To determine the mass of the metre rule, we can use the principle of moments about the pivot point (knife edge).
When the rule is balanced horizontally, the moments about the pivot must be equal on both sides.
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Identify the components:
- Mass of the body \( m = 60 \text{ g} = 0.06 \text{ kg} \)
- Position of the mass from the left end of the rule is \( 29 \text{ cm} \)
- Position of the pivot (knife edge) is \( 44 \text{ cm} \)
- Let the mass of the metre rule be \( M \) (kg). Assuming the mass of the metre rule is uniformly distributed, its center of mass will be at the midpoint, which is \( 50 \text{ cm} \).
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Calculate the moments:
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The distance of the 60g mass from the pivot: \[ d_1 = 44 \text{ cm} - 29 \text{ cm} = 15 \text{ cm} = 0.15 \text{ m} \]
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The moment due to the suspended mass is: \[ \text{Moment}_{\text{mass}} = \text{Force} \times \text{Distance} = (0.06 \text{ kg} \cdot 9.81 \text{ m/s}^2) \times 0.15 \text{ m} \]
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The force due to the mass (weight) is: \[ W_{mass} = 0.06 \cdot 9.81 = 0.5886 \text{ N} \]
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The moment due to the mass is: \[ \text{Moment}_{\text{mass}} = 0.5886 \text{ N} \times 0.15 \text{ m} \approx 0.08829 \text{ Nm} \]
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Calculate the moment due to the mass of the metre rule:
- The distance of the center of mass of the metre rule from the pivot: \[ d_2 = 50 \text{ cm} - 44 \text{ cm} = 6 \text{ cm} = 0.06 \text{ m} \]
- The weight of the metre rule is \( M \cdot g \), where \( g = 9.81 \text{ m/s}^2 \).
- Therefore, the moment due to the mass of the metre rule is: \[ \text{Moment}_{\text{rule}} = (M \cdot 9.81) \times 0.06 \]
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Setting the moments equal: \[ \text{Moment}{\text{mass}} = \text{Moment}{\text{rule}} \] This leads to: \[ 0.08829 = (M \cdot 9.81) \times 0.06 \]
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Solving for M: \[ 0.08829 = 0.5886M \] Dividing both sides by 0.5886: \[ M = \frac{0.08829}{0.5886} \approx 0.150 \text{ kg} = 150 \text{ g} \]
Therefore, the mass of the metre rule is approximately 150 g.
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