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Well of Death: Physics behind it
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[QUOTE="Steve4Physics, post: 6824507, member: 681522"] Hi [USER=682219]@NTesla[/USER]. Can l put my twopence worth in? If you go round a banked curve at a sufficiently high speed, you tend to slide upwards. (You should understand why this happens.) Static friction opposes the upwards sliding by acting downwards. If ##\theta \lt 90^o## and you are going [S]too [/S]fast, you [S]will[/S] may slide upwards because downwards friction isn't big enough to prevent the sliding. Your maximum speed to avoid sliding upwards is ##V_{max}##. Note that if ##\theta= 90^o## there is no tendency to slide upwards (and you should understand why). [B]The equation for ##V_{max}## applies [U]only[/U] when the frictional force is at its maximum (limiting) value of ##\mu N## and is acting downwards - because that’s how the ##V_{max}## equation is derived. [/B] But when ##\theta = 90^0## there is [U]no[/U] downwards friction because there is no tendency to move upwards. Friction will act upwards and it's magntiude will equal ##|mg|##; the equation for ##V_{max}## is inapplicable. _____________ Additional note. It’s not just ##\theta = 90^o## where care is needed. For example if ##\mu = 0.9## and ##\theta = 60^o## then ##\cos(\theta) - \mu \sin(\theta) = \cos(60^o) – 0.9\sin(60^o) = -0.279## which gives a negative value inside the square root. That’s because the ##V_{max}## equation cannot be used because friction doesn’t need to reach its limiting value to prevent upwards sliding for these values of ##\mu## and ##\theta##. The ##V_{max}## equation is therefore inapplicable in this case- and this was signalled by the fact that ##\cos(\theta) - \mu \sin(\theta)## was negative. [/QUOTE]
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