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Anonymous Poster

What am I doing wrong?, springs

04/17/2009 7:51 PM

I am trying to find the efficient force constant k, for two forces k1, k2 in a series. I have seen how you can do it, but I have tried doing it my own way but I keep getting it wrong. Can someone please see what I am doing wrong? I get 2*times the right answer.

http://img27.imageshack.us/my.php?image=k1k2.jpg

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Anonymous Poster
#1

Re: What am I doing wrong?, springs

04/17/2009 7:53 PM

Sorry, I fixed a little detail on the picture here is the new one.

http://img27.imageshack.us/my.php?image=k1k2k.jpg

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Join Date: Jan 2008
Location: since 20 Jan 09, the USSA
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#2
In reply to #1

Re: What am I doing wrong?, springs

04/17/2009 10:57 PM

Here's the question:

An arbitrary force is applied to two springs in series, and the question is, how is the overall spring constant for the two series springs related to the spring constants of the individual springs?

To review, the idea of spring constant is that the spring length deflects a distance (x) that is proportional to the applied force, F. The spring constant k is the constant of proportionality:

(1) F = kx

It is critical to realize that for two springs in series and a force applied, that at equilibrium the two springs must push on each other with the same force as was applied to the system; else they would not be in equilibrium:

(2) F = kx = k1x1 = k2x2, where

(3) x = x1 + x2

Now it simply remains to solve these two equations in two unknowns for the expression for k in terms of k1 and k2:

(4) k = f(k1, k2).

We can solve eqn 2 for either x1 or x2 in terms of the other; arbitrarily we solve for x1:

(5) x1 = k2x2/k1

We can then plug eqn 5 into eqn 3 to get:

(6) x = k2x2/k1 + x2,

which may be simplified into

(7) x = x2 (k2/k1 + 1) = x2 (k2 + k1)/k1

And we know from eqn 2 that

(8) kx = k2x2,

so we can plug eqn 7 into eqn 8 to get:

(9) k x2 (k2 + k1)/ k1 = k2 x2

In eqn 9, the variable x2 cancels on both sides, leaving

(10) k = k2 k1 / (k2 + k1)

From which the more familiar form follows:

(11) 1/k = 1/k1 + 1/k2

Again, the critical realization necessary to solve this problem is that when the springs are in series and a force is applied, that the same force is applied to each spring, and the two springs apply that same force on each other to maintain the new equilibrium state.

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Guru

Join Date: Nov 2007
Location: Sherwood Park, Alberta, Canada
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#3

Re: What am I doing wrong?, springs

04/18/2009 11:02 AM

A diagram helps.

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Bruce
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Anonymous Poster
#4

Re: What am I doing wrong?, springs

04/19/2009 6:21 AM

In my opinion the solution is:

F/k1=x1

F/k2=x2

the total displacement is

X=x1+x2

The resulting K is

K=F/(x1+x2)= F/(F/k1+F/k2)= k1*K2/(k1+k2)

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Commentator

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Location: South Africa Pretoria
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#5

Re: What am I doing wrong?, springs

04/20/2009 2:44 AM

The algebra can be made easier

We know F = k1x1 , F = k2x2 and F = k(x1+x2)

So x1 = F/k1 and x2 = F/k2

So we have F = k(F/k1 + F/k2)

So 1 = k(1/k1 + 1/k2)

So 1/k = (1/k1 + 1/k2)

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Anonymous Poster
#6
In reply to #5

Re: What am I doing wrong?, springs

04/22/2009 10:42 AM

It is interesting that your final equation is similar to the equation for adding electrical resistors that are connected in parallel. We call the equation the "Reciprocal of the sum of the reciprocals". 1/Rt = (1/ R1 + 1/R2)

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Commentator

Join Date: Apr 2007
Location: South Africa Pretoria
Posts: 59
Good Answers: 4
#7
In reply to #6

Re: What am I doing wrong?, springs

04/23/2009 2:23 AM

Yes and for springs in parallel we have the same sort of equation as for resistors in series.

If we fix two springs together in parallel we have the resultant force F = kx

and also F = k1x + k2x for the two strings fixed together

So kx = k1x + k2x

So k = k1 + k2

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