A highway curve with a radius of 430 m is banked properly for a car traveling 115 km/h. If a 1550- kg Porshe 928S rounds the curve at 220 km/h, how much sideways force must the tires exert against the road if the car does not skid?

Porsche is misspelled.

First, figure out the bank angle A from the statement that the car is "propertly banked" for v = 115 km/h. That means no lateral friction force is required to keep the tires from skidding. (At this angle, occupants will also not tend to slide in their seats)
v = 31.9 m/s
m v^2/R = m g tan A
tan A = v^2/(Rg) = 0.241
A = 13.5 degrees

At the higher speed of V = 220 km/h = 51.7 m/s, there is a centripetal force component mV^2*cos A/R up the bank, that must be balanced by friction force F and the weight component in the opposite direction.

m g sin A + F = m V^2/R cos A

Solve for F

you misspelled properly. nice.

To find the sideways force that the tires must exert against the road, we can use the concept of centripetal force. The centripetal force is the force responsible for keeping an object moving in a curved path. In this case, it is the force that keeps the Porsche 928S moving in a circular path around the curve.

First, we need to convert the speeds to meters per second (m/s):

- 115 km/h = (115 * 1000) / 3600 m/s ≈ 31.94 m/s
- 220 km/h = (220 * 1000) / 3600 m/s ≈ 61.11 m/s

Next, we can calculate the centripetal force using the following formula:

Centripetal force = (mass * velocity^2) / radius

Where:
- mass = 1550 kg
- velocity = 61.11 m/s (since the Porsche is traveling at 220 km/h)
- radius = 430 m

Now, let's plug the values into the formula:

Centripetal force = (1550 kg * (61.11 m/s)^2) / 430 m

Simplifying this calculation, we get the centripetal force:

Centripetal force ≈ 135,570 N

Therefore, the tires must exert a sideways force of approximately 135,570 Newtons against the road to prevent the car from skidding.