Weight of gold chain when dropped on a weighing scale

[ShaunPereira]

Homework Statement

A jeweler is holding a gold chain of uniform mass per unit length hanging vertically just above a weighing scale as shown in the figure. He offers to charge the customer for half of the maximum reading of the scale, after he releases the chain. What percentage more than the actual price does the customer pay if he agrees to the offer?

Relevant Equations

F=ma
F=Δp/Δt

The force exerted downwards on the scale by the chain when it is kept on it would be
F_g= Mg
=λLg where λ is the linear mass density

However when the chain is dropped onto the scale it exerts an additional force due to its change in momentum
The force exerted by each part of the chain would be different as each part is at a different height from the scale and hence gains a different velocity and momentum
Hence we resort to integration

Lets take a small elemental mass dm with length dx at a height x from the scale
as this mass falls it gains a final velocity equal to √(2gx)
the momentum of the mass would be √(2gx) dm
the mass comes to rest on impact and hence change in momentum would be √(2gx) dm
This change in momentum would cause an additional force
dF_m= dp/dt
dF_m= √(2gx) dm/dt
dm/dx =λ
Therefore dm=λdx

Substituting and rearranging we get
dF_mdt=√(2gx) λdx

we need to integrate this equation to find F_m

The problem I am facing is what do I take dt to be and for that matter what do I take its integral t to be?
Is it the time taken for the change in momentum. In that case how do I find that ?
OR is it the time taken for the elemental mass dm to fall a distance x
t=√(2x/g)Another way in which I have tried to solve this problem is to take the centre of mass of the chain to be at length L/2 from the scale as it is uniform
the velocity gained by this mass would be v=√(gl)
Therefore the change in momentum being mv= λl(√(gl))

But even here I am faced by the same problem of what to take as time in order to find the force exerted by the chain due to its change in momentum.

 

[Orodruin]

we need to integrate this equation to find Fm

This is incorrect. The force that causes the change in momentum is only present for a link exactly when that link drops onto the scale. Therefore, this force must not be integrated. You do however need to find out what the force is and express it as a function of how much of the chain has fallen onto the scale.

To this you will need to add the force required to keep the links already lying on the scale stationary.

 

[ShaunPereira]

You do however need to find out what the force is and express it as a function of how much of the chain has fallen onto the scale.

I have no idea how to do that.

Also I didn't understand how the force that causes change in momentum could only be present when the chain drops onto the scale
Also which part of the chain are you talking about?
Is it the end closest to the scale(that is not exactly dropped because it is held so close to the scale that it can be assumed that it is on the scale)
or is it the end away from the scale at a distance of L or is it the centre of the chain at a distance L where the mass can be assumed to be concentrated as the chain is uniform?

 

[haruspex]

I have no idea how to do that.

Also I didn't understand how the force that causes change in momentum could only be present when the chain drops onto the scale
Also which part of the chain are you talking about?
Is it the end closest to the scale(that is not exactly dropped because it is held so close to the scale that it can be assumed that it is on the scale)
or is it the end away from the scale at a distance of L or is it the centre of the chain at a distance L where the mass can be assumed to be concentrated as the chain is uniform?

As each link hits the scale it loses momentum. If a link hits each $\Delta t$ and loses momentum $\Delta p$ then it exerts average force $\Delta p/\Delta t$.

 

[Orodruin]

Also I didn't understand how the force that causes change in momentum could only be present when the chain drops onto the scale

If it was present after the chain dropped, the chain would jump off the scale.

 

[ShaunPereira]

If it was present after the chain dropped, the chain would jump off the scale.

True. I didn't mean that though. What I was trying to say is that the force exerted by the scale would be present as long as the chain takes to fall on the scale

The force exerted on a small mass dm would be
v dm/dt
=λvdx/dt
=λv²...( Can we take dx/dt=v? I don't know.)
=λ2gx
also the force that the elemental mass exerts on the scale due to its weight will be
gdm
=λgx

Hence net force =3λgx
but isn't this the force exerted only on the elemental mass and not the total chain?
How do I find the TOTAL force exerted ?

 

[Orodruin]

exerts on the scale due to its weight will be
gdm=λgx

dm is equal to λ dx, not λx. You need to take the full weight of the chain on the scale (this is where you need to integrate - or simply state that the weight of all of the chain that is on the scale is λgx since it has length x).

 

[ShaunPereira]

m is equal to λ dx, not λx. You need to take the full weight of the chain on the scale (this is where you need to integrate - or simply state that the weight of all of the chain that is on the scale is λgx since it has length

Yes. But that would solve only part of the problem. I have stated in the previous post that the force exerted due to change in momentum of elemental mass is 2λgx

But this is only the force on a small mass dm and not on the entire chain

How do I find the total force exerted on the chain due to change in its momentum from this information?
You've previously stated that we wouldn't need integration for that. How else should I then find the total force exerted on the chain due its change in momentum when I have the expression dF_m=2λgx for the elemental mass?

 

[Orodruin]

Only the small part of the chain falling onto the scales in the time $\Delta t$ experiences the change of momentum in this time interval. The other parts of the chain do not, they are either falling freely or have already stopped.

 

[ShaunPereira]

Oh! So do we have to find the force at any instant, which in this case must also be the maximum value ?
The force due to elemental mass dm would be 2λgx. But since we need the maximum value of force it would be 2λgL which means that the instant an elemental mass dm which was originally at height L from the scale touches the scale, transfers its momentum to the scale and comes to rest the force due it would be 2λgL.

Also the force due to total mass would on the scale would be λgl
The scale would at that instant read 3λgL

This could also mean that the scale would from the instant the chain is dropped show different values till an instant where it would be maximum which also happens to be when the last elemental mass dm at the tip of the chain i.e. at length L away from the scale touches it.

 

[haruspex]

Oh! So do we have to find the force at any instant, which in this case must also be the maximum value ?
The force due to elemental mass dm would be 2λgx. But since we need the maximum value of force it would be 2λgL which means that the instant an elemental mass dm which was originally at height L from the scale touches the scale, transfers its momentum to the scale and comes to rest the force due it would be 2λgL.

Also the force due to total mass would on the scale would be λgl
The scale would at that instant read 3λgL

This could also mean that the scale would from the instant the chain is dropped show different values till an instant where it would be maximum which also happens to be when the last elemental mass dm at the tip of the chain i.e. at length L away from the scale touches it.

Exactly.

 

[ShaunPereira]

Exactly.

Thank you.

 

[jbriggs444]

There is a different approach to the problem that can be taken.

Since the chain starts and ends at rest, the average supporting force over the interval of the fall must be equal to its normal rest weight.

Both the support force required for the portion of the chain that has completed its fall and is now at rest and the support force for the incremental chain that is being stopped by the table are quadratic in time. This follows because the amount of already-fallen chain is proportional to displacement and displacement is quadratic in time. Meanwhile the velocity of the falling chain is linear in time and hence both the mass flow rate and the delta v are linear in time. Their product then is quadratic.

So the support force takes the form $F = kt^2$ for some unknown constant $k$.

The average (over time) support force is the integral over time of the instantaneous support force divided by the time interval. If we wave our hands and use a system of units in which the unit of force is the weight of the chain and the unit of time is the interval of the fall then we have that:$$\int_0^1 kt^2 dt = \frac{1}{3}kt^3 = 1$$So we conclude that $k = 3$ and the maximum force of three times the chain's normal weight is achieved at the end of the fall when $t=1$ in our chosen units.