To analyze the problem of a uniform rod suspended from a point, we'll first summarize the given details:
- Mass of the rod, \( m = 120 , \text{g} = 0.12 , \text{kg} \)
- Length of the rod, \( L = 130 , \text{cm} = 1.3 , \text{m} \)
- Suspension point is located \( 80 , \text{cm} = 0.8 , \text{m} \) from one end of the rod.
Center of Mass
Since the rod is uniform, its center of mass is located at its midpoint: \[ \text{Center of Mass (CM)} = \frac{L}{2} = \frac{1.3 , \text{m}}{2} = 0.65 , \text{m} \]
Distance from the Suspension Point
To find moments about the suspension point, we need to know the distances involved:
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Distance from the suspension point to the center of mass (CM): \[ d_{\text{CM}} = 0.8 , \text{m} - 0.65 , \text{m} = 0.15 , \text{m} \]
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The rod hangs vertically, so we will consider the torques acting around the point of suspension.
Torque Calculation
The torque (\( \tau \)) exerted by the weight of the rod about the suspension point can be calculated as: \[ \tau = \text{Weight} \times d = m \cdot g \cdot d_{\text{CM}} \] where \( g \approx 9.81 , \text{m/s}^2 \) is the acceleration due to gravity.
Plugging in the values, we get: \[ \tau = 0.12 , \text{kg} \times 9.81 , \text{m/s}^2 \times 0.15 , \text{m} \] Calculating this value: \[ \tau = 0.12 \times 9.81 \times 0.15 \approx 0.1764 , \text{N} \cdot \text{m} \]
Conclusion
The analysis shows that if any other forces are present, we must consider their contributions to torque as well. However, if the rod is balanced in the vertical position under uniform gravitational force, the tension in the wire must simply counterbalance the weight.
If you have any specific questions or need further calculations, feel free to ask!
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