- #1

geoffrey159

- 535

- 72

## Homework Statement

A chain of mass M and length l is suspended vertically with its lowest end touching a scale. The chain is released and falls onto the scale. What is the reading of the scale when a length of chain, x, has fallen? (Neglect the size of individual links.)

## Homework Equations

Momentum, conservation of energy

## The Attempt at a Solution

Hello, I have checked threads who dealt with this problem but did not understand everything !

After reading different threads, I understand that the problem is to find the normal force from the scale on the chain.

Firstly, the net force on the scale is equal to the time derivative of momentum:

## \frac{dP_x}{dt} = N_x - W_x = N_x - \frac{M}{l}xg##

The change of momentum is

## \left\{

\begin{array}{}

P_x(t) = \frac{M}{l}(l-x(t)) \times (-\dot x(t) )\\

P_x(t+\triangle t) = \frac{M}{l}(l-x(t+\triangle t)) \times (-\dot x(t+\triangle t))

\end{array}

\right. ##

So

##\frac{dP_x}{dt} = -M \ddot x + \frac{M}{l}({\dot x}^2+ x \ddot x ) = \frac{M}{l}g(x-l) +\frac{M}{l}{\dot x}^2

##

because ##\ddot x = g ##.

In order to find ##\dot x^2##, I use conservation of energy on the point mass located at the top of the chain, which is accelerated by gravity:

## \left.

\begin{array}{}

K_0 = 0 ,\ U_0 = \frac{M}{l} g l = Mg \\

K_x = \frac{1}{2} \frac{M}{l}{\dot x}^2, \ U_x = \frac{M}{l} g (l-x) \\

K_0 + U_0 = K_x + U_x

\end{array}

\right\}

\Rightarrow {\dot x}^2 = 2gx \Rightarrow \frac{dP_x}{dt} = 3\frac{M}{l}gx - Mg ##

So

## N_x = \frac{M}{l}gx + \frac{dP_x}{dt} =4 \frac{M}{l}gx - Mg ##

Is that correct?

Last edited: