436_Dynamics 11ed Manual

# 436_Dynamics 11ed Manual - Engineering Mechanics Dynamics...

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Unformatted text preview: Engineering Mechanics - Dynamics Chapter 16 Given rAB sin ( θ 1 ) − rBC sin ( θ 3 ) + rCD sin ( θ 2 ) = 0 ⎞ ⎛ 0 ⎞ ⎛ −rCD cos ( θ 2) ⎞ ⎛ 0 ⎟ ⎛ −rBC cos ( θ 3 ) ⎟ ⎛ 0 ⎞ ⎛ −rAB cos ( θ 1) ⎞ ⎞⎜ ⎜ ⎟⎜ ⎟⎜ ⎜ ⎟⎜ ⎟ ⎜ 0 ⎟ × ⎜ −rCD sin ( θ 2) ⎟ + ⎜ 0 ⎟ × ⎜ rBC sin ( θ 3 ) ⎟ + ⎜ 0 ⎟ × ⎜ −rAB sin ( θ 1) ⎟ = 0 ⎜ ωDC ⎟ ⎜ ⎟ ⎜ ωBC ⎟ ⎜ ⎟ ⎜ ωAB ⎟ ⎜ ⎟ ⎝ ⎠⎝ 0 ⎠⎝ 0 ⎠⎝ 0 ⎠⎝ ⎠⎝ ⎠ ⎛ θ3 ⎞ ⎜ ⎟ ⎜ ωAB ⎟ = Find ( θ 3 , ωAB , ωBC) ⎜ω ⎟ ⎝ BC ⎠ θ 3 = 31.01 deg ⎛ ωAB ⎞ ⎛ −9.615 ⎞ rad ⎜ ⎟=⎜ ⎟ ωBC ⎠ ⎝ −1.067 ⎠ s ⎝ Positive means CCW Negative means CW Problem 16-59 The angular velocity of link AB is ωAB. Determine the velocity of the collar at C and the angular velocity of link CB for the given angles θ and φ. Link CB is horizontal at this instant. Given: ωAB = 4 rad s φ = 45 deg rAB = 500 mm θ = 60 deg rBC = 350 mm θ 1 = 30 deg Solution: Guesses vC = 1 m s ωCB = 1 rad s Given ⎛ 0 ⎞ ⎛ rAB cos ( θ ) ⎞ ⎛ 0 ⎟ ⎛ −rBC ⎟ ⎛ −vC cos ( φ ) ⎞ ⎞⎜ ⎞⎜ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ 0 ⎟ × ⎜ rAB sin ( θ ) ⎟ + ⎜ 0 ⎟ × ⎜ 0 ⎟ = ⎜ −vC sin ( φ ) ⎟ ⎜ ωAB ⎟ ⎜ ⎟ ⎜ ωCB ⎟ ⎜ 0 ⎟ ⎜ ⎟ ⎝ ⎠⎝ 0 ⎠⎝ ⎠⎝ 0 ⎠⎝ ⎠ ⎛ vC ⎞ ⎜ ⎟ = Find ( vC , ωCB) ⎝ ωCB ⎠ ωCB = 7.81 434 rad s vC = 2.45 m s ...
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