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

The equation of the parabola with $(-3, 0)$ as focus and $x + 5 = 0$ as directrix is

The vertex of the parabola is at $(1, 2)$ and its axis is parallel to the $y$-axis. If the parabola passes through $(0, 6)$,then its latus rectum is:

If the normal to the parabola $y^2 = 4ax$ at the point with parameter $t_1$ cuts the parabola again at the point with parameter $t_2$,then:

Let $PQ$ be a double ordinate of the parabola $y^2 = -4x$,where $P$ lies in the second quadrant. If $R$ divides $PQ$ in the ratio $2 : 1$,then the locus of $R$ is:

If a parabola having its axis parallel to the $X$-axis passes through the points $(0, -1)$,$(6, 1)$,and $(-2, -3)$,then the point at which this parabola cuts the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module