# Help!

Greetings, Brilliant!

Happy Holidays!

Boy do I miss this place! But education must... LIVE ON...

So, I'm 10 weeks behind in my AP Physics C work (don't ask why) and have 5 days to make up the work. Sometimes I have questions that cannot be answered by the internet (except, of course, Brilliant ;)), or timely by my instructor. So I thought I'd make this page in hope that someone may answer my call for help when I most need it.

I'll be posting the problems I'm having issues with down below. If you have some time and willingness, please help me out. I'll greatly appreciate any support.

Thanks!

Note by John M.
6 years, 3 months ago

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Day 0. Problem 1.

Status: Resolved

Problem:

In Fig. 8-43, a chain is held on a frictionless table with one-fourth of its length anging over the edge. If the chain has length $L=28$cm and has mass $m=0.012$ kg, how much work is required to pull the hanging part back onto the table?

Solution:

Confusion:

I understand the solution. But, this is not how I originally tackled the problem. I simply calculated the change in poential energy due to the lifting of $1/4$ of the mass of the string. I did that by taking the value of $L/4$ and $m/4$ for $h$ and $m$ in $mgh$, and apparently my answer was wrong.

Either way, I have a conceptual issue: Does the length of the part of the string on the table not matter? Say it had a length of 100cm, and 7cm was still the length that was hanging. According to this solution, the work required would be exactly the same. Does this imply that there is no work required to displace the string across a frictionless table? I think not. This would imply that pushing objects in vacuum space requires no work; not true. So now the question is, how do we calculate that Work?

The way my chapter explains calculating work is to relate it to changes in Mechanical (kinetic, potential, and thermal) energies. Here, I cannot use kinetic since velocity is irrelevant, and I don't see a way of using potential energy either for a string on the table. And since the table is frictionless, there is no thermal energy transfer (as far as classical mechanics and my chapter is concerned).

So, am I missing something, or am I adding too much?

Thank you!

UPDATE:

I've realized why my earlier approach was wrong. I leave it to you to figure it out too if you haven't already. But my confusion about pulling the thing across the table still persists (and moving things in vacuum space in general). Do you think it's the book's simplicity approach?

I do.

:)

- 6 years, 3 months ago

This process of understanding the solution is an extremely valuable aspect of problem solving. Think about the part that tripped you up, and what is the "proper" way to think about it instead. That will help you develop an intuition about how to think about similar problems in future.

Staff - 6 years, 3 months ago

Instead of consider whole L/4 length as you want to do without integration consider the centre of mass of the hanging part which would be at its centre so the displacement of centre of mass would be L/8, so change in potential energy equals

$\displaystyle \dfrac{m}{4}g\dfrac{L}{8} = \dfrac{mgL}{32}$

- 6 years, 3 months ago

Hm that's an interesting approach.

However, I believe it's wrong. The fact that the center of mass approach and the integration approach yield the same result may be a coincidence, but the former approach is conceptually inappropriate:

Here's my justification of the above solution and refutal of your approach:

First of all, we agree on the book's assumption that it takes no work to slide the chain ACROSS THE TABLE. This is simply the book's simplification of the problem in practice of conservation of mechanical energy. So in other words, we neglect any work done past the part at which the chain is being pulled up.

So, as we pull up the rope, with each increment (interval) of pull, the potential energy of the hanging chain changes. This change is associated with the work required to pull the chain up. Now, if we abandon the gravitational potential energy approach for a moment and consider Work as a product of Force times Distance, we see something different. First, we know that force involves mass. It is a GREAT ERROR to treat the hanging chain as a point-mass in a free-body diagram (just as we would treat a block, for example). Can you see why?

Consider this: imagine a heavy rope. You're pulling it up as you stand on a mountain. Now, as you pull up more and more of the rope, does the rope seem to get lighter? Yes it does. This is because as you pull more and more of the rope up and less of it is left to be yet pulled up, the lighter the hanging rope becomes. Hence, less force (and thus less work) is required, per pull, to pull up the rope.

Our example is no different; instead of individual pulls, we consider an exact precision by treating the segments as differentials (infinitesimals). And note the GENIUS representation of the variable mass in terms of height:

$(m/L)dy$

Do you see it? The formatting is confusing, but, if you carefully rearrange you will notice that this is really

$m\left( \frac{dy}{L} \right)$

Do you see it now? Oh yeah ;)

- 6 years, 3 months ago

I think Krishna Sharma is correct. Gravitational potential energy is a state function, which means it does not depend on the path taken. So, we can solve this problem by just considering the potential energy initially, and finally.

