1. The problem statement, all variables and given/known data A plank of wood is lifted off the ground so it has one side on the ground (A) and the rest of the wood at a slope at 15degrees to the horizontal. The length of the wood (AC) is 2.5m. A force is applied at C at 65degrees to the horizontal to the left keeping it in equilibrium. The mass of the wood is 90g and the centre of mass is at B (this was 1.875m from A when the wood was flat, so presuming that it is at the same point?). Show that the lifting force is 649N 2. Relevant equations Moment= Force x Perpendicular distance Total clockwise moments=total anticlockwise 3. The attempt at a solution Well i've tried taking moments about A so: Fsin65 x 2.5 = 90gsin15x1.875 But this doesnt give the right answer. Do I need to consider the reaction force at A? Or friction? Any help at all would be greatly appreciated. Many thanks.