Vector question...(URGENT!)

[siyan]

This is from my geometry text...

There is a string, 200cm long, with the ends fixed 120cm apart on the roof. The tension limit of the string is 300N. A 400N weight is to be hung from some point on the string. How far from the center of the string can the weight be hung before the string breaks?



I've been working on this forever...it seems impossible. I've enlisted the help of both parents (PhD in Mining Engineering and MEng in Civil Engineering) and a really, really smart guy at the University of Toronto, nobody has been able to squeeze out a solution.

This is worth marks, due tomorrow. My hunch is that the teacher assigned the question (he made it up) without first checking the difficulty....

I thought I was missing something, but I find it odd that all the people I've gone to for help haven't come up with anything. And I'm not exactly a slouch in math either...

Still though, I need some help here...and fast. :( :(

-C

[Jim Brown]

I can't quite think of the analytical way to do it, but you can do it quite simply by brute force, and if you were to code it in VB it would be easy to home in on the right spot.

Start with the weight in the middle, then the geometry of the string is an isoscales triangle and you can easily calc the force in the string. I can't remember but I think I got about 200.

At the other extreme, if you put the weight on the string right at the bracket you would get 400 in the string.

So somewhere in between you would get the 300 you're looking for.

Two different techniques for doing this numerically would be

a) start at the centre and move towards a bracket in tiny increments recalculating the geometry and the load in the string each time. Tell it to stop when it gets to 300.

b) do a binary sort of search, where you go halfway from the centre to the bracket and calc there. If it's not bust yet, go forward another 1/2 of what's left. If it broke, come back halfway to where you started. That way you'll home in on the spot from each side, taking ever smaller steps each time.

But I only have a BS and a post- Grad diploma in Engineering so I'll bow to your folks if I'm wrong, and also it's over 20 years since I practised.

[Judd]

At some point the tension in the 'short' side of the string will approach supporting the whole of the weight. The string is too long to be taught on both sides for every value of x (where x is weight's distance from the center).

This occurs when x = 0.36m, making a right angled triangle (the short side of the trinagle is 0.64m). So that answer is something less than 0.36m.

I also wouldn't like to attempt an analytical solution, but find the answer numerically should be relatively easy.

[Judd]

I think if you use the breaking tension as your limit, you should be able to calculate the angles at the load point (e.g. one half of the string under 300N load, the other under 100N load).


(edit - no that's not right - the vertical components of the loads on each side must sum to be 400N, but the tensile load on the breaking side limits at 300N....)

[sql_lall]

I'm not exactly sure how Tension is worked out, but i am guessing that possibly, instead of using VBasic, if you had a good graphics calculator, then possibly you could find the equation, plug it in to one of the Y values, plug in another Y=300 and find the interception.
Its just the first part that i'm not sure how to do, and whether its possible.

[johnpc]

just a few thoughts, if the weight is hung exactly in the middle of
the distended string then the Tension in the string is 200N in each
leg. If you move the weight off center either way then the tension
is no longer shared and the full 400N is imposed on the short leg.
Therefore the string breaks at any distance from center.

[kedaman]

from the figure:

z^2-y^2=x^2=(200-z)^2-(120-y)^2
sin(alfa)=y/z=u/300
u/300=y/z
(120-y)/(200-z)=u/(400-(300^2-y^2))
should be enough to eliminate the variables, btw the tensions on each are not scalars, they add up as vectors |v1+v2|=400N, |v1|=300N