Initially, we can think of it as a block of mass $\frac{m}{4}$ hanging on a string of length $\frac{L}{8}$ (at the position of the center of the mass). Finally, the block has moved up onto the table.

The change in potential energy is $\frac{m}{4} g \frac{L}{8} = \frac{mgl}{32}$. Since there is no change in kinetic energy, this is also equal to the work done to pull the chain.

I think this is the correct and exact answer, and there is nothing wrong in this approach. In fact, this is much simpler than considering the force and distance.

There are no coincidences in this world.

- 6 years, 3 months ago

Yeah you're right.

But can you see the problem from my side of the perspective? It's great that you can defend one side, but getting known to them all is the best way to winning them all. I argue that you cannot treat the chain as a projectile since with each increment of displacement the remainder mass is less than prior to displacement (variable mass), and thus treating it as such would be like simply lifting the chain a certain height as compared to simply pulling the individual remainders of the hanging mass.

What do you argue?

- 6 years, 3 months ago

But according to center of mass, the total mass is assumed to be at that point. Then the problem reduces from extended body to a point. It makes sense to approach that way.

- 6 years, 3 months ago

@John Muradeli , In many of your posts, you have talked about the mass of the hanging part, being reduced. So, according to your "Mountain" example, the rope tends to get lighter and lighter, because less mass is hanging now. But let me ask you, if you are pulling some of the rope upwards, the hanging mass is reduced, no doubt. But, now the left off hanging mass, is at a greater height from the Earth's surface, and hence the potential energy approximately remains constant, because in the product, $U\quad =\quad MgH$. So, clearly when mass decreases, height increases, and so $U$ almost remains same.

- 6 years, 3 months ago

Bro, I did exactly the same mistake when attempting a similar problem.

[Pardon any typo in the following]

My problem's statement was:"A uniform chain of length 'l' and mass 'm' overhangs a smooth table with its two third part lying on the table. Find the kinetic energy of the the chain as it completely slips off the table".

You can solve this problem in similar way (with the help of integration). You must try it. Moreover you will have a revision.

When I looked into the solution, it used calculus. I couldn't figure out why? But you have answered it :"Consider this: imagine a heavy rope. You're pulling it up as you stand on a mountain. Now, as you pull up more and more of the rope, does the rope seem to get lighter? Yes it does. This is because as you pull more and more of the rope up and less of it is left to be yet pulled up, the lighter the hanging rope becomes. Hence, less force (and thus less work) is required, per pull, to pull up the rope."

The solution makes more sense if we think integration as the process of summation. The change in Gravitational Potential energy of the end tip isn't same as that of the part 'just' above it (concept of neighborhood).We have to find them individually and add up all of them to get the accurate answer. For such cases, as you know, we have calculus(integration). You have to pull out the Mathematics from the situation.The part of Mathematics most used in solving problems in Physics is abstraction. I face difficulty at that part, every time.

However, the solution using the concept of center of mass is interesting. It makes sense to me. Using the concept of center of mass would "provide us an approximate answer" not the exact one. But calculus will give the exact one. So after obtaining an answer from any one of the method one must check it using the other. It's always recommended to have two different solutions and then compare them to better the understanding of the processes.

My whole explanation could be 100% wrong, it comes from my belief and experience. And I haven't discussed it yet with anyone.

In problem solving there is a so called "D-I method (D=differential, I= integral)" : as a rule of thumb it isn't totally bad to think of techniques of integration (or differentiation) while dealing with "extended bodies". The rules validity is strengthened by the remark that "not all physical laws are valid for extended bodies but most are valid for a single particle". And extended bodies can be thought of as consisting of collection of single particles (abstraction again).

"The essence of D-I method lies in the following. Suppose that the physical law in the question has the form $K=LM ...(1.1)$

where K, L, and M are physical quantities, with L=cnst being a condition. How should one generalize this law to the case where L is not a constant but a function of M, that is; L=L(M)?

Let us assume an interval dM so small that the variation of M over this interval can be ignored. Thus in the interval dM we can approximately assume that L is constant and hence the law is valid in this interval.Then $dK=L(M)dM ...(1.2)$

Where dK is the variation of K over dM.

Employing the superposition principle (summing the quantities (1.2) over all the intervals of variation of M), we arrive at an expression for K in the form: $K=\int _{ { M }_{ 1 } }^{ { M }_{ 2 } }{ L(M)dM }$

where $\displaystyle { M }_{ 1 }$ and $\displaystyle { M }_{ 2 }$ are the initial and final values of M.

