Question

Find its Tension and Acceleration​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{T= \frac{( m_{1} - m_{2}g \: sin \: \theta ) m_{2} g}{ m_{1} + m_{2} } + 2 m_{2}g \: sin \: \theta}}}$

$\green{\tt{\therefore{a = \frac{( m_{1} - m_{2} \: sin \: \theta)g}{ (m_{1} + m_{2}) }}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given:}}} \\ \tt: \implies Angle \: of \: wedge = \theta \\ \\ \tt: \implies Mass \: of \: blocks = m_{1} \: and \: m_{2} \\ \\ \red{\underline{\bold{To \: Find:}}} \\ \tt: \implies Tension \: in \: the \: string =? \\ \\ \tt: \implies Acceleration \: of \: both \: blocks =?$

• According to given question :

$\bold{For \: mass \: of \: block( m_{1}) } \\ \tt: \implies m_{1}g - T= m_{1}a - - - - - (1) \\ \\ \bold{For \: block \: of \: mass( m_{2})} \\ \tt: \implies \frac{T}{2} - m_{2}g \: sin \: \theta = \frac{m_{2}a}{2} \\ \\ \tt: \implies T - 2 m_{2}g \: sin \: \theta = m_{2}a - - - - - (2) \\ \\ \bold{Adding \: (1) \: and \: (2)} \\ \tt: \implies m_{1}g - 2 m_{2}g \: sin \: \theta = m_{1}a + m_{2}a \\ \\ \tt: \implies( m_{1} - m_{2} \: sin \: \theta)g =( m_{1} + m_{2})a \\ \\ \green{\tt: \implies a = \frac{( m_{1} - m_{2} \: sin \: \theta)g}{ (m_{1} + m_{2}) } } \\ \\ \text{Putting \: value \: of \: a \: in \: (2)} \\ \\ \tt: \implies T - 2 m_{2}g \: sin \: \theta = m_{2} \times \frac{( m_{1} - m_{2}g \: sin \: \theta )g }{ m_{1} + m_{2} } \\ \\ \green{\tt: \implies T= \frac{( m_{1} - m_{2}g \: sin \: \theta ) m_{2} g}{ m_{1} + m_{2} } + 2 m_{2}g \: sin \: \theta}$