What Is the Total Force Exerted by a Falling Chain on the Ground?

[sachin123]

A chain of length L,mass M kept vertical so it just touches ground.A height x of it from bottom falls down.Find the total force exerted by it on ground.
My attempt:
Force due to its weight=Mg
Force due to falling and change in momentum:
consider small element dy at height y
Change in momentum when it falls:M X root(2gy)
on integrating M X root(2gy) within limits 0 and x and dividing the integral by total time taken I get an expression.On adding it to Mg,I don't get the right answer.
Answer is:
3Mgx/L
Can someone help?Thanks

 

[collinsmark]

A chain of length L,mass M kept vertical so it just touches ground.A height x of it from bottom falls down.Find the total force exerted by it on ground.
My attempt:
Force due to its weight=Mg

Only part of the chain is on the ground, not the entire mass M of the chain. The fraction of the chain on the ground is x/L. Multiply that times the total mass to get the mass that happens to be on the ground. Then multiply by g to get the force due to its weight.

Force due to falling and change in momentum:
consider small element dy at height y
Change in momentum when it falls:M X root(2gy)

The above equation has some problems for a couple of reasons. Let's just start over.

The fraction of the length of the chain that is falling onto the ground at any particular time is dy/L. Multiply that times the total mass M to find dm, the differential mass that happens to be falling on the ground at any particular time.

Now find the velocity, v, of piece of the chain as it hits the ground. It will be a function of its initial height above the ground, x.

Once you have dm and v, you can multiply them together to get dp, the differential change in momentum.

Divide both sides of the equation by dt, and now you have the force dp/dt, due to the changing of momentum.

Your result should have a dy/dt term in it. But guess what that is. It's the velocity! And you've already solved for that. Make your substitution, sum everything together, and you have your answer. [Hint: no integration required.]

 

[zorro]

Now find the velocity, v, of piece of the chain as it hits the ground. It will be a function of its initial height above the ground, x.

I think the initial height of the particular element of chain will be 'y'. As there are different chain elements at different heights, there must be some integration involved. Your method will lead to the force exerted by a small element on falling, not the whole 'x' part of the chain.

Make a slight modification in your steps.
Don't divide both the sides of equation by dt. Find the total momentum of the chain P instead.
Then divide it by the time taken by the whole 'x' part of the chain to fall down. You get an expression for total force.

I got it as F=2Mgx/3L

 

[collinsmark]

I think the initial height of the particular element of chain will be 'y'.

That's fine in the end, but eventually one needs to equate x = y.

The only reason I brought the whole y variable into the picture is to use the original poster's terminology.

A more compact approach is to not use the variable y at all, and instead use x for everything.

Without losing any generality, we can redefine x as "the initial height above the ground of a particular section of the rope, dx." Thus when a particular section of rope eventually hits the ground (section dx), there is x amount of rope already on the ground.

Again, my previous hints were done substituting y for x, and dy for dx, but we could have just as well used x and dx from the very beginning.

As there are different chain elements at different heights, there must be some integration involved.

Not in this problem. See below.

Your method will lead to the force exerted by a small element on falling, not the whole 'x' part of the chain.

That's right. We broke up the problem into two parts.

1. Force due to the weight of the particular part of the chain that already happens to be at rest on the ground.
2. Force due to the changing of momentum as section dx (previously dy) hits the ground.

Part 1. is very straightforward. It's just the weight of rope starting at the end of the rope up to length x. This part of the rope isn't changing momentum. It's simply at rest on the ground, and the only force exerted on it by the ground is the normal force due to its weight.

For part 2. we must only consider the instantaneous change in momentum of infinitesimal section of section dx (previously dy). Think of it this way. The part of the chain that is already on the ground isn't changing momentum. It's at rest, sitting there quietly. The part of the chain that hasn't hit the ground yet, isn't experiencing any forces from the ground yet. That part of the chain is in perfect freefall, and the ground doesn't even know it exists yet. The only part of the change that matters is the small section of chain dx that is currently hitting that ground.

The total instantaneous force from the ground is the sum of parts 1 and 2.

Make a slight modification in your steps.
Don't divide both the sides of equation by dt. Find the total momentum of the chain P instead.
Then divide it by the time taken by the whole 'x' part of the chain to fall down. You get an expression for total force.

