Can a Car Safely Turn on a 40-Degree Banked Road at 30 m/s?

[dk2421]

Homework Statement

Will a road banked at 40 degrees allow cars (u=0.6) to safely make a turn (radius=100m) while traveling at 30 m/s (or 67 mi/hr)?

Homework Equations

f = uN
possibly others

The Attempt at a Solution

I'm really not sure how to get started. This is the only problem I don't know how to do. I would really appreciate any help. Thank you.

 

[LowlyPion]

Welcome to PF.

Draw a force diagram on the car.

Gravity is vertically down, so the components of weight can be resolved as vectors along the incline and normal to the incline.

The centripetal acceleration is acting horizontally. Its components can also be resolved along the plane of the incline and normal to it.

If the downward force of gravity and the additional maximum contribution from the friction component is sufficient to prevent the car from slipping then ...

 

[thomate1]

I would approach this problem by first understanding the physics involved in a car turning on a banked road. When a car is turning, there are two main forces acting on it: the centripetal force, which keeps the car moving in a circular path, and the frictional force, which helps the car maintain its grip on the road.

In this scenario, the banked road provides an additional component to the normal force (N), which is the force perpendicular to the surface of the road. This component, known as the banking force, helps to counteract the centrifugal force (F = mv^2/r) of the car as it turns. This means that the car can safely make the turn at a higher speed without slipping off the road.

To determine if a car can safely make a turn on a banked road, we need to consider the forces acting on the car and use the equations of motion to solve for the unknown variables. In this case, we can use the following equations:

F_net = ma (Newton's Second Law)
F_friction = uN (frictional force equation)
tanθ = v^2/rg (equation for the angle of banking force)

Where:
F_net is the net force acting on the car
m is the mass of the car
a is the acceleration of the car
F_friction is the frictional force
u is the coefficient of friction
N is the normal force
θ is the angle of banking force
v is the speed of the car
r is the radius of the turn
g is the acceleration due to gravity (9.8 m/s^2)

Plugging in the given values, we can solve for the normal force (N) and the banking angle (θ):

F_net = ma
F_friction = uN
F_net = F_friction + F_banking
ma = uN + mv^2/r
N = mgcosθ
ma = umgcosθ + mv^2/r
a = ugcosθ + v^2/r

tanθ = v^2/rg
tanθ = (30 m/s)^2/(100 m)(9.8 m/s^2)
tanθ = 0.918
θ = 43.3 degrees

Now that we have calculated the angle of banking force, we can use it to solve for the normal force:

N = mgcosθ
N