# Questions on Moments

Q4.

(a) A vase of weight 8 N, with a base of radius 5 cm and height 25 cm is held at an angle as shown in the diagram.

When it is released it falls back on its base. Explain why (in terms of moments).

(2 marks)

The line of action of the 8N force is on the left of the tipping point at the base of the vase. It therefore gives an anticlockwise moment. This returns the vase to its upright position.

(b) The vase is pushed with a force F at the top rim (see diagram on the right). The force is just large enough to make the vase start to tilt.

(i) Calculate the size of the moment opposing the turning by force F.

Moment = 8 N x 5 cm

Moment = 40 N cm

(2 marks)

(ii) Calculate the size of force F when the vase just starts to tilt

F x 25 cm = 40 Ncm

F = 40/25 = 1.6 N

(2 marks)

(iii) If the base of the vase were wider, a larger force would be needed to make the vase start to tilt. Explain why, in terms of moments.

The moment holding the vase in position is its weight x radius of the vase base. If the vase base radius was increased the moment would be greater you would therefore need to give a larger force as the height would still be the same.

Credit was also given for realising that if the vase was wider it would be heavier so the force would increase too!

(2 marks)

Total 8 marks