Question

The shaft-pulley assembly is supported by the universal joint at O and by a slider bearing at A. The pulley, which has a mass of 13.5 kg, is subjected to the belt tension shown where F = 71 N. The mass of the shaft may be neglected. Determine magnitude of the resultant force at B in N. b) Determine the torque/ moment along the x-axis resulting from the two belt tensions in NM c) Determine the bearing reaction at A in N 200 BI 30 100 300 N F Dimensions in mm M2

Answer 1

Answer(a)- Resultant force at(B)= 48.373N

Answer(b)- torque/moment= 4.8373Nm

Answer(c)- bearing reaction at(A)= 227.844N

Details solutions is given below

Solution Given m= 13.5 kg F = @ ZIN (a) Calculate the resultant force at B in Here distance b/w fixed and universal Joint o is not given * Assume OC = 20mm 20ml **20mm 200mm **? k K 200mm → EfH = 0 = RAH + RCA = 10.71+ 300 Cos 30º = 330.81 RAH +RCH = 330.81 taking EMA -0 1-330.81 x 200 + Rch x 420 =0 RAH = 173.28 N = > RCH =157.53N o) Raw + Rev = 300 Sin30° +13.5x5.51) > Efv =0 282.435 - taking moment about A (EMA =o) - 2820435 x 200 + Rev X 420 134.493 N. RAV =147.9421 -) Rev= - So RA 5 (RAH) + (Raw) – J173028) 7 (147.9421 )! RA = 227.844 N Ang

* Resultant force At 6 (Fe) = (+® + 800 Cosao") - (13.54981 + you sing oo) AI +300 Cosso") - (13.5x3.61 + 300 sinse") FB z FoNet = 48 37 3N Ans 8=0.1m (6.) Calculate the tosque / Moment T = (Foner xr = x8= 48.373 X 6.1 Ans T= 4.8373 N-M (c) Calculate the bearing Reaction at A Calculated Already above RA = 227. 844N Ans