# Challenge problem: The Cable News

• Valerie Park
In summary, the student is trying to find the length of a piece of cable that does not break under its own weight when hung vertically. They set up equations for F = clg and F = m [l1g/(l1 + l2) =L], but it does not work. They eventually find that c is the constant to convert l to mass in the case of the cable hanging vertically and m is the force of tension at its max when l2 causes the cable to break under its own weight.

## Homework Statement

Given the nature of the title of the problem, I need hints to guide me in the right direction. I am sort of lost.
A student has a large quantity of a flexible cable. If she cuts a piece of that cable and hangs it vertically, the longest piece that does not break under its own weight is l. The student then cuts another piece and places it on a horizontal smooth table. A small part of that cable hangs over the edge so that the cable begins to slide down after being released. How long (L) can this piece be so it does not break during this slide?

## Homework Equations

F = ma and F = clg for vertical cable
a = l1g/(l1 + l2 = L) where l1 is the part of the cable on the horizontal table and l2 is the part of the cable hanging off the horizontal table. Similar to pulley and string situation.

## The Attempt at a Solution

I set F = clg and F = m [l1g/(l1 + l2) =L] equal to each other and got L = ml2g/ clg.

What is c, what is m? They should not appear in the final result.
You can assume that l2 (initially) is very small and take the limit for l2 -> 0, the result should still be a finite value.

Valerie Park said:
I set F = clg and F = m [l1g/(l1 + l2) =L] equal to each other
I don't know what clg is and where it comes from, but that does not work.

Where and when do you expect the cable (if it is a bit too long) to break? In other words, where do you get the maximal tension in the cable?

C is the constant to convert l to mass in the case of the cable hanging vertically.

And is the force of tension at its max when l2 causes the cable to break under its own weight similar to the vertical cable?

Valerie Park said:
And is the force of tension at its max when l2 causes the cable to break under its own weight similar to the vertical cable?
Well, yes, that is the condition for breaking, but that was not my question.

At which point does the cable break (if it breaks), and how do you calculate tension at this particular point?

Have you drawn a free body diagram for (a) the part of the cable this is moving vertically and (b) the part of the cable that is still on the table? What are the forces acting on these?

Valerie,

This is a pretty challenging problem. What do you win if you solve it?

Chet

cnh1995

## What is the "Challenge problem: The Cable News"?

The "Challenge problem: The Cable News" is a hypothetical scenario where a cable news network is facing a problem and needs to come up with a solution. It is commonly used as a case study in problem-solving and critical thinking exercises.

## What skills are needed to solve the "Challenge problem: The Cable News"?

To successfully solve the "Challenge problem: The Cable News", one would need strong critical thinking and problem-solving skills. They would also need to be able to analyze data, think creatively, and work well in a team.

## What are some common approaches to solving the "Challenge problem: The Cable News"?

Some common approaches to solving the "Challenge problem: The Cable News" include identifying the root cause of the problem, brainstorming potential solutions, and evaluating the feasibility and effectiveness of each solution. Other approaches may include conducting research, consulting experts, and conducting experiments.

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## How can solving the "Challenge problem: The Cable News" benefit a scientist?

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