Banked Road Physics Problem: Maximum Speed Calculation

[mr1709]

1. Homework Statement
A car is moving along a banked highway on a ramp that is banked at an angle of 14 degrees to the horizontal. The radius of curvature of the bank is 264m and the coefficient of static friction is 0.67. What is the max speed that the car can travel and safely stay on the ramp? (Ans: The max speed to negotiate the turn on a banked curve is 28.5 m/s)

Homework Equations

The Attempt at a Solution

. [/B]
Attempt at solution in uploaded picture. I don't think i did anything wrong...is the solution manual possibly incorrect?

 

[drvrm]

A car is moving along a banked highway on a ramp that is banked at an angle of 14 degrees to the horizontal. The radius of curvature of the bank is 264m and the coefficient of static friction is 0.67. What is the max speed that the car can travel and safely stay on the ramp? (Ans: The max speed to negotiate the turn on a banked curve is 28.5 m/s)

the maximum speed comes around 50 m/s- your

attempt shows 54m/s, so, please check the calculation
i could not get your free body diagram? in my opinion f(.n) = mg .cos(theta) and

mg sin (theta) should act along the slope and frictional force should be mu (s).f(n).

 

[Doc Al]

Your solution looks fine to me. What book is this from?

 

[Doc Al]

in my opinion f(.n) = mg .cos(theta)

I would not make that assumption. (That would hold for the standard block sliding down an incline problem, but not for this problem where the acceleration is horizontal, not parallel to the surface.)

 

[mr1709]

Your solution looks fine to me. What book is this from?

Im not sure. It was a print out sheet my physics teacher gave the class to practice for our upcoming test. Thanks for input

 

[haruspex]

I agree with 53.5 m/s.
There is a slightly easier way to get there.
$mg=N\cos(\theta)-\mu N\sin(\theta)$
$m\frac{v^2}r=N\sin(\theta)+\mu N\cos(\theta)$
Dividing
$\frac{v^2}{gr}=\frac{\tan(\theta)+\mu}{1-\mu\tan(\theta)}$
$=\tan(\theta+\arctan(\mu))$