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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards