# Tension in a string attached to a spring with deformation at a point

I am trying to calculate the tension in a string in the configuration shown in the diagram below. Here is a verbal description of the setup: The string is fixed at one end with the other end connected to a spring. The spring, with constant $$K$$, is limited to only vertical movement. The string begins with a tension $$T$$, which causes the free end of the spring to extend from its resting position $$S_0$$ to position $$S_0+S$$ such that the following relationship is upheld:$$F_s=K\cdot S=T\cdot\frac{h}{\sqrt{D^2+h^2}}$$ A distance $$x$$ along the horizontal from the fixed end of the string, a force $$F$$ is applied vertically downward. This force is such that the string is forced into the configuration shown with a tension $$T_1$$. The spring will naturally extend to accommodate this deformation until the free end is $$h_1$$ above the horizontal line defined by the segment labeled $$x$$. It seems to me that if the string has cross-sectional area $$A$$ and Young's modulus $$Y$$ then it should be possible to determine the tension $$T_1$$ as a function of $$A$$, $$Y$$, $$x$$, $$T$$, $$D$$, and $$h$$. $$S_0$$ is not relevant to the calculation at all. I can calculate $$S$$ using $$S=\frac{T}{K}\cdot\frac{h}{\sqrt{D^2+h^2}}$$ I should be able to calculate $$T_1=K(S+h-h_1)\cdot\frac{\sqrt{(D-x)^2+h_1^2}}{h_1}$$ and from previous problems I know that I can calculate $$T_1=(T+AY)\frac{x+\sqrt{(D-x)^2+h_1^2}}{\sqrt{D^2+h^2}}-AY$$ However, I can't for the life of me find a way to write $$h_1$$ in terms of the other variables so that I can remove it from either equation for $$T_1$$. I think it should be possible since $$h_1$$ is simply an equilibrium point, but I've been unable to work it out myself. I'm probably missing something obvious, but some other eyes on this problem would be greatly appreciated!

I assume the length of the rope remains constant. At first the length is given by:

$$L=\sqrt{D^2+h^2}$$

After applying the force, the length is given by:

$$L=x+\sqrt{(D-x)^2+h_1^2}$$

By assuming $$L$$ is constant, one can derive an expression for $$h_1$$:

$$\sqrt{D^2+h^2}=x+\sqrt{(D-x)^2+h_1^2}$$

$$\left(\sqrt{D^2+h^2}-x\right)^2=(D-x)^2+h_1^2$$

$$h_1 = \pm \sqrt{\left(\sqrt{D^2+h^2}-x\right)^2-(D-x)^2}$$

• Yes I can solve it if I assume the length is constant, but I'm trying to account for the stretching of the string under tension. – FoitGuy May 24 at 1:56