Two masses m_1 and m_2 are connected by a lig | vedclass

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Question

(B) For a system of two masses $m_1$ and $m_2$ connected by a string over a pulley,the tension $T$ in the string is given by:$T = \frac{{2{m_1}{m_2}}}

Vedclass · Accepted Answer

$\frac{{2{m_1}{m_2}gL}}{{YS({m_1} + {m_2})}}$

Vedclass · Answer

(B) For a system of two masses $m_1$ and $m_2$ connected by a string over a pulley,the tension $T$ in the string is given by:$T = \frac{{2{m_1}{m_2}}}{{{m_1} + {m_2}}}g$The stress in the string is defined as the force per unit area:$	ext{Stress} = \frac{T}{S} = \frac{{2{m_1}{m_2}g}}{{S({m_1} + {m_2})}}$According to Hooke's Law,Young's modulus $Y$ is the ratio of stress to strain:$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{	ext{Stress}}{l/L}$Rearranging for elongation $l$:$l = \frac{	ext{Stress} \cdot L}{Y} = \frac{{2{m_1}{m_2}g}}{{S({m_1} + {m_2})}} \cdot \frac{L}{Y}$Therefore,the elongation $l$ is:$l = \frac{{2{m_1}{m_2}gL}}{{YS({m_1} + {m_2})}}$

Two masses $m_1$ and $m_2$ are connected by a light string passing over a pulley as shown in the figure. If the cross-sectional area of the string is $S$,its length is $L$,and its Young's modulus is $Y$,what is the elongation in the string?

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