Question

BONUS (10 Marks) Two smooth wheels A and B are connect by a weightless bar AB and placed on smooth surfaces as shown. The weight of wheel A is WA, and the weight of wheel B is angle 0 at equilibrium is given by WB. Show that the INAL- (WA - WA cot U61= \w6] WA+ WB WR Cot a 0=tan -1 - B A ALA

Answer 1

Hence given smooth balls so frictional force = 0 Given weightlessbar (let compression in bar) F B D of individual balls (Hence concurrent forces no moment creates) From above figure consider equilibrium equations sigma fx = 0 implies N A sin 2 = R cos theta sigma Fy = 0 implies N A cos 2 = R sin theta + W A From implies cot2 = R sin theta + W A/R infinity s theta cot2 = tan theta + (W A/R cos theta) now consider on the ball from figure consider equilibrium equations sigma fx = 0 implies R cos theta = N_beta cos Beta sigma F y = 0 implies N_beta cos beta + R sin theta = w_B implies N_B cos_beta/N_Beta sin 3 = w_B - R sin theta/R cos theta implies cot_Beta = WB/R w s theta - tan theta implies (cot_beta + tan theta) R cos theta = w_B implies R cos theta = WB/tan beta + tan