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

AI Thread Summary
The discussion centers on a problem from Morin's "Introduction to Classical Mechanics" involving a chain released above a scale. Two methods for solving the problem are presented, with Method 1 focusing on the net force equating to the rate of change of momentum of the entire chain, while Method 2 considers the force on the scale, combining the weight of the chain at rest and the force due to the chain's stopping. The confusion arises from why Method 2 includes the weight already on the table, whereas Method 1 does not. Clarifications suggest that Method 1 accounts for the entire chain's momentum, while Method 2 only addresses the portion of the chain that is stopping. Understanding these different perspectives on momentum and force is key to resolving the confusion.
dyn
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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
 
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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
 
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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?
 
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The momentum of the part at rest is zero
 
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