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denpristay [2]
3 years ago
5

A banked highway is designed for traffic moving at v = 88 km/h. The radius of the curve r = 314 m. show answer No Attempt 50% Pa

rt (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g.

Engineering
2 answers:
emmainna [20.7K]3 years ago
6 0

Answer:

a) \tan \theta = \frac{v^{2}}{g\cdot R}

Explanation:

a) The Free Body Diagram of the vehicle and reference axis are included in the image attached below. Equations or equilibrium are presented below:

\Sigma F_{x} = N\cdot \sin \theta = m\cdot \frac{v^{2}}{R}

\Sigma F_{y} = N\cdot \cos \theta - m\cdot g = 0

Tangent of the highway can be found by dividing the first expression by the second one:

\tan \theta = \frac{m\cdot \frac{v^{2}}{R} }{m\cdot g}

\tan \theta = \frac{v^{2}}{g\cdot R}

kherson [118]3 years ago
6 0

Part b) :

Calculate the angle of banking on the highway

Answer:

a)tan \theta = \frac{v^{2} }{rg}

b) \theta = 10.97^{0}

Explanation:

traffic speed, v = 88 km/ h = 88 * (1000/3600) = 24.44 m/s

Radius of the curve, r = 314 m

a) Write an equation for the tangent of the highway's angle of banking.

The traffic should move both in the vertical and horizontal directions

In the vertical direction, it is moving against gravity, therefore

\sum F_{y} = 0\\Ncos \theta - mg =0\\

Ncos \theta = mg..............(1)

For the horizontal movement,

\sum F_{y} = 0\\Nsin \theta - ma =0\\

Nsin \theta = ma................(2)

To get the tangent equation, divide equation (2) bu equation (1)

\frac{N sin \theta}{N cos \theta} = \frac{ma}{mg}

tan \theta = \frac{a}{g}...........(3)

Centripetal acceleration, a = \frac{v^{2} }{r} \\..........(4)

Substituting (4) into (3)

tan \theta = \frac{v^{2} }{r} /g\\tan \theta = \frac{v^{2} }{rg}

b) Angle of banking on the high way

v = 24.44 /s

r = 314 m

g = 9.81 m/s^2

Substituting those values into the result gotten in part a

tan \theta = \frac{24.44^{2} }{314*9.81}\\tan \theta = 0.194\\\theta = tan^{-1} 0.194\\

\theta = 10.97^{0}

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(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1
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Answer:

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The equation is \sigma y = \sigma 0 + k y d^{\frac{1}{2}}

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by putting the above value in the formula we get the \sigma y value that is= 157.5 MPa

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