To solve this problem, we need to balance the torques (moments) acting on the meter rule. Torque is given by the product of the force and the distance from the pivot point (the knife edge).
Here are the steps to solve this problem:
1. **Identify the positions:**
- Knife edge (pivot point) is at the 60 cm mark.
- A 20 N weight is suspended at one end (0 cm mark).
2. **Determine distances from the pivot:**
- Distance of the 20 N weight from the pivot = 60 cm - 0 cm = 60 cm.
- Let \( W \) be the weight of the meter rule.
- Since the meter rule is uniform and its center of gravity is at its center (50 cm mark), the distance of the meter rule's weight from the pivot = 60 cm - 50 cm = 10 cm.
3. **Set up the torque balance equation:**
Since the meter rule is balanced, the clockwise torque should equal the counterclockwise torque around the pivot point.
- Clockwise torque = \( 20 \text{ N} \times 60 \text{ cm} \)
- Counterclockwise torque = \( W \times 10 \text{ cm} \)
Thus:
\[
20 \text{ N} \times 60 \text{ cm} = W \times 10 \text{ cm}
\]
4. **Solve for \( W \):**
\[
1200 \text{ N·cm} = W \times 10 \text{ cm}
\]
\[
W = \frac{1200 \text{ N·cm}}{10 \text{ cm}}
\]
\[
W = 120 \text{ N}
\]
Therefore, the weight of the meter rule is 120 N.
Auniform meter rule balances on the knife edge at60cm.mark when a weight of 20N is suspended at one end.calculate the weight of the meter rule.
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