Question:
Two weights \(w_1\) and \(w_2\) are suspended to the two strings on a frictionless pulley. When the pulley is pulled up with an acceleration \(g\) then the tension in the string is
Two weights \(w_1\) and \(w_2\) are suspended to the two strings on a frictionless pulley. When the pulley is pulled up with an acceleration \(g\) then the tension in the string is
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When pulley accelerates upward with \(a\), use \(g_{\text{eff}} = g + a\).
Updated On: Apr 23, 2026
- \(\frac{4w_1w_2}{w_1 + w_2}\)
- \(\frac{2w_1w_2}{w_1 + w_2}\)
- \(\frac{w_1w_2}{w_1 + w_2}\)
- \(\frac{w_1w_2}{2(w_1 + w_2)}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Atwood machine with pulley accelerating upward. Effective acceleration due to gravity becomes \(g_{\text{eff}} = g + a\) where \(a\) is upward acceleration.
Step 2: Detailed Explanation:
Pulley acceleration = \(g\) upward. Effective \(g_{\text{eff}} = g + g = 2g\).
In Atwood machine: tension \(T = \frac{2m_1m_2}{m_1 + m_2} \times g_{\text{eff}} = \frac{2m_1m_2}{m_1 + m_2} \times 2g\).
\(w_1 = m_1g\), \(w_2 = m_2g\).
\(T = \frac{2(w_1/g)(w_2/g)}{(w_1/g) + (w_2/g)} \times 2g = \frac{2w_1w_2/g^2}{(w_1 + w_2)/g} \times 2g = \frac{2w_1w_2}{g(w_1 + w_2)} \times 2g = \frac{4w_1w_2}{w_1 + w_2}\).
Step 3: Final Answer:
Thus, tension = \(\frac{4w_1w_2}{w_1 + w_2}\).
Atwood machine with pulley accelerating upward. Effective acceleration due to gravity becomes \(g_{\text{eff}} = g + a\) where \(a\) is upward acceleration.
Step 2: Detailed Explanation:
Pulley acceleration = \(g\) upward. Effective \(g_{\text{eff}} = g + g = 2g\).
In Atwood machine: tension \(T = \frac{2m_1m_2}{m_1 + m_2} \times g_{\text{eff}} = \frac{2m_1m_2}{m_1 + m_2} \times 2g\).
\(w_1 = m_1g\), \(w_2 = m_2g\).
\(T = \frac{2(w_1/g)(w_2/g)}{(w_1/g) + (w_2/g)} \times 2g = \frac{2w_1w_2/g^2}{(w_1 + w_2)/g} \times 2g = \frac{2w_1w_2}{g(w_1 + w_2)} \times 2g = \frac{4w_1w_2}{w_1 + w_2}\).
Step 3: Final Answer:
Thus, tension = \(\frac{4w_1w_2}{w_1 + w_2}\).
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