To solve for the weight of the metre rule, we can use the principle of moments, which states that for the system to be in equilibrium, the clockwise moments around any point must equal the counterclockwise moments.
In this case, we can choose the knife edge (pivot point) as our reference point. Given that the knife edge is at the 60 cm mark, we can analyze the forces acting on the metre rule.
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Weight of the suspended load (20 N): This weight is acting downwards at one end of the metre rule, which is at the 0 cm mark. The distance from the pivot (60 cm) to the load (0 cm) is 60 cm.
The moment created by this force is:
\[ \text{Moment}_{\text{load}} = \text{Force} \times \text{Distance} = 20 , \text{N} \times 60 , \text{cm} = 1200 , \text{N cm} \] -
Weight of the metre rule (W): The weight of the metre rule acts at its midpoint, which is at the 50 cm mark. The distance from the pivot (60 cm) to the weight of the metre rule (50 cm) is 10 cm.
The moment created by the weight of the metre rule is:
\[ \text{Moment}_{\text{rule}} = W \times 10 , \text{cm} \] -
Setting the moments equal for equilibrium: \[ \text{Moment}{\text{load}} = \text{Moment}{\text{rule}} \] \[ 1200 , \text{N cm} = W \times 10 , \text{cm} \]
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Solving for W: \[ W \times 10 = 1200 \] \[ W = \frac{1200}{10} = 120 , \text{N} \]
Thus, the weight of the metre rule is 120 N.