posted by Sandhya .
A 600-kg car is going around a curve with a radius of 120 m that is banked at an angle of 25.0° with a speed of 30.0 m/s.The coefficient of static friction between the car and the road is 0.300. What is the force exerted by friction on the car?
Inward force needed =m a = m v^2/R
= 600 (900)/120 = 4500 Newtons inward
down slope component = 4500 cos 25
Force up slope = Ff friction
weight down slope = m g sin 25
Net force down slope = 2488 - Ff
2488-Ff = 4500 cos 25
Ff = 2488 - 4078
Ff = -1590 N so 1590 DOWN the slope
Can that .3 coef of friction provide the 1590 N ?
normal component of weight = 600 * 9.81 * cos 25 = 5334, yes, good, we do not slip up or down