A car with a mass of 900 kg moves along a road at a 89 km/hour at equal speed. On the road is a curve with a radius of 250.
a) What is the force in the curve between the tires and the road if the road is horizontal?
b) What inclination is the best for that car on the road?
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Is that formula:( F=mv^2/r ) right for for a) ?
Yes, you are correct. The formula you mentioned, F = mv^2/r, is indeed the correct formula to calculate the force between the tires and the road when a car is moving in a curve.
To answer part a) of your question, we need to plug in the values given. Here's the breakdown:
- Mass of the car (m): 900 kg
- Speed of the car (v): 89 km/hour
However, before we can continue, we need to convert the speed from km/hour to m/s, as the formula requires the speed to be in meters per second.
To convert km/hour to m/s, you can divide the speed by 3.6. So, 89 km/hour is equal to 89/3.6 = 24.7 m/s approximately.
Now, calculating the force in the curve (F):
F = (m * v^2) / r
F = (900 kg * (24.7 m/s)^2) / 250 m
F = (900 * 24.7^2) / 250
F ≈ 2151.58 N
Therefore, the force between the tires and the road when the road is horizontal is approximately 2151.58 Newtons.
Moving on to part b) of your question: the best inclination for that car on the road. It seems like you may be referring to the banking of the curve. The inclination or banking angle of the road can actually help in providing a necessary centripetal force to keep the car moving in a circular path.
To calculate the best inclination angle, we can use the equation:
tanθ = (v^2) / (g * r)
Where:
- θ is the inclination angle
- v is the speed of the car (in m/s)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- r is the radius of the curve
Plugging in the values:
tanθ = (24.7 m/s)^2 / (9.8 m/s^2 * 250 m)
tanθ ≈ 0.2481
Now, to find the inclination angle θ, we take the inverse tangent (arctan) of 0.2481:
θ ≈ arctan(0.2481)
θ ≈ 13.8 degrees
Therefore, the best inclination or banking angle for the road in order to keep the car on the curve without requiring any additional friction is approximately 13.8 degrees.