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Classical Physics
Mechanics
Normal reaction in banked road circular motion problem
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[QUOTE="jbriggs444, post: 5470366, member: 422467"] Is the angle measured from the horizontal or from the vertical? Is positive for R1, R2 and the centripetal acceleration taken in the sense of an inward net force? I assume yes. Is positive for mg cos(θ) taken in terms of down=positive or up=positive. That would obviously have a key influence on the answer you seek. Since you have adopted coordinates in which the centripetal acceleration is neither aligned with the x-axis or with the y-axis and have restricted your attention to the x axis, one would expect only part of the centripetal force (##\frac{mv^2}{r}##) to be relevant. Yet there is no cos(θ) or sin(θ) factor on the term to account for this. Similarly, there is no cos(θ) or sin(θ) factor on the terms for R1 or R2. So I have to question whether either equation can be correct. Still, a simple argument can be made. If we select a track with a degenerate bank angle so that cos(θ) = 0 then the equations reduce to ##R_1 + R_2 = \frac{mv^2}{r}## and ##R_1 + R_2 = -\frac{mv^2}{r}## Assuming a convention that positive is inward/up then ##R_1## and ##R_2## must be positive. ##\frac{mv^2}{r}## will neccessarily be positive. So the second equation is obviously wrong. Though, as above, that does not mean that the first equation must be right. [/QUOTE]
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Classical Physics
Mechanics
Normal reaction in banked road circular motion problem
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