Understanding Spring Balances and Series Setups

AI Thread Summary
The discussion revolves around understanding the behavior of spring balances in different setups. When a 50g mass is added to a spring balance already reading 100g, the tension in the string reduces, maintaining the reading at 100g. However, confusion arises with two spring balances in series and a 100g mass; the expected readings of 125g and 75g are debated as incorrect. Participants emphasize that for the system to be in equilibrium, the forces must balance, leading to the conclusion that both springs should read the same value in a proper series setup. The conversation highlights the importance of correct interpretations of forces and setups in physics problems.
shyta
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Homework Statement


This was a concept question my lecturer gave us, and although he tried to explain it thoroughly, I still didn't get it. So I was hoping someone could help me out.

There are 2 parts to this question
Firstly a spring balance is attached hanging from a celling using a string. The spring balance is then lightly pulled till it reads 100g. This end of the spring balance is then tied to the floor using a string.


The first question goes as what would happen if a 50g mass is hung on the the end of the spring balance.

The 2nd question goes as what would happen if there are now 2 spring balances in series and a 100g mass is hung on the end.


The Attempt at a Solution


I understand how to do the first question, as the mass is hung the tension in the string which is attached to the floor is reduced. Sort of like a compromising way, thus the reading still remains as 100g.

However, I didn't get the second part. The answer was that the higher spring balance would read 125 and the one below 75g. Can someone explain this to me please?
 
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anyone??
 
There are a number of things that are not clear in your question.

First of all, you said that the spring is attached to the ground, so that it shows up a 100g reading. Then, when u said that u have attached a 50 g mass, is it like that the bottom surface of the mass is attached to a sting and pulled to the ground same like the initial case??
If it is so, then your answer to first question will definitely be 100 g.
Secondly, when u said that you have attached two springs in series, is it pulled to the ground in the same way as of the initial case? And then the mass is attached just like the first case?
Thirdly, where is the 100g mass attached (i mean, it is between the two springs in series or below both of them)?
 
i've attached a diagram illustrating the problems. it should clear up the doubts now.EDIT: oh wait i drew it wrongly. for the initial reading for 2 spring balance in series the total is 100g not 100g each.
 

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That's just wrong. Your lecturer really said this, with this kind of setup?

Draw a set of free body diagrams, one for each spring and one for the weight. The two springs and the weight aren't moving, so the net force on each object must be identically zero. That is not the case for any of the objects if the upper spring balance registers 125 grams while the lower one registers 75 grams.

Try hanging a mass (but not 100 grams) from the upper spring instead. Pick the right mass and you will get the upper spring registering 125 grams and the bottom one 75 grams.
 
hmmm i don't get your drift.
i tried drawing a free body diagram wrt the bottom spring balance

let F1 be the force the top spring balance pulls the bottom spring balance
and F2 be the force the bottom spring balance pulls the weight of the 100g mass and the tension of the string attached to the floor

then
F1-F2 = 0
F1=F2

where F2 = mg + T


Well, it's still logical at this point because mg is 0.1g. and for F2 to read 0.125g, the T must be 0.025g. but how??


my working should be correct. but i got a bit confused while trying to draw FBD for the top spring balance. care to enlighten? (remember to top spring balance is tied to the celling as well.)
 
shyta said:
but how??
That's the point. It doesn't make sense. Take the 100 gram weight, for example. For it to be stationary the net force must be zero. There is a force of 100 gram-force (1 gram-force = g * 1 gram, and old style unit that comes in handy sometimes) directed downward due to gravity. The only upward force available to counteract that is the force 75 gram-force that results from the lower spring, and not necessarily all of that is buoying the weight. The net force on the weight is 25 gram-force or more downward. Similar problems exist for the two springs. So, bzzt wrong.

A set of massless springs connected in series will all register the same value. The only way to get different readings from a set of massless springs is to use something other than a series setup. For example, suspending the mass and the lower spring in parallel from the upper spring and then connecting the lower spring to the ground via a string under tension will do the trick -- but not with a 100 gram weight.
 
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