1. The problem statement, all variables and given/known data A uniform ladder 5m long and of mass 40kg rests with its upper end against a smooth vertical wall and with its lower end 3m from the wall on rough ground. Find the magnitude and direction of the force exerted at the bottom of the ladder. (G=9.81ms^-2) 2. Relevant equations ACM=CM moment=force x distance from the pivot 3. The attempt at a solution w(moment) = (40 x 9.81) x 2.5 =981 Nm clockwise ACM=CM therefore: reaction moment = 981Nm and force = 981/3 =327N anticlockwise Is this right or am i very far off??