Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum 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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See the forum rules here.   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.  Tags angle, banked, roadway Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post jackthehat Kinematics and Dynamics 6 Jul 18th 2015 08:35 AM Pycckuu Kinematics and Dynamics 1 Nov 20th 2010 08:18 PM s3a Kinematics and Dynamics 1 Mar 7th 2010 08:06 PM werehk Kinematics and Dynamics 18 Aug 11th 2009 10:03 PM Air Light and Optics 4 May 13th 2008 05:07 AM