Road Bank Top Speed Calculation | G & \mu

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The discussion focuses on calculating the top speed of a vehicle on a banked road designed for a speed of 60 km/h with a radius of 70 m and a friction coefficient of 0.8. The angle of the bank is initially calculated to be approximately 22.124 degrees, and the forces involved include gravitational force and centripetal force. Participants suggest re-evaluating the angle and the components of the forces acting on the vehicle, particularly how centripetal force relates to the banking angle. Clarification is provided that the centripetal force acts parallel to the ground, not the bank. The conversation emphasizes the importance of accurately resolving forces to determine the correct top speed.
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Advanced Road Bank question

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 won't 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.
 
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dlingo said:

The Attempt at a Solution


First I calculated the angle at which the car won't 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?
 
dlingo said:
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.
 
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