A beam is supported at its ends by supports w | vedclass

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

Question

(A) Let the vertex of the parabola be at the lowest point $O(0,0)$ and the axis be vertical along the $Y$-axis. The equation of the parabola is $x^2 =

Vedclass · Accepted Answer

$2 \sqrt{6} \, m$

Vedclass · Answer

(A) Let the vertex of the parabola be at the lowest point $O(0,0)$ and the axis be vertical along the $Y$-axis. The equation of the parabola is $x^2 = 4ay$.The total span is $12 \, m$,so the ends are at $x = 6 \, m$ and $x = -6 \, m$. The maximum deflection at the centre is $3 \, cm = 0.03 \, m$. Thus,the parabola passes through $(6, 0.03)$.Substituting these coordinates into the equation: $(6)^2 = 4a(0.03) \implies 36 = 0.12a \implies a = \frac{36}{0.12} = 300$.The equation is $x^2 = 4(300)y = 1200y$.We want to find $x$ when the deflection from the top is $1 \, cm$. Since the total deflection is $3 \, cm$,the height from the vertex $O$ is $y = 3 \, cm - 1 \, cm = 2 \, cm = 0.02 \, m$.Substituting $y = 0.02$ into the equation: $x^2 = 1200(0.02) = 24$.Therefore,$x = \sqrt{24} = 2\sqrt{6} \, m$.

$A$ beam is supported at its ends by supports which are $12 \, m$ apart. Since the load is concentrated at its centre,there is a deflection of $3 \, cm$ at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection $1 \, cm$?

Similar Questions

Find the length of the latus rectum of a parabola if $PSQ$ is a focal chord such that $SP = 3$ and $SQ = 2$.

What is the length of the latus rectum of the parabola $x^2 = -y$?

The smallest value of $x^2 - 3x + 3$ in the interval $(-3, 3/2)$ is

The end points of the latus rectum of the parabola $x^2 = 4ay$ are

If the lines $2x + 3y + 12 = 0$ and $x - y + k = 0$ are conjugate with respect to the parabola $y^2 = 8x$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module