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	<title>Comments on: What is the required banking angle for the curve?</title>
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	<link>http://www.klientbanku.swinoujscie.pl/2009/10/what-is-the-required-banking-angle-for-the-curve/</link>
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		<title>By: kuiperbelt2003</title>
		<link>http://www.klientbanku.swinoujscie.pl/2009/10/what-is-the-required-banking-angle-for-the-curve/comment-page-1/#comment-1502</link>
		<dc:creator>kuiperbelt2003</dc:creator>
		<pubDate>Fri, 23 Oct 2009 23:31:19 +0000</pubDate>
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		<description>draw a diagram of the bank, showing the directions of the normal force (perpendicular to the banked road), weight,(straight down), and the centripetal force (to the center of the circle)

we will decompose these forces into x and y components, and apply Newton&#039;s second law

call theta the angle of the bank; then, use geometry principles to show that theta is also the angle between a vertical line and the direction of the normal force

this enables us to write the x and y components of the normal force as:

x component = N sin (theta)
y component = N cos (theta)

the y component must balance the weight, so we have

N cos (theta) = mg  (eq 1)

and the x component must produce the centripetal force, so we have

N sin(theta) = mv^2/r  (eq 2)

divide equation (2) by equation (1) to get:

sin (theta)/cos(theta) = v^2/rg

or tan(theta) = v^2/rg

for the data here, we have

tan(theta) = (16.7m/s)^2/(400m x 9.8m/s/s) 

tan(theta) = 0.07 or theta = 4.05 deg

(note that you should convert km/hr to m/s)</description>
		<content:encoded><![CDATA[<p>draw a diagram of the bank, showing the directions of the normal force (perpendicular to the banked road), weight,(straight down), and the centripetal force (to the center of the circle)</p>
<p>we will decompose these forces into x and y components, and apply Newton&#8217;s second law</p>
<p>call theta the angle of the bank; then, use geometry principles to show that theta is also the angle between a vertical line and the direction of the normal force</p>
<p>this enables us to write the x and y components of the normal force as:</p>
<p>x component = N sin (theta)<br />
y component = N cos (theta)</p>
<p>the y component must balance the weight, so we have</p>
<p>N cos (theta) = mg  (eq 1)</p>
<p>and the x component must produce the centripetal force, so we have</p>
<p>N sin(theta) = mv^2/r  (eq 2)</p>
<p>divide equation (2) by equation (1) to get:</p>
<p>sin (theta)/cos(theta) = v^2/rg</p>
<p>or tan(theta) = v^2/rg</p>
<p>for the data here, we have</p>
<p>tan(theta) = (16.7m/s)^2/(400m x 9.8m/s/s) </p>
<p>tan(theta) = 0.07 or theta = 4.05 deg</p>
<p>(note that you should convert km/hr to m/s)</p>
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