Weird 2-spring problem with VERY limited data

[McFate]

Homework Statement

Two springs are hung in a "v" shape, and a 500g mass is attached to the point of the v.

One spring is observed to stretch 9mm, the other 23mm.

Students are NOT permitted to measure any angles, before or after the weight is attached.

Is it possible to calculate the spring constants of the two springs?

Homework Equations

F = -kx (Hooke's law)

F = k1*x1*sin(θ) + k2*x2*sin(∅) (Vertical component of force if the two springs' angles were known.)

The Attempt at a Solution

The problem doesn't seem solvable to me. If the angle of the springs from horizontal is very small, then the stretching will be the equivalent of many times the weight that is actually applied. At 30 degrees from vertical, EACH spring stretches as much as it would if it had a 500g weight pulling on it vertically. At smaller angles, each stretches further yet.

For example, at 10 degrees for both springs:

F = k1*x1*sin(θ) + k2*x2*sin(∅)
0.5*9.8 N = (k1)*0.009m*.173 + (k2)*0.023m*.173
... the ".173" (sin(10)) means that each stretches about 3x what it would with the entire weight hanging vertically on that spring alone.

If the angle of the springs from vertical is very small, then each will more or less carry a share of the weight, and both will stretch less than they would if they carried the weight alone.

Note that this is an "engineering" class rather than a physics class, so I don't think that the answer has to be rigorous or exact. But as far as I can tell, you could make various assumptions (each spring carries about half of the weight, each spring carries a share in inverse proportion to the amount stretched, both angles about 45 degrees, both angles about 60 degrees, one 45 and the other 60 degrees, etc.) that yield wildly different results.

Is there something obvious that I'm missing, or is this more likely a "make guesses for the things you weren't allowed to measure" sort of problem?

 

[Nesk]

Have you considered the horizontal force component's equilibrium condition?

 

[McFate]

Have you considered the horizontal force component's equilibrium condition?

Thanks for the response.

If I had the angles of the springs, the horizontal tension being equal between the two springs would form the second equation for my two variables (k1, k2):

4.9N = k1*(0.009m)*sin(θ) + k2*(0.023m)*sin(∅)
and
k1*(0.009m)*cos(θ) = k2*(0.023m)*cos(∅)

I have the overall lengths of the two springs (had to measure that before and after the weight to figure out how much they stretched). I'm going to go back to the lab and measure the distance between the pegs. It's not something we were told we couldn't measure, and it will give me the third side of the triangle which will allow me to use those equations.

But the question I had was: IF all I had was the data given originally, is there any solution? (I think the answer is: "no.")

 

[SammyS]

In order to measure the amount of stretch for each spring, you had to measure the un-stretched length of each spring. If you know the distance between the pegs, then you can find all the angles. (That may also be "illegal" in the eyes of the instructor.)

Compare the two configurations:

1.) The locations of the enuds of the springs with the weight hanging.

2.) The locations of the ends of the springs without the weight -- even if you reconstruct this "theoretically" using the measured nu-stretched lengths.​

Both x and y components are important for the equilibrium equations.