Physics

Problem

A uniform chain of length

L

and total mass

M

hangs from a support. The entire chain is pulled upward with a constant acceleration

a

(where

0 < a < g

). Let

x

be the distance measured downward from the support. Which expression gives the tension

T(x)

at position

x

while the chain is accelerating? a)

\displaystyle T(x)=\frac{M}{L}(g+a)x

\displaystyle T(x)=\frac{M}{L}g\,(L-x)

\displaystyle T(x)=\frac{M}{L}(g+a)(L-x)

\displaystyle T(x)=\frac{M}{L}(g-a)(L-x)