Two weights \( w_{1} \) and \( w_{2} \) are suspended from the ends of a light string over a smooth fixed pulley. the pulley is pulled up with acceleration g then the tension in string will be

Question

Two weights \( w_{1} \) and \( w_{2} \) are suspended from the ends of a light string over a smooth fixed pulley. If the pulley is pulled up with acceleration \( g \), the tension in the string will be 1) \( \frac{4 w_{1}+w_{2}}{w_{1}+w_{2}} \) 2) \( \frac{2 w_{1} w_{2}}{w_{1}+w_{2}} x \) \( \frac{w_{1}-w_{2}}{w_{1}+w_{2}} \times \) - \( \frac{w_{1} w_{2}}{2\left(w_{1}+w_{2}\right)} \) A block of mass \( m \) is resting on a smool

Answer 1

Weight of object 1 = \(\frac{Mass}{Gravity} = \frac{m_1}g\)

Weight of object 2 = \(\frac{Mass}{Gravity} = \frac{m_2}g\)

\(m_2 g - T = m_2a - g\)   .....(1)

\(-mg + T = m_1a - g\)  .....(2)

(1) - (2)

\(2m_2g - 2m_1g = m_2 a + m_1a\)

\(a = \frac{2g(m_2 - m_1)}{(m_1 + m_2)}\)

Now tension in string is

\(T = m_1g + m_1a + m_1g = 2m_1g + m_1a\)

Reducing value of mass of object

\(= \frac{w_1}g \left\{2g + \frac{2g(m_2 - m_1)}{(m_1 + m_2)}\right\}\)

\(= w_1 \left\{\frac{2(m_2 + m_1) + 2(m_2 - m_1)}{(m_1 + m_2)}\right\}\)

\(= w_1 \left\{\frac{4\frac{w^2}4}{\frac{w_1}g + \frac{w_2}g}\right\}\)

\(= \frac{4w_1w_2}{w_1 + w_2}\)