The two light pulleys are fastened together and form an integral unit. They are prevented from turning about their bearing at O by a cable wound securely around the smaller pulley and fastened to point A. Calculate the magnitude R of the force supported by the bearing O for the applied 3.5-kN load. I have attached an image of the problem. 1. The problem statement, all variables and given/known data 2. Relevant equations ƩMO = 0 ƩFx =0 ƩFy =0 3. The attempt at a solution First I found the angle between the positive x axis and the cable (AB) θ = arctan(130/285) θ = 24.5196° Then I found the moment about point to calculate the tension (T) in the cable ƩMO = 0 0 = (3.5 kN)(200mm) -(130mm)T T = 5.3846 kN Then, knowing that the pivot point, O, has a force in the x direction (Rx) and a force in the y direction (Ry) I calculated these: ƩFx= 0 0 = -Rx + Tcos(24.5196) Rx = 4.899 kN ƩFy = 0 0 = -3.5 kN + Ry -Tsin(24.5196) Ry = 5.7305 kN Thus the magnitude R is: R = sqrt((4.899 kN)^2 +(5.7305 kN)^2) R = 7.53914 kN It says my answer is wrong and I can't see where my mistake is. Help would be appreciated.