A 10m long uniform beam is pivoted in its centre. A 30kg point mass is placed on one end of the beam. Where must a 50kg mass be placed in order to balance the beam?

For this question, we will use moments. A moment is defined as:Moment = force x perpendicular distance from pivotHere, the moment of the 30kg mass which acts anti-clockwise through the pivot. The moment is:M_30kg = 30g x 5_mThe force is 30g as F=ma with g being the acceleration under gravity. The perpendicular distance is 5m as the beam is pivoted in the centre and the mass is placed at the end of the beam.The moment of the 50_kg mass is:M_50kg = 50g x DWhere D is the distance from the pivot. Since we know for the beam to be balanced, the clockwise moment must be equal to the anti-clockwise moment, we can say:M_30kg = M_50kg30g x 5 = 50g x DWe can cancel out the g factor as it is present on both sides of the equation.30 x 5 = 50 x DD = (30 x 5)/50 = (30 x 1)/5 = 3mSo the mass must be placed 3m from the pivot which is also 8m from the end which the 30kg mass is placed on.