find the direction and magnitude of frictional force acting on a vehicle. if vehicle moves with speed 5 m /s (assume that friction is sufficient to prevent slipping
Answers
A turn of radius 20m is banked for a vehicle with the mass of 200kg going at a speed of 10m/s. The frictional force acting on the vehicle, if vehicle moves with speed 5m/s is 688N in outward direction or away from center of circular path.
• Given that : r = 20m, m = 200kg, v = 10m/s and v' = 5m/s. Let g = 9.81m/s^2
• As the turn is banked for 10m/s so, so angle of inclination of the bank = tan^-1 (v^2/rg)
• So, angle of inclination, x = 27°
• We know that as the speed of vehicle decreases, the force of friction acting upwards to the inclined bank increases.
• Now, we will find the magnitude of this force by given formula-
force of friction, f = m*g*sinx -(m*v^2*cosx)/f = 667.977N ≈ 668N
• As the value of frictional force is positive. Hence, the direction of f is away from the center of circular path.