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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 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.

Respuesta :

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

The normal forces on the car at points A, B and, C are equal to 11,183.4 N, 14,016.33 N, and 14,809.54 N respectively.

Given to us

Mass of the car = 1140 kg

Speed of the car = 71.8 km/h = 19.94444 m/s

What is the Normal Force on the car at point A?

We can find the normal force on the car at point A by equating all the vertical forces at point A,

[tex]\sum F_y=0\\\\\text{Weight of the car} = Normal\ force\\\\m \cdot g = N\\\\N = 1140 \times 9.81\\\\N = 11,183.4\rm\ N[/tex]

Hence, the normal force on the car at point A is equal to 11,183.4 N.

What is the normal force exerted by the road on the car at point B?

We know that according to Newton's second law of motion,

We can write centripetal acceleration as,

[tex]a = \dfrac{v^2}{R}[/tex]

Calculate the net vertical forces,

[tex]\rm Normal\ force -\text{ weight of the vehicle} = m \times a[/tex]

[tex]N - (m \cdot g) = m \cdot a\\\\N = (m \cdot a )+(m \dot g)\\\\N = m(a+g)\\\\N = m(\dfrac{v^2}{R}+g)\\\\N =1140(\dfrac{19.9444^2}{160}+9.81)\\\\N = 14,016.32565\rm\ N \approx 14,016.33\ N[/tex]

Hence, the Normal force on the car at point B is equal to 14,016.33 N.

What is the normal force exerted by the road on the car at point C?

We know that according to Newton's second law of motion,

We can write centripetal acceleration as,

[tex]a = \dfrac{v^2}{R}[/tex]

Calculate the net vertical forces,

[tex]\rm Normal\ force -\text{ weight of the vehicle} = m \times a[/tex]

[tex]N - (m \cdot g) = m \cdot a\\\\N = (m \cdot a )+(m \dot g)\\\\N = m(a+g)\\\\N = m(\dfrac{v^2}{R}+g)\\\\N =1140(\dfrac{19.9444^2}{125}+9.81)\\\\N = 14,809.544832\rm\ N \approx 14,809.54\ N[/tex]

Hence, the Normal force on the car at point C is equal to 14,809.54 N.

Learn more about Forces:

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