To analyze the situation presented, we have a uniform rod with the following characteristics:
- Mass (m): 120 g = 0.12 kg (converted to kg for standard unit consistency)
- Length (L): 130 cm = 1.3 m (also converted to meters)
- Suspension point: 8 cm from one end of the rod
First, we can find a few key dimensions and properties related to the rod:
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Position of the center of mass (CM): a uniform rod has its center of mass at its midpoint. For a 1.3 m long rod, the center of mass will be at: \[ \text{CM position} = \frac{\text{Length}}{2} = \frac{1.3 , \text{m}}{2} = 0.65 , \text{m} , \text{from one end} \]
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Position of the suspension point: The rod is suspended 8 cm from one end, which is: \[ \text{Suspension point} = 8 , \text{cm} = 0.08 , \text{m} \]
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Distance from the suspension point to the center of mass:
- The distance from the suspension point to the center of mass is: \[ d = 0.65 , \text{m} - 0.08 , \text{m} = 0.57 , \text{m} \]
Torque and Equilibrium
If the rod is in equilibrium, the sum of torques about the suspension point must be zero. The gravitational force acts downwards at the center of mass. The torque (\(\tau\)) generated by the weight (force due to gravity, \(F_g = m \cdot g\)) is given by:
\[ \tau = F_g \cdot d \]
Where:
- \(F_g = m \cdot g = 0.12 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 1.1772 , \text{N}\)
- \(d = 0.57 , \text{m}\)
Calculating the torque about the suspension point:
\[ \tau = 1.1772 , \text{N} \cdot 0.57 , \text{m} \approx 0.6703 , \text{Nm} \]
Summary
This system's forces and torques must balance for static equilibrium. If you have specific analyses or calculations in mind (such as finding tensions in the wire if the rod were to rotate or any other detail), please let me know!
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