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प्रश्न
Two forces each of 5N act vertically upwards and downwards respectively on the two ends of a uniform metre rule which is placed at its mid-point as shown in the diagram. Determine the magnitude of the resultant moment of these forces about the midpoint.
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उत्तर १
Moment of force F1 = 5 ⨯ 25 = 125
Moment of force F2 = 5 ⨯ 25 = 125
Since both the forces are acting in anticlockwise direction, the resultant moment of force is the sum of the two moments.
∴ Resultant moment of force = 125 + 125 = 250
उत्तर २
M = `"F" × vec("d")` (or) F × S
M = `5xx50/100 "m"`
= 5 × 0.5
= 2.5 Nm.
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