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Old Jan 3rd 2018, 01:10 AM   #1
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Interesting problem

Hello,

1-st chapter (Axially loaded members) of the book.

Problem:
The long cable is suspended vertically and loaded with its own weight. What is the maximum possible Length of that cable at which it won't break, if the cable is made of steel with Ultimate tensile strength of 21,000 kg/cm2, density 7,85 g/cm3 ?
the answer in the book is 26,900 m, but i can't understand what is the algorithm of getting that answer

thank you

Last edited by ferdigo; Jan 3rd 2018 at 01:21 AM.
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Old Jan 3rd 2018, 01:58 AM   #2
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which point in the cable experiences the most tension ?

Clearly the top most part where it is connected to the support .

at this point the whole weight of the cable is pulling down A x L x density x g (A is cross sectional area of the cable)

this is breaking force , and must equal tensile strength x A

Tensile strength is measured in Newtons /Area NOT Kg / Area

The questioner means Kgweight (in earth gravity) ....omitting g in the first equation will get the right answer ..

This is typical of sloppy inaccuracies found in school books and MSM articles .. this is by design , see this video


Last edited by oz93666; Jan 3rd 2018 at 02:43 AM.
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Old Jan 3rd 2018, 03:25 AM   #3
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Thank you for the answer
* A kilogram-force per centimetre square (kgf/cm2), often just kilogram per square centimetre (kg/cm2), or kilopond per centimetre square is a deprecated unit of pressure using metric units. It is not a part of the International System of Units (SI), the modern metric system. 1 kgf/cm2 equals 98.0665 kilopascals.

The meaning is next. The position of the cable is vertical. One end of the cable is fixed, another is free. The is no additional load applied to the cable.
The question is how long should be the cable to rich the breaking limit under its own weight? What is the maximum length? And there is an answer in the book. But I can't imagine the way to solve it only with this amount of data. No area, no other specific condition.

Last edited by ferdigo; Jan 3rd 2018 at 03:27 AM.
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Old Jan 3rd 2018, 07:20 AM   #4
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internal resistance

Electromotive force of a current source equals to 12 V, while the voltage at its poles is 10 V. Calculate the internal resistance of the current source, if the current in the circuit is 1 A
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Old Jan 3rd 2018, 07:37 AM   #5
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Originally Posted by iverigame View Post
Electromotive force of a current source equals to 12 V, while the voltage at its poles is 10 V. Calculate the internal resistance of the current source, if the current in the circuit is 1 A
Probably it needs only one formula I = U/R (Ohm's law). Form here 1A = (12-10)/R ; R = 2 Ohms.

Something like that ,but you'd better post it inside a proper forum Branch to get more accurate answer.
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Old Jan 3rd 2018, 12:53 PM   #6
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Originally Posted by ferdigo View Post
Thank you for the answer
* A kilogram-force per centimetre square (kgf/cm2), often just kilogram per square centimetre (kg/cm2), or kilopond per centimetre square is a deprecated unit of pressure using metric units. It is not a part of the International System of Units (SI), the modern metric system. 1 kgf/cm2 equals 98.0665 kilopascals.

The meaning is next. The position of the cable is vertical. One end of the cable is fixed, another is free. The is no additional load applied to the cable.
The question is how long should be the cable to rich the breaking limit under its own weight? What is the maximum length? And there is an answer in the book. But I can't imagine the way to solve it only with this amount of data. No area, no other specific condition.
You should listen to Oz93666, he has nailed the answer for you.
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Old Jan 3rd 2018, 07:56 PM   #7
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Originally Posted by oz93666 View Post
which point in the cable experiences the most tension ?

Clearly the top most part where it is connected to the support .

at this point the whole weight of the cable is pulling down A x L x density x g (A is cross sectional area of the cable)

this is breaking force , and must equal tensile strength x A

Tensile strength is measured in Newtons /Area NOT Kg / Area

The questioner means Kgweight (in earth gravity) ....omitting g in the first equation will get the right answer ..

This is typical of sloppy inaccuracies found in school books and MSM articles .. this is by design , see this video

Originally Posted by studiot View Post
You should listen to Oz93666, he has nailed the answer for you.
Thank you people
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