. The diagram in fig.1.34 shows a uniform metre rule weighing 100 gf, pivoted at its centre O

The diagram in fig.1.34 shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf 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 100 gf weight should be placed to balance the metre rule.

Best Answer

(i)Total anticlockwise moment about O

= 150 gf x 40 cm=6000 gf cm

(ii)Total clockwise moment about O,

=250 gf x 20 cm= 5000 gf cm

(iii) The difference of anticlockwise and clockwise moment= 6000- 5000= 1000 gf cm

(iv) From the principle of moments,

Anticlockwise moment= Clockwise moment

To balance it, 100 gf weight should be kept on right hand side so as to produce a clockwise moment about the O. Let its distance from the point O be d cm. Then,

150 gf x 40 cm=250 gf x 20 cm +100 gf x d

6000 gf cm= 5000 gf cm + 100 gf x d

1000gf cm =100 gf x d

So, d= on the right side of O.