......

[edwardraff]

 

[burnedout]

OK, I'm probably wrong on this, but here goes:

Given:

V= 35 m/s
M=1000 kg
MUs= 0.50
g=9.8 m/s^2
r=?
theta = ?

First I found the maximum radius on a flat curve:

MUs(0.5) = (V^2/R)/g = (1225/R)/9.8. R = 250 m

Then find tan theta. tan theta = (V^2)/rg = 1225/2450 = 0.5. tan -1 (0.5) = 26.56 degrees.

Check: (mg/cos theta)*sin theta = m*(v^2/r)

9800/.894 * .447 = 4900 = 1000*(1225/250)

I may be wrong. There are lot's of physics gurus running around on ATOT. So one of them may be so kind as to verify my calc.

 

[edwardraff]

 

[burnedout]

I didn't know the radius before you edited. That makes it easier.

tan theta = v^2/rg

tan theta = (35 m/s)^2/(50*9.8) = 2.5. tan -1 theta = 68.19 degrees.

Again, I may be wrong.

 

[edwardraff]

 

[edwardraff]

 

[Gnurb]

I remember this problem from Physics 1. The answer comes down to:

tan(theta)=mu(coefficient of friction).

 

[edwardraff]

<< I remember this problem from Physics 1. The answer comes down to:

tan(theta)=mu(coefficient of friction). >>

Could someone explain how this works

 

[edwardraff]

ttt

 

[edwardraff]

ttt

 

[burnedout]

edwardraff: I think I got it. Here is how I believe it is done (with disclaimer, of course).

Given of course:

V= 35 m/s
M=1000 kg (Important! For this problem, consider mass just 1 kg instead of 1000 kg)
MUs= 0.50
g=9.8 m/s^2
r=?
theta = ?

1st: tan theta = (35 m/s)^2/(50*9.8) = 2.5. tan -1 theta = 68.19 degrees (This is the minimum frictionless bank)

2nd: MUs = 0.50 = tan -1 theta = 26.56 degrees. (This is the friction)

3rd: 68.19 - 26.56 = 41.63 degrees.

4th: Check: For y: Fn = mg/(cos theta - MUs sin theta) = 9.8 / ((0.747) - 0.50 (.664)) = 9.8/.415 = 23.61 m/s

5th: v = SquareRoot of (50m{(23.61 m/s2) sin(41.63) + 0.5(21.61 m/s2)cos(41.63)} = 1224. Square root of 1224 ~ 35 m/s

I think that's it. But we need a guru. Anyway, I never give up until I have the solution.

 

[MereMortal]

I am always leery of providing solutions to persons who do not show that they have even attempted a solution to the problem. I find it better to only help as much as necessary, otherwise the person gains little. But as burnedout has aready provided a solution for you, I will continue.

You have three forces in the problem, the weight of the car, W, the normal force from the road, N, and the frictional force, f. Since N and f are parallel/perpendicular to the road surface, it is useful to decompose the forces into horizontal and vertical components, based on the angle of inclination from the horizontal, theta.

There is no acceleration in the vertical, so a force balance gives:
N*cos(theta) - W - f*sin(theta) = 0

In the horizontal direction, the normal force and friction provide the centripetal acceleration to keep the car from slipping:
N*sin(theta) + f*cos(theta) = Fc = m*v^2/R


Using the fact that f = mu*N and W =m*g both these equations can be solved for the unknown normal force, and set equal. This gives

sin(theta) + mu*cos(theta)
--------------------------------------- = v^2/(g*R)
cos(theta) - mu*sin(theta)


Note that :

1) The mass drops out. It would be a tremendous difficulty if engineers had to worry about the mass of the vehicle when determining banking angle.

2) As mu ---> 0, the equation reduces to tan(theta) = v^2/(g*R)

3) As theta ---> 0, the equation reduces to mu = v^2/(g*R)

4) The equation can be solved for theta, or as burnedout has shown, you can arrive at the answer more simply by taking the frictionless banking angle and subtracting off the equivalent angle due to friction being the sole contribution to the centripetal acceleration.


Edit: Found a typo, minus sign in horizontal force eqn. changed to plus sign.

 

[burnedout]

Thanks MereMortal. I just learned something cool from the way you just presented that.

 

[MereMortal]

Well, I was bored, so I decided to work out the connection between the solution I provided and the one burnedout provided.

They are equivalent since

sin(theta) + mu*cos(theta)
--------------------------------------- = v^2/(g*R)
cos(theta) - mu*sin(theta)

can be rewritten as

v^2/(g*R) - mu
--------------------------- = tan(theta)
1 + mu*v^2/(g*R)


This can be written in the form tan(A - B):

tan(arctan(v^2/(g*R))) - tan(arctan(mu))
------------------------------------------------------------ = tan(theta) = tan(arctan(v^2/(g*R)) - arctan(mu))
1 + tan(arctan(mu))*tan(arctan(v^2/(g*R)))

so

theta = arctan(v^2/(g*R)) - arctan(mu)

The angle is just the difference between the frictionless angle and the equivalent angle due purely to friction.

 

[burnedout]

I just cut and pasted how you worked that out. Great job and again, Thanks!

 

[MereMortal]

Fixed a typo in the horizontal force equation. Had a minus where there should have been a plus.