# Question for anyone with Morin's "Introduction to Classical Mechanics"

• dyn
In summary: It is zero. The net force on it is its weight, which is balanced by the normal force of the scale. So the rate of change of momentum of that part of the chain is zero. The rate of change of momentum of the moving part of the chain is what is equated to the force on the moving part of the chain.In summary, the conversation discusses a problem from a physics book about a chain on a scale. Two different methods are presented for solving the problem, one using a free-body diagram approach and the other focusing on the force on the scale. There is confusion about how the two methods are equivalent and why one method requires the addition of the weight already
dyn
Hi.
There is a worked example in this book on P168-169 titled "Chain on a scale". Two different ways of obtaining the solution are shown. I am confused about the 2 different methods.
Method 1 equates the rate of change of momentum of the chain to the net force on the chain giving F.
Method 2 equates the rate of change of momentum to part of the force called ##F_(dp/dt)## which is then added to ##F_(weight)## to give F.
I am confused about why method 2 needs the addition of the weight already on the table but method 1 doesn't even though both methods equate forces to rates of change of momentum.
Any help would be appreciated.
Thanks

I have seen quite a few interesting problems from that book posted here. I do not own the book and I don't know how many PF members do. Your question is one of strategy. If you posted the statement of the problem and more details on the two methods, perhaps someone might be able to explain how the two are equivalent. Otherwise wait until someone who has access to the book responds.

Thanks for replying. The question is regarding a chain held vertically above a scale/table and then released and the force exerted by the scale/table. I will wait and hope that someone has the book. It is an excellent book

dyn said:
Hi.
There is a worked example in this book on P168-169 titled "Chain on a scale". Two different ways of obtaining the solution are shown. I am confused about the 2 different methods.
Method 1 equates the rate of change of momentum of the chain to the net force on the chain giving F.
Method 2 equates the rate of change of momentum to part of the force called ##F_(dp/dt)## which is then added to ##F_(weight)## to give F.
I am confused about why method 2 needs the addition of the weight already on the table but method 1 doesn't even though both methods equate forces to rates of change of momentum.
Any help would be appreciated.
Thanks
In don't have the book. But it sounds like two different reference frames: free falling vs. rest frame of the table. In the free falling frame the weight force effectively transforms away.

dyn said:
Method 1 equates the rate of change of momentum of the chain to the net force on the chain giving F.
Method 2 equates the rate of change of momentum to part of the force called ##F_(dp/dt)## which is then added to ##F_(weight)## to give F.
I am confused about why method 2 needs the addition of the weight already on the table but method 1 doesn't even though both methods equate forces to rates of change of momentum.
Hi Dyn.

Just a little suggestion if you want people to respond: It is helpful to print the question. I found a copy online and the problem is stated as follows:

"Example (Chain on a scale): An “idealized” (see the comments following this
example) chain with length L and mass density σ kg/m is held such that it hangs
vertically just above a scale. It is then released. What is the reading on the scale, as a
function of the height of the top of the chain?"

The author then goes on to explain two different approaches.

In the first approach, he takes the free-body-diagram approach, determines the net force on the chain and sets that equal to dp/dt. So dp/dt is equal to the normal force of the scale on the chain + the force of gravity on the chain. That dp/dt is the rate of change of momentum of the whole chain, some of which is acclerating downward and some of which is accelerating upward (the part that is stopping on the scale).

In the second approach he focuses on just the force on the scale. That force is the weight of the part of the chain that is at rest on the scale + force on the scale due to the stopping of the chain. The latter is not the same as the dp/dt in the first approach because it just takes into account the part of the chain that is stopping, not the part of the chain above the scale that is accelerating downward in free-fall. (That part could have been better explained).

AM

PeroK, kuruman and Delta2
Andrew Mason said:
Hi Dyn.

"Example (Chain on a scale): An “idealized” (see the comments following this
example) chain with length L and mass density σ kg/m is held such that it hangs
vertically just above a scale. It is then released. What is the reading on the scale, as a
function of the height of the top of the chain?"

The author then goes on to explain two different approaches.

In the first approach, he takes the free-body-diagram approach, determines the net force on the chain and sets that equal to dp/dt. So dp/dt is equal to the normal force of the scale on the chain + the force of gravity on the chain. That dp/dt is the rate of change of momentum of the whole chain, some of which is acclerating downward and some of which is accelerating upward (the part that is stopping on the scale).

In the second approach he focuses on just the force on the scale. That force is the weight of the part of the chain that is at rest on the scale + force on the scale due to the stopping of the chain. The latter is not the same as the dp/dt in the first approach because it just takes into account the part of the chain that is stopping, not the part of the chain above the scale that is accelerating downward in free-fall. (That part could have been better explained).

AM
Thanks for your reply. I'm still confused because in Morin's 1st method he states that the momentum of the entire chain just comes from the moving part

dyn said:
I'm still confused because in Morin's 1st method he states that the momentum of the entire chain just comes from the moving part
What do you think the momentum of the part at rest is?

PeroK
The momentum of the part at rest is zero

## 1. What is Morin's "Introduction to Classical Mechanics" about?

Morin's "Introduction to Classical Mechanics" is a textbook that covers the fundamental principles and laws of classical mechanics, including topics such as kinematics, dynamics, energy, and momentum.

## 2. Is this textbook suitable for beginners?

Yes, this textbook is suitable for beginners as it starts with the basics and gradually builds upon them. It also includes many examples and practice problems to help students understand the concepts.

## 3. What makes Morin's "Introduction to Classical Mechanics" different from other textbooks on the subject?

One of the main differences is that this textbook focuses on the development of physical intuition and problem-solving skills rather than just memorizing equations. It also includes many real-world examples and applications to help students see the relevance of the concepts.

## 4. Are there any prerequisites for using this textbook?

Some basic knowledge of physics and calculus is recommended, but the textbook does not assume any prior knowledge of classical mechanics.

## 5. Can this textbook be used for self-study or is it better suited for a classroom setting?

This textbook can be used for both self-study and in a classroom setting. It includes many exercises and problems for self-study and also provides a comprehensive coverage of the subject for classroom use.

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