You could do this with the exception that instead of dividing by the total time, you need to differentiate to find the instantaneous change in momentum. By dividing by the total time you're finding the average change in momentum, which would give you the average total force that the ground exerts. This problem is asking you for the instantaneous total force exerted by the ground (at least that's how I interpret it [and if one interprets it this way the "given" correct answer is correct]).

[Edit: I concede that the original problem statement isn't very clear on this. But it's the only interpretation that gives the correct answer. I interpret the problem statement as requesting, "give the total instantaneous force exerted on the chain by the ground, as a function of x, where x is the original height of the particular section of chain presently hitting the ground (which is the same thing as saying that x is the length of chain presently on the ground at the particular time."]

So what you could do, is integrate to find the total change in momentum and then differentiate (with respect to t) to find the instantaneous change in momentum, giving you what you started with. It's kind of a useless step. If you already have dp/dt, there's no reason to integrate (to find p) and differentiate again (to find dp/dt), because you already have dp/dt from the start and that's what you're looking for.

I got it as F=2Mgx/3L

Sorry, not quite correct.

The correct answer is given in the original post, F = 3gMx/L.

 

[zorro]

I did not see the correct answer in O.P. I thought that we have to find the average force exerted which is not as per the answer. Your solution is perfect

P.S. - I like the way you solve problems - by drowning in them

 

[sachin123]

Thanks a lot collinsmark
and Abdul.
I thought we had to find the total force exerted.
Suppose that were the case,how would you find the total force(average) due to gravity.We would have to integrate right?Can you give me an expression?I couldn't find it.

 

[zorro]

Suppose that were the case,how would you find the total force(average) due to gravity.We would have to integrate right?Can you give me an expression?I couldn't find it.

There is no avg. force due to gravity (no instantaneous either).The force which I found out (F=2Mgx/3L) is just the average force exerted on hitting the ground. It does not include the force due to its weight (gravity). For that add up Mgx/L to it, which is the force due to gravity.

Did you find out the expression for dp as Collinsmark explained? If yes then integrate both sides of the equation with appropriate limits to find out total momentum P. Divide it by the total time taken by 'x' part of the chain to fall. You will get an expression for avg. force exerted.

 

[sachin123]

That I found. But what about the gravity? Force due to it keeps varying as length of chain on ground varies.

 

[collinsmark]

That I found. But what about the gravity? Force due to it keeps varying as length of chain on ground varies.

Yes, that's right. The instantaneous force varies as a function of x, the length of chain that is presently on the ground. And the independent variable x is already in your equation, F = 3Mgx/L.

I believe the problem statement calls it the "total" force, because it's asking for the sum of the force due to the chain's weight (that already happens to be on the ground) and the force due to the changing of momentum.

As Abdul points out, one could calculate the time averaged force, which is the instantaneous force averaged over the period of time that it takes for a fixed length of chain to drop. But the problem doesn't seem to be asking for that. I assume that the problem is asking for the instantaneous force.

But remember this: force does not accumulate over time (momentum, perhaps yes. But not force). If you push on a wall with both hands, with a given force for 1 second, it's the same force even if you push on the wall for 10 seconds. It's called the "total" force because you're pushing on the wall with both hands. But it doesn't matter how long you push, if you're only being asked to find the force.

 

[sachin123]

Alright.But how do I find a time averaged force due to gravity?(though not for this problem as you say)

 

[collinsmark]

Alright.But how do I find a time averaged force due to gravity?(though not for this problem as you say)

In general:

• First, find the total time of whatever it is you're looking for (the time it takes for some specific section x to hit the ground [hint: use kinematics]). I'll call this parameter T.
• Express the instantaneous force in question as a function of time, t.
• Then calculate the time averaged force, f_ta:

f_{ta} = \frac{\int_0^T f(t)dt}{T}

Back this particular problem.
Since you already know the overall instantaneous force, you can use the above method directly on the

F(y) = 3Mgy/L​

But of course, that equation is a function of y. You first need to express it as a function of t. How does y relate to t?
Also, you'll need to calculate the overall time, T. This parameter T will be a function of x.

Or, you could break up the problem into two parts, like we did for the instantaneous force.

• Use the above method to find the time averaged weight of the chain.
• Use Abdul's method to find the time averaged force due to momentum. (Rather than converting everything to be a function of time, you can integrate with respect to y to find the total change of momentum, and then divide that by T. See Abdul's previous post for details.) Alternately, you use the instantaneous force due to the change in momentum, convert that to a function of t (instead of y), integrate that over the time interval 0 to T, then divide by T
• Then sum both parts together.

You should get the same answer using either method.