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3 years ago

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A car 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 f

lat 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.

1 answer:

Alinara [238K]3 years ago

4 0

Answer:

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

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