Question

Question 4: (learning outcomes 5, 6) (10 Marks) Bar AB has the angular motions shown. Determine the velocity of the slider block C at this instant. (use instantaneous center of zero velocity equations only)

Answer 1

In the figure the dimensions & angles are not clear, hence , the angle I found from the image is 48° & 600 respectively. ue of AB = 4 radls AB=0.5m BL=1m If the dimensions (Datas) are different, change it allordingly. allo Solusion By instantanion Center Method it will be lie Perpendiculed to the direction of linea velocity of the link. So the point I is the instantanious leader where it is perpendicular to y qv, (Directions)

des horizontal, from action of ye will be perpendicules to link AB & slider (c) slides horizon direction will be horizontale & its direction will be f W = 4 vello 45° A BY V = AB X WAB B this leng = 0.5 x 4 = 2 m/s to It Drawing a horizontal line from Point B which meets the point D'on IL From A BDC BC= lm Sin 30= BD BC BD = I sin 30 = 120.5 (os 30 = D = Dc= 1 Costo 0.866 m

From D BDI Sin 45 = 3 IB= 0.70 ID (os 45= IB 0. Tot = ID = ID= 0.5m 0707 By the Method of instantanious Center, the point I have zero velocity at this Pesticular instant and we find the length of IBG IC. TB: 0.70 Ic= ID+DC = 0.5+ 0.866 366 We have ₂ = 2 m/s = TB x u loc > 2 =0.707 x 1 V can be calculated from both sides at A&I= 1:0. W = 2.8288 wad's

Velocity at a v - Tc xube 1366 x 2:32. - 3864 m/s