Calculating Force on Particle in Spring Setup

[AriAstronomer]

Homework Statement

A particle is attached between two identical springs on a
horizontal frictionless table. Both springs have spring
constant k and are initially unstressed. (a) If the particle
is pulled a distance x along a direction perpendicular to
the initial configuration of the springs, as in Figure
P7.66, show that the force exerted on the particle by the
springs is:
F = -2kx(1 - x/(sqrt[x^2 + L^2])

I've attached a printed screen that includes Figure P7.66.

Homework Equations

The Attempt at a Solution

So I know new stretched length is sqrt(x^2 + L^2), forces in y direction cancell, x is multiplied by 2.
New stretched length is L' = sqrt(x^2 + L^2) - L.
F = -kx = -2k[sqrt(x^2 + L^2) - L].
Now here's where I get stuck. It seems like to get the right answer, I need to multiply this by a factor of x/sqrt(x^2 + L^2)...why?
Thanks,
Ari

 

[method_man]

Until now, you have done a good job.

Now, look at the solution. The result is a vector force. It's a component of a red force in my photo, i.e. that is a horizontal component of a spring force.

 

[AriAstronomer]

Alright, I understand that it's only a horizontal component that survives, so is this x/sqrt(x^2 + L^2) like a unit vector thing (vector/length)? If so, what context of the question tells you "oh, I should replace this unit vector with vector/length, instead of keeping it as is..". If this is not a unit vector thing still not sure where this extra term is coming from.

 

[method_man]

F = -kx = -k[$\sqrt{x^{2}+L^{2}}$ - L]--->Force of one spring

$\sqrt{x^{2}+L^{2}}$-->stretched length of a spring

$\frac{x}{\sqrt{x^{2}+L^{2}}}$--->cosine between x and stretched length

-k[$\sqrt{x^{2}+L^{2}}$ - L] * $\frac{x}{\sqrt{x^{2}+L^{2}}}$ i.e.

-kx[$\sqrt{x^{2}+L^{2}}$ - L] * $\frac{1}{\sqrt{x^{2}+L^{2}}}$ --->force of one spring

 

[AriAstronomer]

Ahh, of course. I figured just slapping an i hat on was sufficient. I should have realized it was still involving both x and y until a cosine is taken. Thanks a lot method_man.