 900(icos 50° +jsin 50“) = 5785i + 689.4j (mm), The velocity of point B is «Ms >< l'B/A = V8 V3 = ~98,35i ~ 367.Ij (mm/s). The velocity of point C V(' = VB + (03c X 1773 i — 367.lj + (2)(k x (578.5i + 689.4j)). —98.3 J 0.
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0., V('  1477.2i + 790j (mm/s) Problem 17.162 If the crankshaft AB is turning in the
counterclockwise direction at 2000 rpm (revolutions per
minute), what is the velocity of the piston? Solution: The angle of the crank with the vertical is 45°. The Equate expressions for vB and separate components:
angular velocity of the crankshaft is
—296.2 = 4.7961030 2
a) = 2000 (l) = 209.44 rad/s.
60 —296.2 —_ uc — l.4l4w3c. The vector location of point B (the main rod bearing) r3 =
2(—isin 45° +jcos 45°) = .4l4(—i +j) in. The velocity of point B
(the main rod bearing) is Solve: vc = —383.5j (in/s) = —32j (ft/s)
(03c = —6l.8 rad/s. i j k
vB=wxr3M=L4l4w 0 0 l
—l l O = —296.2(i +j)(in/s).
From the law if Sines the interior angle between the connecting rod
2 5
and the vertical at the piston is obtained from —,—— = , , from
smb’ stn45°
which
25'n 45°
0 = sin"( '5 )=16.43°. The location of the piston is rc = (2 sin 45° + 5cos(I)j = 6.21j (in.).
The vector mm = r3 — rc = —t.414i— 4.796j (in.). The piston is
constrained to move along the y axis. In terms of the connecting rod
the velocity of the point B is i j k
VB = VC + (03c X l‘B/c = vcj + 0 0 war.
4.414 —4.796 0 = vcj + 4.7%de — l.4l4w3c j (in/s). v“= Irma}, (4.: 363m w: mu“ w— Hf“  “ —::I' __,6.
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 Spring '08
 Elliott

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