A uniform metre rule of weight 10 gf is pivoted at its 0 mark.

(i) What moment of force depresses the rule?

(ii) How can it be made horizontal by applying a least force?

Figure shows a uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40gf at the 10 cm mark and a weight of 20 gf at the 90 cm mark.

(i) Is the metre rule in equilibrium? If not, how will the rule turn?

(ii) How can the rule be brought in equilibrium by using an additional weight of 40gf?

When a boy weighing 20 kgf sits at one end of a 4m long see-saw, it gets depressed at its end. How can it be brought to the horizontal position by a man weighing 40 kgf.

The diagram shows a uniform meter rule weighing 100 gf, pivoted at its centre O. Two weights 150gf and 250gf hang from the point A and B respectively of the metre rule such that OA = 40 cm and OB = 20 cm.

Calculate:

(i) the total anticlockwise moment about O,

(ii) the total clockwise moment about O,

(iii) the difference of anticlockwise and clockwise moment, and

(iv) the distance from O where a 100gf weight should be placed to balance the metre rule.