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# Figure shows a uniform metre rule weighing 100 gf provided with a knife edge at its centre.  Two weight 150 gf and 250 gf hang from the metre rod as shown.  Calculate1. The total anticlockwise moments about O.2. The total clockwise moments about O,3. The difference of anticlockwise and clockwise moments about O and4. The distance from 0 where a 100 gf weight should be placed to balance the metre rule

Asked by Topperlearning User 4th June 2014, 1:23 PM

The diagram of the arrangement of the weights is as shown in the figure.

1. The anticlockwise (positive) moments the weight of 150 gf is = 150 gf x 40 cm = 6000 gfcm

2. The clockwise (negative) moments of the weight of 250 gf is = -250 gf x 20 cm = -5000 gfcm, i.e., clockwise

3. The difference of moments = (6000 - 5000 ) gfcm = 1000 gfcm

4. To balance this 100 gfcm anticlockwise moment, 100 gf weight must be at a distance x

= on the right side of O.

Answered by Expert 4th June 2014, 3:23 PM
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