Physics Help Forum Angle of a Banked Roadway
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 Sep 30th 2017, 12:18 PM #1 Junior Member   Join Date: Sep 2017 Posts: 3 Angle of a Banked Roadway While reading the chapter for uniform circular motion, one of the examples details finding the angle of a banked roadway given the maximum speed and radius. The equations for forces in horizontal and vertical directions are: (1) ncosθ=mg (2) nsinθ=mv^2/r where n is the normal force Now, my question is why do we divide equation 2 by equation 1? I can't seem to wrap my head around the logic of doing this. Thanks
Sep 30th 2017, 05:25 PM   #2

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 Originally Posted by VectorVictor While reading the chapter for uniform circular motion, one of the examples details finding the angle of a banked roadway given the maximum speed and radius. The equations for forces in horizontal and vertical directions are: (1) ncosθ=mg (2) nsinθ=mv^2/r where n is the normal force Now, my question is why do we divide equation 2 by equation 1? I can't seem to wrap my head around the logic of doing this. Thanks
It's simply to eliminate the n in the two equations.
$\displaystyle \frac{n~sin( \theta )}{n~cos( \theta )} = tan( \theta )$

-Dan
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 Oct 1st 2017, 08:00 AM #3 Junior Member   Join Date: Sep 2017 Posts: 3 Ahh that makes a lot of sense. Just using that simple identity to get rid of normal force and mass.

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