Coefficient of Friction for 25° Banked Curve

[MikeH]

Hey everyone, I could use some help with this problem.

What coefficient of friction is required to make a 25 degree banked curve safe at 90. km/h, if without friction it is safe at 60. km/h?

Any help would be appriciated.

 

[ahrkron]

Hi MikeH,

Can you post some of you thoughts about it?
What information do you have?

Basically, the case without friction will allow you to obtain the missing piece from the statement. Can you see what that "missing piece" is?

 

[MikeH]

I found the radius of the curve by Fc=Nsin25 / mg=Ncos25 which leaves you with Tan25= Fc/mg.
Replace Fc with (mv^2)/r
which allows the masses to cancel
leaving tan25 = (v^2)/gr
rearrange for r
r= 3.65

I got this far.

Then for the second part I drew a free body diagram but I don't know where to add my friction and what to calculate next.

 

[ahrkron]

I may be wrong, but I got 10.54m for r.

As for the role of friction, imagine, in your diagram what is the effect of having a higher speed. mv²/r is the force you need to keep the car in that curve. When v=60km/h, things are fine, but now v=90km/h, which means you need to push the car into the curve a little more.Here's where friction enters.

 

[MikeH]

Were my equations right or did you use something different.
Though I did find one mistake with my calculations, I forgot to square the speed.

tan25 = (v^2)/gr
rearrange for r
r = (v^2)/g tan25
r = (16.7)^2/(9.8)tan25
r = (278.9)/(4.57)
r = 61m
This seems quite large though. How did you get your answer?

The friction part.
After your post it seems more evident that where the speed increases it will fall off the upper part of the curve therefore the friction must be pushing down the curve. I separated all my forces into the horizontal and vertical components and I think that I need to make the horizontal equal to the Fc and the vertical components equal to zero. I'm not sure where that puts me but that is what other sites that I've looked on say to do.

 

[ahrkron]

I was using 25km/h instead of 60

Ok, so now the radius looks reasonable
(think about a curve with 3m, 10m and 60m radius).

Yes, the friction will point "down" (it will be parallel to the surface of the curve).

Instead of horizontal and vertical components, try components parallel and perpendicular to the asphalt (sp?), since the friction will be one of them.

 

[MikeH]

0 = N - Fgtan25
0 = N - (9.8*m)tan25
N = 4.57m m = mass
u = coefficient of friction
Then

Fc = Ff + Fg*tan25
(mv^2)/r = uN + 4.57m
(mv^2)/r = u(4.57m) + 4.57m

Divide by m to cancel out m

(v^2/r) = 4.57u + 4.57
(25)^2/61 = 4.57u +4.57
10.25 = 4.57u +4.57
5.68 = 4.57u
u = 1.24
Obviously there is a mistake since u cannot be >1.
Where did I go wrong?

 

[ahrkron]

Hi MikeH,

I'm sorry I was out of touch for a while.

Why do you write "0 = N - Fgtan25" ?

Isn't it cos[/color] instead of tan[/color]? (I'm assuming you use N for the normal force)

Then, I think you're missing a subtle point:

The centripetal force needed to keep the car in the curve points horizontally[/color] towards the center of curvature; i.e., not along the 25deg slope.

This means that the friction needs to be decomposed in two parts; only the component along the horizontal will contribute to keeping the car in the trajectory we want.

i.e., the usual expresion for friction Fc = uN gives you the component along the surface. In this case, you need to multiply by cos25 to get the amount of friction that will "push" the car in the right direction (i.e., towards the center of curvature). Similarly with the weight.

Alternatively, you can keep using the friction and the weight components in the plane of the curve, but use the right Fc instead (i.e., the force you need in that plane so that, when decomposing into components along x and y, give you mv^2/r in the horiz direction).

I get something like 0.8 for the friction, but (as I already proved) I can be quite sloppy .

 

[MikeH]

WOW!
I'm getting 0.76 !
My mistakes.

1--I forgot to calculate the force of gravity pulling down on the car which pulls it down the curve.

2--I drew the triangle of the Fg to find the Force perpendicular to the slope with the right angle in the wrong spot thus giving me the Fgtan25

3--I calculated Fc as parallel to the slope.

Thank you so much for your help. It's greatly appriciated.