Question

Two unequal masses are connected by a very light string over a clamped, light, smooth pulley. Find the acceleration of the system and the tension in the string.

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Answer

To solve for the acceleration and tension in a system where two unequal masses ( m_1 ) and ( m_2 ) are connected by a very light string over a clamped, light, smooth pulley, we can follow these steps:

Step-by-Step :

1. Accelerations and Tensions in the System

  • Assume ( m_1 > m_2 ). Thus, ( m_1 ) will move downward, and ( m_2 ) will move upward, both with acceleration ( a ).

  • The forces acting on ( m_1 ) and ( m_2 ) will create the following equations (considering downward as positive for ( m_1 ) and upward as positive for ( m_2 )):

    • For mass ( m_1 ): [ m_1 g - T = m_1 a \quad \text{(Equation 1)} ]

    • For mass ( m_2 ): [ T - m_2 g = m_2 a \quad \text{(Equation 2)} ]

2. Finding Acceleration

  • Add Equations (1) and (2): [ (m_1 g - T) + (T - m_2 g) = m_1 a + m_2 a ] [ m_1 g - m_2 g = (m_1 + m_2) a ] [ (m_1 - m_2) g = (m_1 + m_2) a ]

  • Solve for ( a ): [ a = \frac{(m_1 - m_2) g}{m_1 + m_2} ]

3. Finding Tension

  • Substitute ( a ) back into Equation (1): [ T = m_1 g - m_1 a ]

  • Substitute ( a = \frac{(m_1 - m_2) g}{m_1 + m_2} ): [ T = m_1 g - m_1 \left( \frac{(m_1 - m_2) g}{m_1 + m_2} \right) ]

  • Simplify: [ T = m_1 g \left( 1 - \frac{m_1 - m_2}{m_1 + m_2} \right) ] [ T = m_1 g \left( \frac{m_1 + m_2 - (m_1 - m_2)}{m_1 + m_2} \right) ] [ T = m_1 g \left( \frac{2 m_2}{m_1 + m_2} \text{ (common factor of } m_1 + m_2 \text{ gets cancelled) }\right) ] [ T = \frac{2 m_1 m_2 g}{m_1 + m_2} ]

Final Expressions:

  1. Acceleration ( a ): [ \boldsymbol{a = \frac{(m_1 - m_2) g}{m_1 + m_2}} ]

  2. Tension ( T ): [ \boldsymbol{T = \frac{2 m_1 m_2 g}{m_1 + m_2}} ]

These expressions provide the acceleration and tension in the system given two unequal masses connected by a string over a pulley.


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