Pregunta
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The horizontal uniform rod shown below has length 0.60 m and mass 2.0 kg . The left end of the rod is attached to a vertical support by a frictionless hinge that allows the rod to swing up or down. The right end of the rod is supported by a metal wire that makes an angle of \( 30^{\circ} \) with the rod. A 0.50 kg block is also attached to the right end of the rod as shown below. Calculate the tension in the metal wire.

Ask by Hampton Davies. in the United Arab Emirates
Mar 08,2025

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The tension in the metal wire is 29.4 N.

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To find the tension in the metal wire supporting the rod system, we can analyze the torques around the hinge (the left end of the rod) to ensure that the system is in static equilibrium. ### Steps to Calculate Tension: 1. **Identify Forces:** - The weight of the rod, \( W_{rod} = m_{rod} \cdot g = 2.0 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} \). - The weight of the block, \( W_{block} = m_{block} \cdot g = 0.50 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 4.905 \, \text{N} \). 2. **Location of Forces:** - The weight of the rod acts at its center, which is at \( 0.30 \, \text{m} \) from the hinge. - The block's weight acts at the end of the rod, which is at \( 0.60 \, \text{m} \) from the hinge. 3. **Setting Up the Torque Equation:** - Let \( T \) be the tension in the wire. The torque due to the tension works at the end of the rod and needs to be balanced by the torques due to the weights. - Torque due to weight of the rod: \[ \tau_{rod} = W_{rod} \cdot \frac{L}{2} = 19.62 \, \text{N} \cdot 0.30 \, \text{m} \] - Torque due to weight of the block: \[ \tau_{block} = W_{block} \cdot L = 4.905 \, \text{N} \cdot 0.60 \, \text{m} \] - Torque due to tension in the wire: \( \tau_{tension} = T \cdot L \cdot \sin(30^\circ) = T \cdot 0.60 \cdot 0.5 \) 4. **Setting Up the Equation:** - The sum of clockwise torques must equal the sum of counterclockwise torques around the hinge. Thus: \[ T \cdot 0.60 \cdot 0.5 = 19.62 \cdot 0.30 + 4.905 \cdot 0.60 \] 5. **Plugging in Numbers:** - Calculate the torques: - \( W_{rod} \) torque: \( 19.62 \cdot 0.30 = 5.886 \, \text{N m} \) - \( W_{block} \) torque: \( 4.905 \cdot 0.60 = 2.943 \, \text{N m} \) - Total clockwise torque: \[ 5.886 + 2.943 = 8.829 \, \text{N m} \] - Now set up the equation: \[ T \cdot 0.60 \cdot 0.5 = 8.829 \] 6. **Solving for Tension \( T \):** \[ T = \frac{8.829}{0.60 \cdot 0.5} = \frac{8.829}{0.30} \approx 29.43 \, \text{N} \] ### Final Answer: The tension in the metal wire is approximately \( 29.43 \, \text{N} \).

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