Question

The diagram in the figure shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf hang from the points A and B 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 moments,

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

Answer 1

(i) Total anticlockwise moment about O = 150 gf $\times$ 40 cm = 6000 gf cm

(ii) Total clockwise moment about O = 250 gf $\times$ 20 cm = 5000 gf cm

(iii) Difference = 6000 gf cm - 5000 gf cm = 1000 gf cm

(iv) Let the 100 gf be placed at a distance x to the right of point O (to balance the extra anticlockwise moment). From the principle of moments,

150 $\times$ 40 = 250 $\times$ 20 + 100 $\times$ x

$\Rightarrow$ 1000 = 100 x

$\Rightarrow$ x = 10 cm