Question

The block at C is moving downward and the link AB has an angular velocity of 3 [rad/s] and angular acceleration of 1 [rad/s2] in the clockwise direction at the instant shown. 3 ft AB = 3 [rad/s] =1 [rad/s? ] UC d AB A 30° B 2 ft Copyright 2016 Pearson Education, Inc. All Rights Reserved (1) Determine the angular velocity of the link BC. (2) Determine the velocity of point C. (3) Determine the angular acceleration of the link BC. (4) Determine the acceleration of point C.

Answer 1

Solution:

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given sign convention A ve Ctre ZAB I krad/s² WAB =-3 k rad/s ZAB = 2 i ft roc que 3 cosso it 3 sin3o j) ft = 0 (x component of relocity of accn of point -c is zero CA=&A=0 (fixed point) x ① velocity of Bi (92 - TA to ABX r AB В. = o+(-3 k) x (21) 6 ft/ Velocity of C V = UBTW Bay RBC =(-6j) to BC k) x (3 cos 30 it3 sinoj) c =(-6 +300 BC Cos 30) j - 3 W Basho l ---(1) kxi=j, kxj=-1

sos سالا (0) = -3 WBC =-23-181 - 2.598 dB cjtl. 5 LBC ? sino WBC Ans (I) Now, from ean Oc=(-6+3 (6) cos30) j - 3 (o) sin3o í ů= -6 | ft/s Ans-(2) (3) Acceleration at 3, qo=aaTQABXrAB =o+(-k) x (2i) - (-3² (2) % =(-2;-181) ff Acceleration at C, a=ab ta betroa w schroe (-23-181) +(-aBck) x (320530 It3 sin 30 j) - (035 Assuming doc (TBC=-dock) - RAB S ان) - in clockwise direction ac =(-18 +1.5&c) i + (-2-2.5982) j ----(2) al

Now, cx= -1871.54BC=0 2. doc=18 12 radce 15 LBC = -12 k rada BC Ans (3) (4) From ean (2) a = (-18 + 15 (12)) {+ (-2-(2.598X(2)); ас = -33.18 ĭ ft/s2 Ans (4)