# What is the magnitude of resultant force on a car on a banked curve?

• brsclownC
In summary: So in summary, your equation for centripetal force is the mv^2/r and this should equal the horizontal component of the normal force on the car. Vertical component of normal force and gravity would cancel out.
brsclownC
Homework Statement
A race car travels 47 m/s around a circular track of radius 182 m which is banked at an angle of 21 degrees. Assuming the track is frictionless, what is the magnitude of the resultant force on the 1300 kg driver and his car if the car does not slip?
Relevant Equations
Fc = mv^2/r
Fc = mgtan(21)
Ok so I think that the equation for centripetal force is the mv^2/r and this SHOULD equal the horizontal component of the normal force on the car. Vertical component of normal force and gravity would cancel out. However, when I input the numbers into the equations I don't get equivalent values.

(1300)(47^2)/182 = 15.8 kN
(1300)(9.8)(tan(21)) = 4.9 kN

What am I doing wrong?

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brsclownC said:
Homework Statement: A race car travels 47 m/s around a circular track of radius 182 m which is banked at an angle of 21 degrees. Assuming the track is frictionless, what is the magnitude of the resultant force on the 1300 kg driver and his car if the car does not slip?
Homework Equations: Fc = mv^2/r
Fc = mgtan(21)

Ok so I think that the equation for centripetal force is the mv^2/r and this SHOULD equal the horizontal component of the normal force on the car. Vertical component of normal force and gravity would cancel out. However, when I input the numbers into the equations I don't get equivalent values.

(1300)(47^2)/182 = 15.8 kN
(1300)(9.8)(tan(21)) = 4.9 kN

What am I doing wrong?
Go back to basics. Write the equations for vertical and horizontal force balance as you described them. (In the equations you posted you seem to have gone further, dividing by the cos.)
Use symbols, not numbers. (Never plug in numbers until the end... many benefits.)

Have in mind that since the car is doing circular motion, the result of the combined force of gravity and the normal force from track will be the centripetal force.

We usually analyze gravity to a component vertical to the incline and a component parallel to the incline, but here you have to analyze the normal force from the track to a horizontal component (which horizontal component will be equal to centripetal force) and a vertical component (which will cancel out gravity).

## 1. What is the definition of a banked curve?

A banked curve is a curved section of road or track that is tilted or angled in such a way that it helps a vehicle to turn without relying solely on friction.

## 2. How is the magnitude of resultant force on a car on a banked curve calculated?

The magnitude of resultant force on a car on a banked curve is calculated using the formula F = m * (v^2 / r), where F is the resultant force, m is the mass of the car, v is the velocity of the car, and r is the radius of the curve.

## 3. What factors affect the magnitude of resultant force on a car on a banked curve?

The magnitude of resultant force on a car on a banked curve is affected by the mass and velocity of the car, as well as the angle of the banked curve and the radius of the curve.

## 4. How does the angle of the banked curve affect the magnitude of resultant force on a car?

The angle of the banked curve affects the magnitude of resultant force on a car by determining the amount of centripetal force required to keep the car moving in a circular path. A greater angle will require a smaller centripetal force and therefore result in a smaller resultant force on the car.

## 5. How does the speed of the car affect the magnitude of resultant force on a banked curve?

The speed of the car affects the magnitude of resultant force on a banked curve by increasing the centripetal force needed to keep the car in a circular path. As the car's speed increases, the resultant force also increases. This is why cars must slow down on sharp turns to avoid losing control due to a large resultant force.

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