620 Ans. Ans. aA = {-3.00 i + 1.75 j} ft>s2 aA = -1i - 1.25 j

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91962_06_s16_p0513-0640
6/8/09
3:00 PM
Page 620
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z
16–139. The man stands on the platform at O and runs out
toward the edge such that when he is at A, y = 5 ft, his mass
center has a velocity of 2 ft>s and an acceleration of 3 ft>s2,
both measured relative to the platform and directed along
the positive y axis. If the platform has the angular motions
shown, determine the velocity and acceleration of his mass
center at this instant.
v ⫽ 0.5 rad/s
a ⫽ 0.2 rad/s2
O
y ⫽ 5 ft
vA = vO + Æ * rA>O + (vA>O)xyz
x
vA = 0 + (0.5k) * (5j) + 2j
vA = {-2.50i + 2.00j} ft>s
#
aA = aO + Æ * rA>O + Æ * (Æ * rA>O) + 2Æ * (vA>O)xyz + (aA>O)xyz
Ans.
aA = 0 + (0.2k) * (5j) + (0.5k) * (0.5k * 5j) + 2(0.5k) * (2j) + 3j
aA = -1i - 1.25j - 2i + 3j
aA = {-3.00i + 1.75j} ft>s2
Ans.
620
A
y
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