# Chain falling in a scale

geoffrey159

## 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:

Gold Member
What is N_x in your force equation (dp/dt)? I think there may be an inconsistency in that statement.

geoffrey159
Hello, ##N_x## is the normal force from the scale on the chain

Gold Member
Ok, so then what's W_x?

geoffrey159
It is the weight of the chain on the scale after length x has fallen

geoffrey159
No obviously I'm wrong, the net external force in the system chain-scale is ## N_x - Mg ## and not ## N_x - \frac {M}{l}gx##, so ## N_x =3 \frac {M}{l}gx##,

Gold Member
Haha, yea sorry about that, I forgot how a scale worked for a minute. I just went through and double check you, it looks good.

Gold Member
The chain that's still in the air doesn't exert a weight force on the scale.

geoffrey159
yes, but it must be taken into account because it is an external force in the system chain-scale

Gold Member
Gravity accelerates the chain at a constant velocity. If you drop two balls, one on top of the other (neglecting drag) does on exert a force on the other?

Gold Member
Also, if you replace your mass/unit length with the entire mass, wouldn't the scale always read more than the weight of the chain?

geoffrey159
I think that in my first attempt I have cheated with the relationship between net external force and momentum.
I am using momentum of the chain-scale system, but don't take into account the weight of the falling chain, but it is an external force.
So without cheating,
## N_x = 3 \frac{M}{l}gx## and not ## N_x = 4 \frac{M}{l}gx - Mg ##

Gold Member
Doesn't the falling chain have mass (m/l)(l-x)? The chain that's already on the scale has 0 momentum. What do you mean you didn't take into account the weight of the falling chain? Maybe I'm missing something here.

geoffrey159
The definition is net external force = dP/dt.
I used the momentum of the chain-scale system, there is a part that is on the scale at rest, and a part falling.
But I only took into account the weight of the chain on the scale, which is not correct in this system, because the falling part contributes to total external force.

geoffrey159
A way to be 100% sure that It was wrong at first attempt is that ##N_x < 0## for ## 0 \le x \le \frac{l}{4}##, which is not admissible.

Homework Helper
Gold Member
The definition is net external force = dP/dt.
I used the momentum of the chain-scale system, there is a part that is on the scale at rest, and a part falling.
But I only took into account the weight of the chain on the scale, which is not correct in this system, because the falling part contributes to total external force.
You can get there by looking at the whole system, but it's unnecessarily complicated. All the weigh pan cares about is the speed and density of the chain hitting it.
At time t after release, with what velocity is the chain hitting the pan? What is the rate of change of momentum on those links?

If you want to look at the whole system, that will include gravity acting on the falling part of the chain, and the rate of change of momentum of that part of the chain. These two will balance, which is why you can ignore them.

Gold Member
From where I'm sitting, I see the scale can read the part that's sitting on it, at rest, definitely. Also, there will be some portion of the chain that has recently come into contact with the scale, and the scale must exert a force greater than the weight force of that segment to slow it down. That's newtons second law. But I for some reason am not seeing how the top most point of the chain, or the point under that, or... will exert a force on the scale or vice versa.

geoffrey159
You can get there by looking at the whole system, but it's unnecessarily complicated. All the weigh pan cares about is the speed and density of the chain hitting it.
At time t after release, with what velocity is the chain hitting the pan? What is the rate of change of momentum on those links?

If you want to look at the whole system, that will include gravity acting on the falling part of the chain, and the rate of change of momentum of that part of the chain. These two will balance, which is why you can ignore them.

I'm trying, but I don't understand your reply. Can you restate it? Do you agree with the solution ?