Question

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 72 mt. long is supported by a vertical wires attached to the cable. The longest being 30 mts. and the shortest being 6 mts. Find the length of the supporting wire attached to the road-way 18 mts. from the middle. 

Answer 1

Let AOB be the cable [O is its lowest point and A, B are the highest points]. Let PRQ be the suspension bridge suspended with PR = RQ = 36 mts. PA = QB = 30 mts (longest wire) OR = 6 mts (shortest wire) Therefore, PR = RQ = 36 m. We take O as origin of coordinates at O, X-axis along the tangent at O and Y-axis along RO . So the equation of the cash is x² = 4ay for some a > 0. 

∴ B = (36, 24) and B is a point on x² = 4ay. 

We have (36)² = 4a × 24.

⇒4a = 36×36/24=54mts  

If RS = 18 m and SC is the vertical through S meeting the cable at C and the X-axis at D.

Then SC is the length of the required wire. 

 Let SC = l mts, then DC = (l – 6) m. 

∴ C = (18, l – 6) which lies on x² = 4ay