1. The problem statement, all variables and given/known data A shaft rotates at 1500rpm between 2 bearings, Ra and Rb. Ra force is 5 Kn and Rb is 3 Kn. A single mass is to be used to balance the shaft, to give reactions of zero. Mass to be placed at 90 degrees to shaft and at 200 mm radius. What is the required mass and position? 2. Relevant equations I have done the following: For equilibrium, the bearing reactions = 8 Kn so balance weight must be 8 Kn too. BUT, as the weight is non central (200 mm radius (and we can then assume shaft diameter at 400 mm)), So, Ra = mass x position length / shaft length = 8 x position length / 200 x 1000 = 5 KN Ra = 1000 / 8000 = 0.125 Ra = 8 x 0.125 / 0.2 x 1000 = 5 Kn Then taking moments about Rb gives 3 Kn which is correct. 3. The attempt at a solution Knowing Ra = 5 Kn, Ra x 0.2m - 8 Kn x distance = 0 for equilibrium So, Ra x 0.2 m - 8 kn x unknown distance = 0 Ra = 8 x 125 / 200 x 1000 = 5 Kn ∴ mass should be 8 Kn 125 cm from Rb. BUT, have I neglected the angular rotation wr? As the weight is 200 mm out from shaft centre? And if so, then Mc Rc = mr So I get Mc Rc = 8 Kn x 0.2 m = 1.6 Kg. So final answer weight of 1.6 Kg should be placed at 200 mm radius (outer edge of shaft at 180 degrees to bearing reactions) at position 125 cm from Rb (bearing B of 3 Kn). I apologise if this is messy but I'm struggling a bit here and pointers or a "yes well done" would be simply fantastic, I really appreciate the time you guys spend helping! Thank you!