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Mass attached by two springs

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1. The problem statement, all variables and given/known data
The following problem is an example from the book ' Berkely - Waves by Frank S. Crawford Jr '.

Mass 'M' slides on a frictionless surface. It is connected to rigid walls by means of two identical springs, each of which has zero mass, spring constant 'K' and relaxed length 'Ao' .
At equilibrium position , each spring is stretched to length 'A' and thus each spring has tension 'K(A-Ao)' at equilibrium.

Find the total force acting on the mass in the z direction when it is stretched to the right.
note: in the pictures it is written 'a' instead of 'A' , I used 'A' just for convenience
upload_2018-2-8_0-30-5.png


Solution : Fz = -2K(z-A)
Derivation:
upload_2018-2-8_0-34-25.png


2. Relevant equations
Hooke's law : F = -k(x-xo)

3. The attempt at a solution
I don't fully understand the derivation to the solution in terms of coordinates, Especially the second term: ' +K(2A-z-Ao) '

Here's my attempt: I redrew the third image as follows:

upload_2018-2-8_1-11-0.png


Solving:

Approach 1:
I used newtons second law on the mass : Fz = -k(z-Ao)+k(z-(2A-Ao) )

Explanation:
'-k(z-Ao)' because of the left spring , the displacement of the mass from the left spring's relaxation point is 'z-A0'

'+k(z-(2A-Ao) )' because of the right spring ( which is compressed) . the displacement is as it is because 'z' is the coordinate of the mass and ' 2A-Ao ' is the relaxation coordinate of the right spring. hence , I wrote ' z - (2A-Ao) '

However,this approach seems false because: Fz=-k(z-Ao)+k(z-(2A-Ao) ) =-kz+kAo+kz-2Ak+kAo=2k(Ao-A)
and this does not match the solution, so after that I tried another approach:

Approach 2:

Fz = -k(z-Ao)-k(z-(2A-Ao) )

Explanation:

everything's the same as in approach 1 , but I only changed sign on the second expression, from
'+k(z-(2A-Ao) )' to '-k(z-(2A-Ao) )'
using this approach I get the right answer:
Fz = -k(z-Ao)-k(z-(2A-Ao) ) = -2k(z-A)


So the questions are:
1.why was my first approach incorrect? ( isn't the sign on the second term suppost to be positive? [ because the right spring moves the mass to the right ] )
2. even though eventually I got to the right answer using approach 2, was this a 'correct' way to solve the problem?
3. was I right to make the displacement corresponding to the first spring as 'z-Ao' and the displacement regarding to the second spring as 'z-(2A-Ao) ' ?
 

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dRic2

Gold Member
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I'll be brief because I'm not very good with english (sorry). I'll consider just the left spring because you seem to have problem with that one only:

## F = -k(z-(2a-a_0)) = -k(x-x_0) ## with ##x = z## and ##x_0 = 2a-a_0##.

Since ##x-x_0 < 0 → F = +k(x_0-x) = +k((2a-a_0)-z) ## and you get the formula from the book

Hope it's clear

isn't the sign on the second term suppost to be positive
Yes but only if you switch ##x## with ##x_0##, because the general formula is ##F=-k(x-x_0)##, but in this very case ##x-x_0<0## so you can change the sign
 

haruspex

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isn't the sign on the second term suppost to be positive? [ because the right spring moves the mass to the right ]
The initial extension of the right spring is a-a0 to the left, so exerts a force K(a-a0) to the right.
To this is added a displacement (z-a) to the right, so adds a force K(z-a) to the left.
Summing, a force K(a-a0) -K(z-a) = K(2a-z-a0) to the right.
 
73
1
I'll be brief because I'm not very good with english (sorry). I'll consider just the left spring because you seem to have problem with that one only:

## F = -k(z-(2a-a_0)) = -k(x-x_0) ## with ##x = z## and ##x_0 = 2a-a_0##.

Since ##x-x_0 < 0 → F = +k(x_0-x) = +k((2a-a_0)-z) ## and you get the formula from the book

Hope it's clear



Yes but only if you switch ##x## with ##x_0##, because the general formula is ##F=-k(x-x_0)##, but in this very case ##x-x_0<0## so you can change the sign
The initial extension of the right spring is a-a0 to the left, so exerts a force K(a-a0) to the right.
To this is added a displacement (z-a) to the right, so adds a force K(z-a) to the left.
Summing, a force K(a-a0) -K(z-a) = K(2a-z-a0) to the right.

So I thought about it, and I think the second term : ' -k(z-(2A-Ao) ) ' is negative because the right spring is compressed , this means a force to the right acts on this spring to compress it, and by newtons third law, it exerts the same magnitude of force but to the negative 'z' direction on the mass , hence the force from this spring on the mass is 'Force on mass from right spring= -k(z-(2A-Ao) ) '
Would you say that my reasoning is correct?
 

haruspex

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So I thought about it, and I think the second term : ' -k(z-(2A-Ao) ) ' is negative because the right spring is compressed , this means a force to the right acts on this spring to compress it, and by newtons third law, it exerts the same magnitude of force but to the negative 'z' direction on the mass , hence the force from this spring on the mass is 'Force on mass from right spring= -k(z-(2A-Ao) ) '
Would you say that my reasoning is correct?
Looks ok.
 

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