Thus, the D-I method consists of two parts. In the first we find the differential (1.2) of the sought quantity. This is done in the majority of the cases by partitioning the object into parts so small that they can be considered as being particles or by partitioning a large time interval into time intervals dt so small that in the course of each interval the process being investigated may be thought of as approximately uniform (or steady-state).

In the second part of the method we sum (or integrate). The most difficult aspect here is to choose the correct integration variable and determine the limits of investigation. To determine integration variable we must analyze in detail on what quantities the differential of the sought quantity depends and which variable is most important. This variable is usually chosen as the integration variable, and the other variables are expressed as functions of the integration variable. As a result the differential of the sought quantity assumes the form of a function of integration variable. The next step is to determine the limits of integration. Evaluation of the finite integral yields the numerical values of the sought quantity. "

- 6 years, 3 months ago

Oh boy, a long ride, eh? :O

Well, let me comment on your claim that some math in physics is abstraction. True, but not for most part. The actual manipulation of given equations and derivation of more complex formulations is abstract, oh yeah. But the actual fundamental ideas described by equations can be perfectly explained, and often their derived quantities. That's why I spend hours (not literally) staring at equations; in the end, I have the perfect picture.

So, let me finish off the explanation of the previous problem. Basically, the idea behind it is this: as a segment of the rope of length y gets pulled up, segment $(L-y$) is what remains. The total mass remaining hanging will be the whole minus the portion displaced, or, $\left(\frac{\Delta y }{L}\right )m$. Now since initially 3/4 of the rope has already been lifted, we are left with $L/4$ of the rope to work with at the start. As the rope gets pulled all the way up, in the end, $0L$.

This is EXACTLY what our integral does.

- 6 years, 3 months ago

As for the part of abstraction: the quantities in the formula will have physical interpretation. Other wise it is pretty much useless or wrong.

However, in the process of application of the Mathematical ideas.It's abstraction that rule: words like, "suppose", "point", "interval" . etc. And I was talking about these things.

- 6 years, 3 months ago

ooh man I guess you won't stop modifying your manifesto, now will ya? sorry, won't be draining much more of your time;

okay look I was playing a dirty trick; in reality, an important part of my reasoning is wrong. I kinda did that on purpose to see if people would catch the mistake, but everybody seems to be hooked on their own versions of the story rather than trying to debunk mine. So I tell you [not just Math Philic]: Where is the mistake in my reasoning?

You've got 12hrs. Then I'll reveal the answer... if it's necessary at that time.

Cheers

- 6 years, 3 months ago

may be you were focusing on the the system consisting of the full chain whereas your question asks about only the hanging part:"...how much work is required to pull the 'hanging part' back onto the table?"

- 6 years, 3 months ago

SOLUTION:

Let $y$ be the height by which the chain has to be lifted.

Let $m$ be the mass of the chain.

Let $L$ be the total length of the chain.

Then, $\frac{y}{L}m$ will represent the portion of the mass remaining hanging. This is variable mass - that is, as we pull more and more of the rope, less mass remains to be yet pulled.

Then, the change in gravitational energy will be as follows:

$-\displaystyle\int_{-L/4}^{0}{\left( \frac{y}{L}m \right)(g)dy}$

That is, as the chain gets lifted by a height segment $dy$, the total remaining height will be $(y-dy)$, which is cleverly accomplished by the limits of the integral (from starting length $L/4$ to 0). (There is a negative in front of the intergal due to the fact that change in grav. potential energy is the negative of work. Ex.: As a tomato descends, the gravitational force does positive work on it, increasing its kinetic energy (and decreasing its potential energy).)

And I shall get back to dispute (or recognize) the 'center of gravity' concept once I'm done with that chapter, which is the chapter I'm doing right now.

Cheers

- 6 years, 3 months ago

my brain is going to explode

and so is my clock

I'll come back to this in the next decade

thanks all!

- 6 years, 3 months ago

Lol..Good Luck!!:)

- 6 years, 3 months ago

eh trust me my mind can never rest if I don't understand a concept in physics. Spent 3 hours, got the answer (I hope... not quite sure about representing height as $dy$) but yeah I think I should have just LET IT GOOO...

- 6 years, 3 months ago

Haha...And i thought i was the only one who could not let it go!! Lol...:D

- 6 years, 3 months ago

a uniform chain of length l and mass m is lying smooth table one fourth

- 5 years, 3 months ago