Math wrong with physics springs problem

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The discussion centers on a physics problem involving the compression of springs in series and parallel configurations under a 5 kg load. The initial confusion arises from the assumption that the spring constant (k) is equal to 1, leading to the same compression distance for both two and four springs in parallel. Calculations confirm that for four springs in parallel, the compression is indeed 1.25 cm, while for two springs, it is 2.5 cm, which differs from the earlier assumption. The correct approach involves calculating the equivalent spring constant based on the series deflection before determining the parallel deflection. Overall, the thread emphasizes the importance of accurately calculating spring constants and understanding their effects on compression.
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Homework Statement


Problem:A force of 5kg compresses the spring in series 10cm. What will be the total distance that the springs in parallel are compressed.
Givens: S1,S2,S3,S4 are in parallel being forced by a 5kg weight.

Homework Equations


F=-kx but k is not given so it is assumed to be 1


The Attempt at a Solution



(1) S1 and S2 in series with 5 Kg load: S1 gets the 5Kg load and S2 gets the 5 Kg Load so S1+S2 = 10Kg total load. F=kX-->10Kg=(1)(X)--> X=10cm

Example given with solution
(2)S1 and S2 in parallel with 5 Kg load:S1 and S2 get the 5Kg Load so there is a split among them. S1+S2 = 5kg-->5Kg=(1)(X)-->X=5cm

This is where my issue is
(3)S1,S2,S3,S4 are in parallel with a 5Kg load: S1 and S2 and S3 and S4 get the 5Kg load so there is a split among them. S1+S2+S3+S4=5Kg-->5Kg=(1)(X) = 5cm.

How can having two springs in parallel(2) and four spring in parallel(3) be the same?
I have a feeling my math in wrong so please help me if posible.
 
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Why did you assume that k = 1? Why not calculate it?
Do you know the equations for calculating k equivalent for both parallel and series springs?
Be sure to include units when calculating spring constants.
 


tvavanasd said:
Why did you assume that k = 1? Why not calculate it?
Do you know the equations for calculating k equivalent for both parallel and series springs?
Be sure to include units when calculating spring constants.

If K is not given then your saying K is relevant then ok...
Spring in Series: Keq=1/Keq=1/K1+1/K2
Spring in Parallel: Keq=K1+K2

Equivalent Spring Constant (Parallel)[in my problem]
F=F1+F2+F3+F4
F=-k1X-k2X-K3X-k4X
F=-(k1+k2+k3+k4)X
5=-(4)X
X=1.25cm for 4 springs in parallel now ?

For series initial problem:
F=-(k1k2/(k1+k2))X
5=-(1/2)X
X=10 which matches the given answer

I am assuming K=1 still because it is not given.
 


Yes, 1.25 cm is correct for the 4 springs compressed in parallel.
Basically, you have checked, and confirmed that if k = 1 kg/cm the series deflection matches that in the question. Luckily, your assumption was correct.
A better way of approaching the problem would be to calculate the k value of the spring based on the series deflection, and then calculate the parallel deflection.
 


tvavanasd said:
Yes, 1.25 cm is correct for the 4 springs compressed in parallel.
Basically, you have checked, and confirmed that if k = 1 kg/cm the series deflection matches that in the question. Luckily, your assumption was correct.
A better way of approaching the problem would be to calculate the k value of the spring based on the series deflection, and then calculate the parallel deflection.

This is where this is all stem from...

http://www.psychometric-success.com/aptitude-tests/mechanical-reasoning-tests-springs.htm

1. A force of 5 Kg compresses the springs in series 10cm.What will be the total distance that the springs in parallel are compressed? 2 springs
A) 10cms
B) 2.5cms
C) 5cms
D)7.5 cms

Answer to Question
1. C – The total force will be divided equally between the 2 springs in parallel. Since the force is divided in half, the distance moved will also be halved. The springs in series were compressed 10 cms, so the springs in parallel will be compressed 5 cms.

It should be 'B' right!

F=F1+F2
F=-k1X-k2X
F=-(k1+k2)X
5=-(2)X
X=2.5cm for 2 springs!

:bugeye:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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