1.

The diagram in Fig 1.38 shows a uniform metre rule weighing 100gf, pivoted as its centre O. two weights 150 gf and 250 gf hang from the metre rule as shown.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»

(i) Total anticlockwise moment about O

= 150 gf × 40 cm = 6000 gf cm

(ii) Total clockwise moment about O,

= 250gf × 20 cm = 5000 gf cm

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

(iv) From the principle of moments,

Anticlockwise moment = Clockwise moment 

To balance it, 100gf 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,

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

6000gf cm = 5000gf cm + 100gf × d

1000gf cm = 100 gf × d

So, d= 1000gf cm/100gf= 10 cm on the right side of O.


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