Spring Balance Problem.?

Question

I have 1 spring balance with me(k1). I suspend 10 kg. It shows 10kg.

So far so gud. No big Deal.

Now i suspend this arrangement to another spring(k2). Now both show 5kg.
Q. Why is this so. Y dont they give anyreading totalling to 10. y only 5 and 5.

Q. Why does it not happen that spring1(k1) show 10kg. and the other spring show (10+weight of spring1)
as if we draw fbd of 2nd spring it has 10kg weight and spring 1 below it.


i hope u got what im trying to ask

OK SORRY. MY QUESTION IS WRONG. THE FIRST ANSWER IS CORRECT. NO MORE REPLIES PLEASE.

Answer 1

If whole arrangement (k1 + 1st spring) is suspended by 2nd spring, you WILL get 10+weight of spring 1, like you said.

Answer 2

I assume both springs have the same spring constant and are the same length. Actually from your description, the results tend to support that.

Do a thought experiment. Suppose you had a single spring of K=K1=K2 that was the total length S1 + S2. You now hang the 10kg mass and the spring elongates (Delta L = Weight / K). It isn't just the end of the spring that elongates, but the spring elongates equally everywhere along its length. Suppose you had measured how much the single long spring had elongated at its mid-span. It would have elongated one-half of the total elongation, DeltaL / 2. For the spring with K that elongates DeltL/2, the result would read only 5kg AT THAT POINT. Hooking the 2 springs in series is the same as using a single equivalent spring.

For an axial loaded structure, Delta = F*L/A*E. The effective spring rate is K = F/Delta = A*E/L. For A and E the same, the spring rate is inversely proportional to its length. Halve the length of the structure and the spring rate doubles (it gets stiffer).