1. The problem statement, all variables and given/known data A flagpole of mass 25kg and 2.5m in length is held in a horizontal position by a cable shown in the diagram. The centre of gravity of the flagpole is at a distance of 1.5m from the fixed end. Determine: a the tension T in the cable; b the vertical component of the force at the fixed end of the pole. 2. Relevant equations moment = Fx x = d sinθ ac/w moment = c/w moment Fy = F sinθ 3. The attempt at a solution I attempted part a and think I have got the answer, however I would be grateful for somebody to verify this for me. If we call the distance between the fixed end of the flagpole to its centre of gravity x1 and its weight F1 then moment1 = F1x1 = 25 x 9.81 x 1.5 = 367.875 Nm. If we then call the length of the flagpole d and the tension T F2, and we know that x2 = d sinθ, then moment2 = F2 x 2.5 sin30 = F2 x 1.25 and therefore F2 = moment2 / 1.25. The principle of moments states that when an object is in equilibrium , the anticlockwise moments equal the clockwise moments and therefore moment1 = moment2. Therefore the tension T = 367.875 / 1.25 = 294.3 N. Where I am really stumped is part b. I know how to work out the vertical component of a force - Fy = F sinθ - that's pretty easy. However when the question asks for the force at the fixed end of the pole i'm not quite sure what that is and thus I can't work out the vertical component. If anybody could enlighten me of what that force is I'd be very pleased.