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Old Sep 30th 2017, 12:18 PM   #1
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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
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Old Sep 30th 2017, 05:25 PM   #2
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Originally Posted by VectorVictor View Post
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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Old Oct 1st 2017, 08:00 AM   #3
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Ahh that makes a lot of sense. Just using that simple identity to get rid of normal force and mass.
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