Last updated at May 29, 2018 by Teachoo

Transcript

Example 18 A beam is supported at its ends by supports which are 12 metres 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? Beam is always facing upwards with the axis vertical Since, the axis is positive y-axis, its equation is x2 = 4ay First we find coordinates of point B Given Width of beam = 12 m Hence, AB = 12 m So, BC = 𝐴𝐵2 = 122 = 6 m Also, there is a deflection of 3 cm from centre So, OC = BD = 3 cm OC = BD = 3 cm = 3100 m Hence point B is B(6, 𝟑𝟏𝟎𝟎) Now, Since point B(6, 3100) lies on the parabola Putting x = 6, y = 3100 in equation x2 = 4ay (6)2 = 4a 3100 36 = 3𝑎 25 325 a = 36 a = 36 × 253 a = 12 × 25 a = 300 m Now, we need to find how far from the centre is the deflection 1 cm Hence RQ = 1 cm, We need to find OP QP = 3 cm – 1 cm = 2cm = 2100 m Let OP = x So, coordinates of point Q is Q(x, 𝟐𝟏𝟎𝟎) Since point Q lies on parabola it will satisfy the equation of parabola Equation of parabola is x2 = 4ay Putting x = x & y = 2100 m & a = 300 m x2 = 4 (300) 2100 x2 = 1200 × 2100 x2 = 24 x = 24 = 6 ×4 = 26 𝑚 Thus, the required distance is 2𝟔 𝒎

Chapter 11 Class 11 Conic Sections

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.