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enter image description hereA mass m hangs on 3 ropes symmetrically with equal base angles. The base angles are 45 degrees. So the middle rope Tension is vertical to the load mg. How can one find the delta l of the middle and the side rope? i know that the two side ropes will have the same load and therefore the same change in length.

I did the sum of vertical forces and now end up staring at the question on how to move further to solve for tensions s1 and s2 also the change in lengths.

Also E and A are given for the calculation of delta l.

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Let the vertical distance between the ceiling and the point where the 3 ropes meet be $L+\Delta L$. The original length of the side rope is $L\sqrt{2}$. From the pythagorean theorem, the new length of the side rope is $\sqrt{L^2+(L+\Delta L)^2}=\sqrt{2L^2+2L\Delta L+(\Delta L)^2}$. If we linearize this with respect to $\Delta L$, we obtain $L\sqrt{2}(1+\frac{\Delta L}{2L})$. So the change in length of the side rope is $$L\sqrt{2}(1+\frac{\Delta L}{2L})-L\sqrt{2}=\frac{1}{\sqrt{2}}\Delta L$$So the tensile strain in the side rope is $$\epsilon=\frac{(L/\sqrt{2})}{(L\sqrt{2})}=\frac{1}{2}\frac{\Delta L}{L}$$So the tension in the side rope is $$T_{SR}=\frac{EA}{2}\frac{\Delta L}{L}$$ The tension in the vertical rope segment between the ceiling and the point where the three ropes meet is: $$T_{VR}=EA\frac{\Delta L}{L}$$ So the tension in the side rope is half the tension in the vertical rope segment.

The remainder of the analysis is straightforward, and just involves doing the equilibrium force balance.

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  • $\begingroup$ what does linearize mean? how do you do that? $\endgroup$
    – Sherlock Homies
    Commented Apr 22, 2017 at 8:22
  • $\begingroup$ i did the calculations and came up with the middle rope tension as mg/(sqrt2+2) $\endgroup$
    – Sherlock Homies
    Commented Apr 22, 2017 at 11:55
  • $\begingroup$ You linearize by neglecting a term such as $(\frac{\Delta L}{L})^2$ as being insignificantly small compared to unity and compared to $(\frac{\Delta L}{L})$ $\endgroup$
    – Chet Miller
    Commented Apr 22, 2017 at 12:06
  • $\begingroup$ I get twice your value for the tension in the middle rope. $\endgroup$
    – Chet Miller
    Commented Apr 22, 2017 at 12:12
  • $\begingroup$ is it because i forgot the otherside somewhere in my calculations? I got the relation between delta l1 and l2 as follows: delta l1 = sqrt2*delta l2 by using the sides as ratios correspondingly. $\endgroup$
    – Sherlock Homies
    Commented Apr 22, 2017 at 12:20
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You have to use equilibrium of forces in vertical direction. It has the form:

$S_{middle} + S_{left}cos \alpha + S_{right}cos \alpha - mg = 0$.

Here, $\alpha$ is the angle between the ceiling where ropes hanging and the side ropes. It also holds from horizontal equilibrium:

$S_{left}sin \alpha - S_{right}sin \alpha = 0$.

The tensions you compute with the formula $S = \frac{EA}{L}\Delta l$ with the length of ropes $L$ (without a weight). You can express the change in lengths by using elementary trigonometric relations; you can express the length change of the side ropes in dependence on the length change of the middle rope; it holds:

$\Delta l_{middle} = \Delta l_{Side} sin \alpha$

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  • $\begingroup$ your answer is better to understand. can you tell me how the sides actually relate to one another in simple trig? $\endgroup$
    – Sherlock Homies
    Commented Apr 22, 2017 at 8:04

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