# What Happens to the Tensions in a Cut String Problem?

• Vibhor
In summary, the conversation discusses a problem involving blocks connected by strings and a spring. The bottom right string is cut, and the resulting tension in the other strings is calculated. There is some disagreement over the correct answer, with options given as 21g/4 and 20g/4. The conversation also delves into the assumptions made about the strings and their behavior in this scenario. Ultimately, one method of calculating the acceleration of the system results in a tension of 21g/4.
Vibhor

## Homework Statement

Since the image is not clear ,the blocks on the left are 1 Kg(top) and 5kg(bottom) connected by a spring. On the right are 1Kg(top) and 2Kg(bottom) .The bottom right string is cut .

## The Attempt at a Solution

Suppose T1 is the tension in string AB , T2 in string CD, T3 in EF and GH , T4(=kx) in the spring .

Just after AB is cut T4 (i.e spring force) does not change . Tension in strings CD become T2' and in string EF and GH becomes T3' .

Writing force balance equation for the two 1 Kg blocks and 2Kg block ,

For the left 1Kg block , 6g - T3' = 1a

For the right 1Kg block , T3' -T2' - 1g = 1a

For the right 2Kg block , T2' - 2g = 2a

Solving , I get T3' = 21g/4 , but this isn't an option .

Thanks

#### Attachments

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The form of the offered answer d strongly suggests a misprint. Why would they not specify it as 5g?

Vibhor
haruspex said:
The form of the offered answer d strongly suggests a misprint. Why would they not specify it as 5g?

Thanks . But the answer given is option c) .

Option d) i.e 20g/4 is obtained under the assumption that the two blocks on the left (connected by the spring) have same acceleration just after the string is cut .

Do you agree that just after the string is cut , the acceleration of 5Kg block on left is 0 whereas that of both the 1Kg blocks is 3g/4 ?

Last edited:
Yes, I agree with those accelerations.
Vibhor said:
Option d) i.e 20g/4 is obtained under the assumption that the two blocks on the left (connected by the spring) have same acceleration just after the string is cut .
Doesn't that error give 4g?

Vibhor
haruspex said:
Doesn't that error give 4g?

Yes , that gives 4g .

Could you help me understand why tensions T2 and T3 should change when the string is cut ? Why doesn't T2 and T3 remain same just after the string is cut ?

Vibhor said:
Yes , that gives 4g .

Could you help me understand why tensions T2 and T3 should change when the string is cut ? Why doesn't T2 and T3 remain same just after the string is cut ?
Well, why would they? Masses are accelerating. If T3 stayed the same the top left mass would not move. T2 had to match the weight below it plus the tension in the string which has now been cut.

Vibhor
Vibhor said:
Yes , that gives 4g .

Could you help me understand why tensions T2 and T3 should change when the string is cut ? Why doesn't T2 and T3 remain same just after the string is cut ?

This problem to some extent exposes the assumptions you make when dealing with a "light, inextensible" string. If you think of the strings as slightly extensible (hence slightly stretched), then the behaviour of the string and spring should be the similar at the instant the system is released. Both should instantaneously retain their current tension until the system has had time to move.

The assumption for the inextensible string, however, is that this change happens instantaneously (or, at least, in a very short time). Whereas, the assumption for the spring is that this takes time to happen (i.e. it's not instantaneous).

haruspex and Vibhor
Do you agree with the results in the OP and post#3 ?

Vibhor said:
Do you agree with the results in the OP and post#3 ?

@haruspex is never wrong!

To add something, another way to do it is to calculate the acceleration of the 3 masses. The 5kg mass is instantaneously out of the equation and replaced by the 5g force in the spring. So:

Net force on the 3-body system (down to the left): ##F = 5g + 1g - 3g = 3g##
Total effective mass ##M = 4kg##.
Acceleration of 3-body system ##a = 3g/4##

Therefore:

##6g - T = 3g/4##
##T = 21g/4##

Vibhor
PeroK said:
@haruspex is never wrong!

I agree

PeroK said:
To add something, another way to do it is to calculate the acceleration of the 3 masses. The 5kg mass is instantaneously out of the equation and replaced by the 5g force in the spring. So:

Net force on the 3-body system (down to the left): ##F = 5g + 1g - 3g = 3g##
Total effective mass ##M = 4kg##.
Acceleration of 3-body system ##a = 3g/4##

Therefore:

##6g - T = 3g/4##
##T = 21g/4##

Nice !

PeroK said:
@haruspex is never wrong!
Steady on there! Perhaps I do as well as the Captain of HMS Pinafore.
What, never?
No, never.
What, never??
Well, hardly ever!

## 1. What is tension in a string?

Tension in a string refers to the pulling or stretching force applied to the string. It is the force that keeps the string taut and prevents it from breaking or becoming slack.

## 2. How is tension measured in a string?

Tension is typically measured in units of newtons (N) or pounds (lbs). It can be measured using a tension meter or by using Hooke's law, which states that the tension in a string is directly proportional to the force applied to it.

## 3. What factors affect the tension in a string?

The tension in a string is affected by several factors, including the material and thickness of the string, the force applied to it, and the length and angle at which the string is pulled. Other factors such as temperature and humidity can also impact the tension in a string.

## 4. How does tension affect the vibrations of a string?

Tension plays a crucial role in the vibrations of a string. As the tension increases, the frequency and pitch of the vibrations also increase. This is why tightening the strings on a guitar or violin produces higher-pitched sounds.

## 5. Can tension in a string be too high?

Yes, tension in a string can be too high. If the tension exceeds the maximum limit for the string, it can result in the string breaking. Additionally, excessive tension can cause the string to become stiff and affect the sound quality or playability of an instrument.

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