1. The problem statement, all variables and given/known data Three forces are applied to a locked pulley. Determine the magnitude and direction of the resultant of the three forces and the perpendicular distance from the axle of the pulley to the line of action of the resultant. **I'm unable to upload a photo of the pulley at the moment, but here's a description that should give you all you need to know: The two cords hanging off of the pulley are 2 feet apart (so the inner circle around the pully has a diameter of 2 feet). The cords are both 20 degrees above the horizontal, and are each in the second quadrant (-x, +y direction). There is a 120 lb force on one cord, and a 40 lb force on the other. The outer circle around the pulley is 4 feet in diameter, and there is a 90 lb force hanging off of it in the 4th quadrant (+x, -y). Overview: Fa= 120 lb at 20 degrees above horizontal (-x,+y) Fb=40 lb at 20 degrees above horizontal (-x, +y) Fc=90 lb at 90 degrees below horizontal (+x,-y) 2. Relevant Equations Perpendicular axis from the axle of the pulley to the line of action of the resultant: d=M/R 3. The attempt at a solution I already found the resultant by finding Rx and Ry and finding the magnitude. The resultant is 154.4 lb at 13.2 degrees above the horizontal. I'm having trouble with the second part, specifically finding the moment. -120cos(20)(1)-40cos(20)(1)+(90)(2)=30 30/154.4=.192 in 1 and 2 are the radii. The answer should be .647 in... I've tried every combination of force components to get the moment to be 100, but I'm out of luck (the moment is 100 when the distance is .647). I'd appreciate any help!