Question

part of a racing track is to be designed for a curvature of 72m. We are not recommending vehicles to drive faster than 216km/h. with what angle should the road be tilted.? by what height will it outer edge be w.r.t the inner edge if track is 10m wide?

Answer 1

Answer:

Part of a racing track is to be designed for a curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angle

should the road be tilted? By what height

will its outer edge be, with respect to the

inner edge if the track is 10 m wide?

by Rupali1810 4 weeks ago

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Answers

kelkarakshay21

Kelkarakshay21Ambitious

Answer:

Ө = tan inverse( 5 )

Ө = 78.69 degree

h = 9.805 meter

Explanation:

Given data :-

radius of curvature r = 72m

vmax = 216 km/hr = 216 x (5/18 ) = 60 m/s

width of tack = w = 10 m

What we have to find out :-

1) angle should the road be tilted i.e angle of banking = Ө= ?

2) height of outer edge h = ?

Solution :-

1)The angle of banking is given by,

Ө = ( / rg )

= tan^{-1} ( 60^{2} / 72 x10 ) g = 10 m/s^{2}

= (3600 /720 )

Ө = ( 5 )

Ө = 78.69 degree

2) now,

h = w (sin Ө)

h = 10 (sin 78.69 degree )

h = 10 ( 0.9805 )

h = 9.805 meter

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