Wednesday

April 16, 2014

April 16, 2014

Posted by **Ryan** on Wednesday, October 14, 2009 at 10:39am.

- physics -
**bobpursley**, Wednesday, October 14, 2009 at 11:08amWell, my first thought is that of road engineers: Banked curves beg speeders to speed. One cannot on a public highway encourage racing.

Now, the question.

Friction down the incline= mg*.8*cosTheta

so gravity down the incline cant be greater than that.

(1) mgSinTheta<mg*.8*cosTheta

or TanTheta<.8

Now at 60km/hr (16.7m/s) centripetal force will have two components: up the plane (mv^2/r * CosTheta) and normal to the plane (mv^2/r*sinTheta). We still have the up the plane and normal components of the weight of the car.

So, the force up the plane < retarding friction force.

mv^2/r*CosTheta -mgSinTheta < .8(mgCosTheta+mv^2/r * SinTheta)

Divide both sides by cosTheta

mv^2/r -mg TanTheta<.8mg+.8mv^2/r tanTheta

solve for tan Theta in (1)put that into the equation and solve for r.

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