## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$T = \frac{y~m~g}{L}$
At any position $y$, the tension $T$ in the rope holds up the weight of the section of rope that is below the position $y$. At any position $y$, the mass of the rope that is below the position $y$ is equal to ($\frac{y}{L}~m$). The tension $T$ at position $y$ is equal to the weight of this mass. Therefore; $T = \frac{y}{L}~m~g$ $T = \frac{y~m~g}{L}$