Acar with a mass of 1140 kg is traveling in a mountainous area with a constant speed of 71.8 km/h. the road is horizontal and flat at point a, horizontal and curved at points b and c. the radii of curvatures at b and c are: rb = 160 m and rc = 125 m. calculate the normal force exerted by the road on the car at point a. tries 0/20 now calculate the normal force exerted by the road on the car at point b. tries 0/20 and finally calculate the normal force exerted by the road on the car at point c.

point A  N = 11172 N , point B  N = 13486 N  and point C  N = 14801 N

Explanation:

Let's use Newton's second law on the Y axis, where the car is in equilibrium

a) Horizontal and flat road

Point A

     N -w = 0

     N = W = mg

     N = 1140 9.18

     N = 11172 N

b) Horizontal and curved road

In this case, when applying Newton's second law, we have centripetal acceleration, due to the curve

     N -W = m a

     a = v² / R

     N = mg + m v² / R

     N = m (g + v² / R)

Point B

     Rb = 160 m

We reduce the speed to SI units

     V = 71.8 km / h (1000m / 1km) (1h / 3600s) = 19.95 m / s

     N = 1140 (9.8 + 19.95²/160)

     N = 1140 11.83

     N = 13486 N

B point

Rb = 125 m

     N = 1140 (9.8 + 19.95²/125)

     N = 1140 (12.98)

     N = 14801 N

the curve is assumed 90º

less than one you can't create energy

1 the 4 is sig