What is the distance Y in meters after the weight is hung from point A

Question

--- |--------------------------------z----------------------------------------/|C --------------                   Z = 10m                                    

  x  |                                                                                                         |                             X = 2m

---B|\                                                                                                       Y                       at point A a weight is placed of 588 N

           \                                                                                                    |          A wire  between BAC of a total length of 15m

               \A                                                                                 |-----------------

Answer 1

What is the distance Y in meters after the weight is hung from point A

I had to guess at the arrangement of the wire and position of the points A, B, C. You question was marked as belonging to Geometry answers, but if so, then why include the size of the weight hanging at A? Or did that mean it was a mechanics question? If so then you would need the stiffness constant for the wire. I took the simplest choice and assumed it was a geometry question and the following diagram is what I think the layout of things should be. Please correct me if I’m wrong.

To find the length h

Using Pythagoras.

h^2 = l1^2 – 4       h^2 = l2^2 – 100

Therefore, l2^2 = l1^2 + 96 -------------------- (1)

From l1 + l2 = 15, then l1^2 = (15 – l2)^2 = 225 – 30l2 + l2^2

Substituting for l1^2 = 225 – 30l2 + l2^2 into (1),

l2^2 = 225 – 30l2 + l2^2 + 96

30l2 = 225 + 96 = 321

l2 = 107/10, giving

l1 = 43/10

Now we can get h

h^2 = l2^2 – 100 = 10.7^2 – 100 = 114.49 – 100

h^2 = 14.49

h = 3.8066 m

Answer 2

It took me a while to interpret the diagram, but I came up with the following picture (finger-drawn on my iPad) (not to scale).

The tensions in the wire labelled T1 and T2 have to counterbalance the weight. BA+AC=15m and T1 and T2 are directly proportional to the lengths of AB and AC. The perpendicular distance Y is the distance AD between A and the 10m horizontal reference line EC at the top of the picture. 

We know that, since there is no net horizontal force, T1sinBAD=T2sinCAD, so ABsinBAD=(15-AB)sinCAD; ED+DC=10, so, since ED=ABsinBAD=DC=(15-AB)sinCAD, ABsinBAD=5=(15-AB)sinCAD. sinBAD=5/AB, sinCAD=5/(15-AB). Therefore, cosCAD=sqrt(1-25/(15-AB)^2), cosBAD=sqrt(1-25/AB^2).

Y=AD=2+ABcosBAD=(15-AB)cosCAD

ABcosBAD=sqrt(AB^2-25) and (15-AB)cosCAD=sqrt((15-AB)^2-25).

Y=2+sqrt(AB^2-25)=sqrt((15-AB)^2-25); the angles are not needed; we just need to find AB;

squaring both sides: 4+4sqrt(AB^2-25)+AB^2-25=(15-AB)^2-25=225-30AB+AB^2-25

4sqrt(AB^2-25)=221-30AB, because underlined terms cancel out;

squaring again: 16(AB^2-25)=48841-13260AB+900AB^2;

884AB^2-13260AB+49241=0, from which AB=8.23994 or 6.76006. These values are in fact AB and AC, adding up to 15. But only AB=6.76006 satisfies the equation for Y above, so Y=6.54955m (2+sqrt(AB^2-25) or sqrt((15-AB)^2-25)). Y=6.55 metres to 2 dec places.