## Homework Statement

There is a road bank designed for 60km/hr (16.7 m/s) with a radius of 70m and no friction. What is the top speed with a mew of 0.8$$\mu$$ ?
G=gravitational constant
V=16.7m/s

## Homework Equations

Tan$$\theta$$=V^2/(R*G)
Fc=(M*V^2)/R

## The Attempt at a Solution

First I calculated the angle at which the car wont move with no friction and I got 22.124
Then I started filling in what I knew:
Fnormal:=(G*sin($$\theta$$))+(Fc*Sin(90-$$\theta$$)) The component of gravity added to a component of Centripetal force.
then:
Ffriction:=Mew*Fnormal
Fc:=Ffriction+the component of fnormal pushing back against the Fc.
(M*V^2)/R=Mew*((G*sin($$\theta$$))+(Fc*Sin(90-$$\theta$$)))the component of fnormal pushing back against the Fc.
But I don't know if I'm on the right track or completely wrong.

Last edited:

gneill
Mentor

## The Attempt at a Solution

First I calculated the angle at which the car wont move with no friction and I got 22.124
Then I started filling in what I knew:
Fnormal:=(G*sin($$\theta$$))+(Fc*Sin(90-$$\theta$$)) The component of gravity added to a component of Centripetal force.

You might want to recheck where the angle is. I suspect you want $$M g cos(\theta) + F_c sin(\theta)$$

then:
Ffriction:=Mew*Fnormal
Fc:=Ffriction+the component of fnormal pushing back against the Fc.
(M*V^2)/R=Mew*((G*sin($$\theta$$))+(Fc*Sin(90-$$\theta$$)))the component of fnormal pushing back against the Fc.
But I don't know if I'm on the right track or completely wrong.

See if you can find the components of -centripetal force (well, I suppose we could call it centrifugal force if we whisper softly) and gravitational force that lie along the road surface.

There is one thing I don't understand, Is the Fc parallel to the ground or the bank?

gneill
Mentor
There is one thing I don't understand, Is the Fc parallel to the ground or the bank?

Fc is in the same direction as the radius vector for the moving object describing the circle. In this case, it's parallel to the ground.