a curve of radius 139 m is banked at an angle of 24 degrees. At what speed can it be negotiated under icy con

Answer 1

Grav. accel inward = gsin(24deg)
Cent. accel = v^2/r
These two must be equal for no skidding on frictionless surface.
When they are equal, v=sqrt(rgsin(24deg))

Answer 2

Ac=centripetal accleration = v^2/r Ac.p =Ac in direction parallel to floor = Ac cos theta Ag = Gravitational rigidity = -g Ag.p = Ag in direction parallel to floor = Ag sin theta the motor vehicle won't slip whilst the forces stability out, i.e., Ac.p + Ag.p = 0 Ac cos theta + Ag sin theta = 0 v^2/r cos 40 5 - gsin45 =0 v^2/r = g tan 40 5 v = sqrt(g r tan45) v = sqrt(9.80 one * 100m * a million) v = 31.3 m/s