For the banked curve, the horizontal component of the normal force toward the ce
ID: 1694860 • Letter: F
Question
For the banked curve, the horizontal component of the normal force toward the center of the curve reduces the need for friction to prevent skidding. In fact, for a circular curve banked at a given angle, there is one speed for which no frictional force is required--the centripetal force is supplied completely by the inward component of the normal force.(a) Show that the banking angle ? for which no friction is required for a car to negotiate a circular curve safely is given by tan ? = v2 / gr, where v is the car's speed and r is the radius of curvature. (b) How would the banking angle differ for a more massive truck? (c) Show how friction affects the situation.
Explanation / Answer
N sin theta = m v^2/R centripetal force N cos theta = m g weight of car a)tan theta = v^2 / (R g) b) since m is not in the equation the angle does not depend on the mass c) f = u N cos theta if friction is present N sin theta + f cos theta = m v^2 / R N cos theta + f sin theta = m g [sin theta + u cos^2 theta] / [cos theta + u sin theta cos theta] = v^2 / (R g) If theta were zero then u = v^2 / (R g) and the car could negotiate the turn even on a perfectly flat curve