How to Calculate Tension in a Uniform Rope with a Constant Force Applied?

[Knightycloud]

Homework Statement

How to find the tension in a string with a weight. The string is not a light one it has a we say 1kg weight. How to find the tension at the both ends and the middle if it is equally pulled at both sides horizontally?

Homework Equations

F = ma

The Attempt at a Solution

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[PeterO]

Homework Statement

How to find the tension in a string with a weight. The string is not a light one it has a we say 1kg weight. How to find the tension at the both ends and the middle if it is equally pulled at both sides horizontally?

Homework Equations

F = ma

The Attempt at a Solution

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If the string is equally pulled at each end, the tension will be constant all the way through.

You only get a different effect if you use a heavy string to accelerate an object across a surface for example.

EG: suppose you have a 9kg mass connected to a 1kg rope, and a 10 N force is applied to the free end of the rope.

You are applying a 10 N force to a 10 kg load, so should achieve accelersation of 1 ms^-2

The end of the rope that you are pulling will have a tension of 10 N

Down the other end, the rope is accelerating a 9kg mass at 1 ms^-2 so clearly the tension there is only 9 N.

Half way along the rope, the Tension will be 9.5 N

 

[Knightycloud]

OH thanks, That really made it clear. So, if we say that a heavy string is freely hanged in a roof, how is the tension?

 

[PeterO]

OH thanks, That really made it clear. So, if we say that a heavy string is freely hanged in a roof, how is the tension?

OK suppose we have a 1m rope, of mass 1kg from the roof.

We could have a similar set-up with a series of 10 x 0.1 kg masses connected by an ideal, mass-less string.

I am hoping you can calculate the tension in each if the 11 segments of string.

[the set up - reading from the roof down, is string - mass - string - mass - string - mass - ... - mass - string.

The top string shows what the tension would be at the top of the rope.
The second string shows the tension 10 cm down the rope - where you have 0.1 kg of rope above and 0.9 kg below.
The third string is a little further down, where there is 0.2 kg above, and 0.8 kg below.

The bottom string shown you the tension in the bottom of the rope.

 

[kushan]

Just make the hanging mass 0
At bottom no tension and at top the weight of rope

 

[Knightycloud]

hm, Got it. Thanks! :D

 

[PeterO]

hm, Got it. Thanks! :D

Every situation with a heavy string can be analysed by creating your own situation with small masses connected by ideal strings taking the place of the heavy string.

Usually you only have to "replace" the string with a couple of masses as the question will be quite specific.

For example, two masses of 2 kg and 3 kg are connected by a heavy string of mass 1kg, and 1m in length.
The system is accelerated across a frictionless surface by a pulling force of 12 N applied to the 3 kg mass.

(a) What is the acceleration of the system?
(b) What is the tension in the string at a point 35 cm from the 3 kg mass?

 

[Knightycloud]

a) 2 ms^-2
F = ma
12 N = (2 + 3 + 1) a

b) 5.2 N
Tension at the both ends : 4 N and 12 N
8 * (65/100)

Am I correct?

 

[PeterO]

a) 2 ms^-2
F = ma
12 N = (2 + 3 + 1) a

b) 5.2 N
Tension at the both ends : 4 N and 12 N
8 * (65/100)

Am I correct?

Acceleration is indeed 2 ms^-2

Tension at th 2kg end of the string is 4N [at that point the string only has to accelerate the 2 kg mass.

Tension at the other end should be 6N [at that point we need a force to accelerate a 2kg mass plus a 1 kg string]

Note that the tension in the string will steadily drop from 6N to 4N as we move through the length of string.

As such, the tension 35 cm from the 3kg mass will be 5.3 N
the 5.3 is 4N [needed fro the 2kg mass] plus 2*65/100, the fraction of the 2N needed to accelerate the whole string, that is necessary to accelerate the remaining 65 cm of string [I hope that makes sense]

As a matter of interest, notice the great difference that occurs in the string if the 12N force had been applied to the 2N mass instead.
The overall acceleration would still be 2, but the tensions would be 8N, at the 2kg mass end, 6N and the 3kg mass end, and 3.7N at a point 35 cm from the 3kg mass.
The tension in the string would still have that 2N "gradient" along its length [from 8N to 6N rather than 6N to 4N we had in the first situation]

I trust this increases your understanding of heavy strings.

 

[Knightycloud]

hm, Yea. I thought the 12N force will be the same as the beginning of the string. Now I got it. I really appreciate this knowledge you've given. I never expected such a vast explanation. I'll never forget the theory now! :D

 

[Mainak Gupta]

can any 1 solve this for me?

a uniform rope of length L is pulled by a constant force F.what is the tension in the rope at a distance 1m from the end ,where the force is applied?

 

[PeterO]

can any 1 solve this for me?

a uniform rope of length L is pulled by a constant force F.what is the tension in the rope at a distance 1m from the end ,where the force is applied?

Could you solve this if the rope had a length of 1 m? 2m? 3 m? 5m? 7m?
If so, do a couple of those and establish the pattern you can use to get the answer for length L m.

Note: The answer is quite different if there is a mass attached to the other end of the rope!