To determine the maximum load that the elevators can carry in the new building, we can use the following steps:
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Determine the maximum safe tension: The maximum tension the cables can withstand is given as 19.3 kN. To find the safe tension (two-thirds of the maximum), we calculate: \[ T_{\text{safe}} = \frac{2}{3} \times 19.3 , \text{kN} = \frac{2 \times 19.3}{3} , \text{kN} \approx 12.867 , \text{kN} \]
Converting this value to Newtons (since \(1 , \text{kN} = 1000 , \text{N}\)): \[ T_{\text{safe}} \approx 12867 , \text{N} \]
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Use the formula for tension in the elevator cables: The tension in the elevator cable can be expressed as: \[ T = m_{\text{total}} \cdot g + m_{\text{total}} \cdot a \] where:
- \(m_{\text{total}} = m_{\text{empty}} + m_{\text{load}}\)
- \(m_{\text{empty}} = 510 , \text{kg}\) (mass of the empty elevator)
- \(m_{\text{load}}\) is the mass of the load we want to determine.
- \(g = 9.81 , \text{m/s}^2\) (acceleration due to gravity)
- \(a = 2.28 , \text{m/s}^2\) (maximum acceleration)
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Set up the equation: The tension can be rearranged to solve for the total mass: \[ T_{\text{safe}} = (m_{\text{empty}} + m_{\text{load}}) \cdot g + (m_{\text{empty}} + m_{\text{load}}) \cdot a \] Simplifying this: \[ T_{\text{safe}} = (m_{\text{empty}} + m_{\text{load}}) \cdot (g + a) \]
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Substituting the numbers: \[ 12867 , \text{N} = (510 , \text{kg} + m_{\text{load}}) \cdot (9.81 , \text{m/s}^2 + 2.28 , \text{m/s}^2) \] \[ 12867 , \text{N} = (510 , \text{kg} + m_{\text{load}}) \cdot 12.09 , \text{m/s}^2 \]
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Solve for \(m_{\text{load}}\): \[ 12867 = (510 + m_{\text{load}}) \cdot 12.09 \] Dividing both sides by 12.09: \[ \frac{12867}{12.09} = 510 + m_{\text{load}} \] \[ 1069.07 \approx 510 + m_{\text{load}} \] \[ m_{\text{load}} \approx 1069.07 - 510 \approx 559.07 , \text{kg} \]
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Result: The maximum load that the elevator can carry is approximately 559.07 kg.
Thus, rounding this value for practical purposes, we can specify the maximum load as 559 kg.
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