Solving HW Geometrically Nonlinear Problem

Apparently your springs started life lined up horizontally, attached to the roof at the outside ends, and now have W hanging down from their common intersection point, right? Why not use

sin(alpha) = sqrt((L+x)^2 - L^2)/L

for your second equation, with which you can eliminate alpha and, at least in principle, solve for x? Looks to be a 4th degree which you could solve numerically if you know L and k. Engineers like that stuff, right?

--Lynn

Typo alert:

should be

sin(alpha) = sqrt((L+x)^2 - L^2)/(L+x)

--Lynn

So the long leg of the triangle should be L = (L+x)*cos(alp). That gives a 2nd equation.

Solve one equation for x and plug into the other equation. Maybe there's some trig identities to use after that. I don't see anything at the moment.

You can't use small angles on problem, right?

That seems fine, based on your definitions.

What do you mean you "tried to solve a triangle"? What did you do with this triangle.

Try writing down an expression for cos(alpha). That gives you a relationship between alpha and x. Then you can do a substitution of one for the other and solve.

What you said above also gives this relationship. Notice that by the Pythagorean Theorem:

L^2 + sin(alpha)^2 * (L+x)^2 = (L+x)^2 L^2 = (L+x)^2 * [1 + sin(alpha)^2]

But doing what I said will get you there quicker.

- Randy

Hint: If you do differential equations: Let 2a be distance between ponts of attachment, let L,L+x be y, y+dy and let F be force in spring. There are 3 situations dF = k dy ; y = a sec(alpha) , differentiate this ; and W = 2F sin(alpha).. Now tire again to solve it further eliminating dy,dF to get (W,alpha,F,y) relations with initial value of y as L ...