To solve this problem, we will use the principle of moments, which states that for an object to be in equilibrium, the sum of the clockwise moments about any pivot point must equal the sum of the counterclockwise moments about that same point.
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Identify the pivot point: The meter rule is pivoted at 30 cm from end A. Therefore, the distance from A to the pivot point (P) is 30 cm, and the distance from B to the pivot point is 70 cm (since the total length of the meter rule is 100 cm).
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Identify the forces acting on the meter rule:
- A weight (5 N) is hung at point A (0 cm from A).
- Let the weight of the meter rule be \( W \) N, which acts at its center of mass (50 cm from A).
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Calculate the moments:
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The moment due to the 5 N weight at point A (which creates a clockwise moment) about point P is given by: \[ \text{Clockwise Moment} = \text{Force} \times \text{Distance from pivot} = 5 \text{ N} \times (30 \text{ cm} - 0 \text{ cm}) = 5 \text{ N} \times 30 \text{ cm} = 150 \text{ N cm} \]
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The weight of the meter rule, \( W \), acts at its center, which is 50 cm from point A. Therefore, the distance from P to the center of the meter rule (50 cm) is: \[ \text{Distance from pivot to center of meter rule} = 50 \text{ cm} - 30 \text{ cm} = 20 \text{ cm} \] The counterclockwise moment due to the weight of the meter rule is: \[ \text{Counterclockwise Moment} = W \times 20 \text{ cm} \]
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Set the moments equal to each other for equilibrium: \[ \text{Clockwise Moment} = \text{Counterclockwise Moment} \] \[ 150 \text{ N cm} = W \times 20 \text{ cm} \]
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Solve for W: \[ W = \frac{150 \text{ N cm}}{20 \text{ cm}} = 7.5 \text{ N} \]
Therefore, the weight of the meter rule is \( 7.5 \text{ N} \).
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