The cable of a uniformly loaded suspension br | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola,while its vertical axis

Vedclass · Accepted Answer

$9.11$

Vedclass · Answer

(A) The vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola,while its vertical axis is taken along the positive $y-$ axis.Here,$AB$ and $OC$ are the longest and the shortest wires,respectively,attached to the cable. $DF$ is the supporting wire attached to the roadway,$18 \, m$ from the middle.Here,$AB = 30 \, m$,$OC = 6 \, m$,and $BC = \frac{100}{2} = 50 \, m$.The equation of the parabola is of the form $x^{2} = 4ay$ (as it is opening upwards).The coordinates of point $A$ are $(50, 30-6) = (50, 24)$.Since $A(50, 24)$ is a point on the parabola,$(50)^{2} = 4a(24)$.$\Rightarrow a = \frac{50 	imes 50}{4 	imes 24} = \frac{625}{24}$.$	herefore$ Equation of the parabola is $x^{2} = 4 	imes \frac{625}{24} 	imes y$,which simplifies to $6x^{2} = 625y$.The $x-$ coordinate of point $D$ is $18$.Hence,at $x = 18$,$6(18)^{2} = 625y$.$\Rightarrow y = \frac{6 	imes 324}{625} = \frac{1944}{625} = 3.1104 \, m$.$	herefore DE = 3.11 \, m$ (approx).$DF = DE + EF = 3.11 \, m + 6 \, m = 9.11 \, m$.Thus,the length of the supporting wire attached to the roadway $18 \, m$ from the middle is approximately $9.11 \, m$.

Similar Questions

$PQ$ is any focal chord of the parabola $y^2=32x$. The length of $PQ$ can never be less than: ............ $unit$

The length of the normal chord to the parabola $y^2 = 4x$,which subtends a right angle at the vertex,is

What is the point of intersection of the latus rectum and the axis of the parabola $y^2 - 4y - 2x - 8 = 0$?

The diameter of the parabola $y^2 = 4ax$ which bisects the chords parallel to $y = mx + \alpha$ is:

For the parabola represented in the parametric form by $x=t^2+t+1$ and $y=t^2-t+1$,the length of the latus rectum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module