Tension in a String: Why is the Scale Reading 10N?

[Harsh Bhardwaj]

[Source]
Both these masses are of 1kg each and the scale reads in Newtons.
The intuitive answer for the reading of the scale is about 20N because it is being pulled by 10N from each side.
However, this is clearly not true as the scale reads about 10N. Why is it so?
I believe that I am misled by thinking of action reaction pairs working on one body.

 

[NFuller]

Looking at just one of the masses, the downward pull of gravity must be balanced by the tension of the string above it since the mass is not moving. Since there is 10N of force pulling the mass down there must also be 10N of force from the string pulling it upward. Thus the tension in the string is 10N.

 

[kuruman]

Here is another way of looking at this:
Suppose I put a screen blocking your view of the left pulley. I then carefully cut the string and tie it to the post on which the pulley is clamped without you knowing what is going on. I remove the screen to reveal the string being tied to the post. What would you say the tension in the string is now? What has changed from the previous situation?

 

[berkeman]

What happens if you change the right weight to 2kg? 0.5kg?

 

[Harsh Bhardwaj]

Here is another way of looking at this:
Suppose I put a screen blocking your view of the left pulley. I then carefully cut the string and tie it to the post on which the pulley is clamped without you knowing what is going on. I remove the screen to reveal the string being tied to the post. What would you say the tension in the string is now? What has changed from the previous situation?

I have got it now. The spring balance will measure the tension in the string no matter what the weights are. Even if one of the weight is removed and the string is tied to a pole the tension will be same. The spring balance will still read about 10N. Actually, I was confused that a force of 10N is acting on each end of the spring balance. Then I thought what happens when we use a spring balance normally(i.e. vertically), even then there is a force of 10N on both ends of the balance but only the downward force of 10N on the bottom part of the spring can cause expansion and hence the balance shows 10N.

 

[kuruman]

The spring balance will measure the tension in the string no matter what the weights are.

Yes, and as long as the two weights on either side are equal and there is no acceleration of the masses, the tension will be equal to one of the weights.

 

[Harsh Bhardwaj]

What happens if you change the right weight to 2kg? 0.5kg?

The system is an Atwood's machine and unequal masses produce acceleration. Check the attachments to see how I worked out the Atwood's machine.
So T will be as derived in the attachments.

 

[kuruman]

Correct. Note that your expression for T gives T = mg when the masses are equal, m₁ = m₂ = m.

 

[Harsh Bhardwaj]

Correct. Note that your expression for T gives T = mg when the masses are equal, m₁ = m₂ = m.

I saw a beautiful demonstration of the m1 = m2 = m case. There the masses remained stationary no matter where they were placed along the string(but on opposite sides). The acceleration was always 0.
In the case m1 != m2, there was always an acceleration.

 

[kuruman]

There the masses remained stationary no matter where they were placed along the string(but on opposite sides).

You have to view this with a grain of salt. The string connecting the masses has mass, so if there are uneven lengths on the two sides, there will be some acceleration. If there isn't, it's because of friction at the pulley.

 

[Harsh Bhardwaj]

You have to view this with a grain of salt. The string connecting the masses has mass, so if there are uneven lengths on the two sides, there will be some acceleration. If there isn't, it's because of friction at the pulley.

You are very much connected to the reality of the setup. I was just thinking about idealistic bookish assumptions of massless and nonextensible strings and light pulleys without any friction. This approximately worked in that setup because of friction(as you said). Thanks for pointing out. :)