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http://courses.edx.org/static/content-mit-mrev~2013_Summer/problems/MIT/rayyan/Physics801/Figures/Spring%20constants.png!

Can someone please help me in solving this question:

What is the effective spring constant for the system of the two springs, perfect pulley, and string shown on the left for it to be modeled by just one spring (constant $k_{eff}$) as shown on the right? Use only the variables $k_1$ and $k_2$ in your answer.

I have tried a lot for solving this question. This is what I tried:

From the fig. (1) : $$\tag{1}F = k_2x_2.$$ Also, the net force on $k_1$ spring is 2F.

So, $$\tag{2} 2F = k_1x_1.$$

Also, from fig. (2), $$F = k_{eq} x.$$ Now, I cannot understand where I am going wrong or how to proceed! Please help me.

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closed as too localized by Brandon Enright, Chris White, akhmeteli, Qmechanic Jun 15 '13 at 19:02

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Can someone please correct the image code (I don't have much idea about that) ? –  Kushashwa Ravi Shrimali Jun 13 '13 at 14:23
1  
I've added the homework tag. In the future, please add this tag yourself on questions that are homework. –  Ben Crowell Jun 13 '13 at 15:00
    
Tried fixing the image tag but get "Failed to upload: format not supported" (or wording to that effect). Not sure what's up, png should be fine right? –  Michael Brown Jun 13 '13 at 15:13
    
Thanks @BenCrowell, I will remember it now. –  Kushashwa Ravi Shrimali Jun 14 '13 at 0:30

2 Answers 2

up vote 0 down vote accepted

Well first to solve the problem, let us assume force $F$ is applied to the end of rope as shown.
Now as the tension is uniform in the rope around the pulley, the extension in spring 2 is $f/k_2$. For the first spring the extension is that which is caused by $2F$ and is $2F/k_1$

Now as the point where the force actually applied will go down, its downwards displacement would be $F/k_2+2*2F/K_1$ (the factor of two is because if the spring one extends by $y$ the displacement of the point of force is $2y$)

Now just write extension as $x_t=F/K_2+4F/K_1$, remove the force common and take the remaining part on the otherside, and that part is your equivalent $K$

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OK so I proceed like this: $x_t = \frac{f}{K_2} + 4\frac{F}{K_1}$ $$$$ In second figure, $F = K_{eq}x_{t}$ $\implies \frac{F}{x_t} = K_{eq}$ $\implies K_{eq} = \frac{K_1K_2}{K_1 + 4K_2}$ Thanks a lot for your help. Sorry I don't have the reputation to vote up your answer. –  Kushashwa Ravi Shrimali Jun 14 '13 at 0:25
    
oh that's all right! –  Satwik Pasani Jun 14 '13 at 2:12

The problem is more of an equivalent force problem:

$ 2F = f$ where $f=k_1 x$ and $F=k_2 y$ and $y=2x$, so that substituting back in and moving one of the F's to the RHS. $$F = f-F = k_1 x-k_2 y = k_1 x-2k_2 x = k_\mathrm{eq} x$$

Thus $k_\mathrm{eq} = k_1-2k_2$

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