Calculating Maximum Speed on Banked Curve: Centripetal Force

[swimstar]

A highway curves to the left with radius of
curvature R = 46 m. The highway’s surface
is banked at θ = 26 so that the cars can take
this curve at higher speeds.
Consider a car of mass 1018 kg whose
tires have static friction coefficient µ = 0.57
against the pavement.
The acceleration of gravity is 9.8 m/s/s.

Ho fast can the car take this curve without skidding to the outside of the curve?
Answer in units of m/s.

Here is a link for the picture: https://quest.cns.utexas.edu/student/assignments/problem_pdf?courseuserassignment=10797960

It is question #4.

I tried to do this by setting the equations for Normal Force and Frictional Force equal and solving for V.
Therefore I got,

{ (v2/R)cos(theta) - mgsin(theta) } = { u (v2/R)*sin(theta) + mgsin (theta))

However, I am unable to get the correct answer.

Help is needed as soon as possible. Thank you.

 

[pgardn]

A highway curves to the left with radius of
curvature R = 46 m. The highway’s surface
is banked at θ = 26 so that the cars can take
this curve at higher speeds.
Consider a car of mass 1018 kg whose
tires have static friction coefficient µ = 0.57
against the pavement.
The acceleration of gravity is 9.8 m/s/s.

Ho fast can the car take this curve without skidding to the outside of the curve?
Answer in units of m/s.

Here is a link for the picture: https://quest.cns.utexas.edu/student/assignments/problem_pdf?courseuserassignment=10797960

It is question #4.

I tried to do this by setting the equations for Normal Force and Frictional Force equal and solving for V.
Therefore I got,

{ (v2/R)cos(theta) - mgsin(theta) } = { u (v2/R)*sin(theta) + mgsin (theta))

However, I am unable to get the correct answer.

Help is needed as soon as possible. Thank you.

The diagram to not show up.

Is it possible for you to draw a free body diagram showing the forces on the car so we might be able to work through the equations you will need based on the diagram?

 

[graphene]

The link asks for a login.

 

[ehild]

I tried to do this by setting the equations for Normal Force and Frictional Force equal and solving for V.
Therefore I got,

{ (v2/R)cos(theta) - mgsin(theta) } = { u (v2/R)*sin(theta) + mgsin (theta))

.

Why should be the normal force equal to the frictional force?
You mix acceleration and force in your equation. How did you get it?

ehild