मराठी

The diagram in Fig. 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

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प्रश्न

The diagram in Fig. 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 :

  1. the total anticlockwise moment about O,
  2. the total clockwise moment about O,
  3. the difference of anticlockwise and clockwise moments, and
  4. the distance from O where a 100 gf weight should be placed to balance the metre rule.

Figure shows a uniform metre rule weighing 100 gf, pivoted at its centre 'O'. Two weights of 150gf and 250 gf hang from the metre rule as shown. Calculate:

  1. Total C.W. moment about 'O',
  2. Total A.C.W. moment about 'O',
  3. Difference of C.W. and A.C.W. moments.
  4. The distance from 'O' where a 100 gf weight should be suspended to balance the metre scale.

संख्यात्मक
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उत्तर

i. Anticlockwise moments = 150 × 40 = 6000 gf

ii. Clockwise moments = 250 × 20 = 5000 gf

iii. Difference of anticlockwise moments and clockwise moments = 6000 − 5000 = 1000 gf

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

6000 gf cm = 5000 gf cm + 100 gf × d

1000 = 100 × d

d = 10 cm on the right side of O

∴ 100 gf wt. should be placed 10 cm to right of O.

Distance from O = 60 − 50 = 10 cm right side of O.

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