Question

b) At the instant shown in the Figure Q1, rod OA is horizontal, and slider B is moving with a velocity of 20 cm/s to the left and acceleration of 10 cm/s2 to the right. Determine the velocity and acceleration of point A. (CO2P03:15 Marks) 40 cm 50 cm 40 cm B + Figure Q1

Answer 1

A we see, BC - = JAP - AC² = $50% -40% so, AB = -30% -40

Let o A rotates with Wo kand AB rotates with W K AB Then, VB a (woxo VAT VAB oh ) + WARY AB) [wo R xl-402)] + { WAS EX(-301-408) -40 Wo +(-30 WAR WARI) + 40 AB I 40 WAB i - (40 Wo of 30 WAB) } But VB = -20 î (giver) 40 W AB i - (400o +30W AB)} Co J - 201 cm comparing 20= 40W AB - WAB --- and 3 8 V 4000 + 30WAB 20 - Wo= -38 ( 7 ) VA = Worx (-40%) = 3 R* (-408) -15 joniere 40 so

Let Then, x and & AB be angular acceleration of OA and AB. a on = (doxoa) - Wohn [xok X (-40%)] - (3 ²1-401) -40xoj + a AR - (LASXAB)_(WAR AB) - [o ao R X 1-301-4}} - (-2) 1-300-407 -BOX ARJ + 40 Karit 7.51 toj + 5.625 i and AOA + CAB + > 10 AB-409) 10 l () (5.625 +7.5 +40XAB (ro-30x, comparing 5.625 +7.5 +40XAB - 3.125 =-0.078 40 10-30XAB 400o =0 & & AB - and Co 10- 30(-01075) 40 0.3085. So, OLO A 11 12.34 J -40 (0.3085 j + 5.625 = 5.625 î - 12.34 -153 5625 9 - 12.349 confect Thus chlsee VA?