Friction Driven
Posted in September 3, 2009 ¬ 1:45 pmh.admin
Friction Driven
At any speed> v, a car slips when on a route that is driven at an angle theta to the horizontal cross slope
At any speed is greater than V, a car driven if slip on a horizontal circular path with radius r by a distance which is at an angle theta from the horizontal. Is u is the coefficient of friction between the tires of the car and the surface of the road show that u (= v ² - rg tan theta) / (rg + v tan ² theta). I know that u = F / R, The result u = mrw ² / mg = mv ² / r (mg). As tan theta have come about, though, and not any other value as, say, sin theta and cos theta, especially when it comes to the centrifugal force? All statements are on this particular aspect'd greatly appreciated.
The hard to explain without figure .. i hope u get my statement here. Of roll attitude, it means that road is not horizontal. So, you have to draw an inclined plane on the way (toward the center of the curve inclination). Now is the normal reaction of the road through the mass, but not mg mg cos theta. Well, the centrifugal force must be resolved, and u will get a load of sin ^ 2 mrw Theta such as the force normal traffic, the two u and u get the normal reaction mg cos theta sin theta ^ 2 + MRW To Get the frictional force, u need to parallel to the centrifugal pull parallel to the road and weight component of the road. this wud be mrw ^ 2 cos theta - mg sin theta. Divide these values u and get the friction coefficient. u hv to multiply both numerator and denominator by r / cos theta receive the required response
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