Banked Curve question


by hollystella
Tags: banked, curve
hollystella
hollystella is offline
#1
Oct11-06, 04:34 PM
P: 6
Hey all,

I'm new here and would appreciate some help with this question about banked curves.

A curve with a 120-m radius on a level road is banked at the correct angle for a speed of 20 m/s. If an automobile rounds this curve at 30 m/s, what is the minimum coefficient of static friction neeed between tires and road to prevent skidding?

I wasn't quite sure how to attack this problem, especially since I wasn't given the mass of the object...

Advanced thanks!
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Mindscrape
Mindscrape is offline
#2
Oct11-06, 06:16 PM
P: 1,877
What makes you think that there is a mass dependency? What have you done so far? Have you made a free body diagram? If so, then what equations and ideas have you gotten out of it? I only ask these questions because we're unsure about what you're unsure about.
hollystella
hollystella is offline
#3
Oct11-06, 08:17 PM
P: 6
The problem was:
A curve with a 120-m radius on a level road is banked at the correct angle for a speed of 20 m/s. If an automobile rounds this curve at 30 m/s, what is the minimum coefficient of static friction neeed between tires and road to prevent skidding?

I drew a free body diagram with normal force n perpendicular to the surface, and because it was a banked road, there is a vertical component n*cos(angle) and n*sin(angle). I also put in w = mg pointing down in the vertical direction. The static friction force is pointing towards the center of the circle. After that, I didn't know where to go. I tried using F = ma, but then I didn't know what the value of m was. I calculated the angle using the relationship tan(angle) = (v^2)/(Rg) and got angle = 18.8 degrees, thinking that might play into the problem somehow. After that, I didn't know where to proceed. I know that static friction f[s] = mu[s]*n, but I wasn't sure how to get n because I didn't have a value for mass, or how to get the static friction force value because I couldn't use Newton's 2nd equation.

Any help will be greatly appreciated.

OlderDan
OlderDan is offline
#4
Oct11-06, 09:41 PM
Sci Advisor
HW Helper
P: 3,033

Banked Curve question


I'm going to edit/annotate your text with blue
I drew a free body diagram with normal force n perpendicular to the surface, and because it was a banked road, there is a vertical component n*cos(angle) and a horizontal component n*sin(angle). I also put in w = mg pointing down in the vertical direction. The static friction force is pointing towards the center of the circle but tilted downward, parallel to the surface of the track. After that, I didn't know where to go. I tried using F = ma, but then I didn't know what the value of m was. I calculated the angle using the relationship tan(angle) = (v^2)/(Rg) [I don't know where this comes from] and got angle = 18.8 degrees, thinking that might play into the problem somehow. After that, I didn't know where to proceed. I know that static friction f[s] = mu[s]*n, but I wasn't sure how to get n because I didn't have a value for mass, or how to get the static friction force value because I couldn't use Newton's 2nd equation.

Any help will be greatly appreciated.
You have some good thoughts here. This is a two step problem. First the track is designed so that at a speed of 20m/s, no friction is needed. The only forces acting are gravity and the normal force. The sum of all (two) forces acting causes the acceleration. We know the acceleration is horizontal because the car does not move up or down, and we know that the acceleration is toward the center of a circle with magnitude v^2/r (centripetal acceleration) Use the force components you have identified to find n in terms of mg; then find the horizontal force and set it equal to the known magnitude of a centripetal force. All the forces are proportional to the mass of the auto, so that mass will divide out. This part of the problem allows you to solve for the angle of bank of the track.

Now change the speed and add the friction force. Resolve it into horizontal and vertical components. Follow the reasoning above, but include the additional force and use the known angle from the first step. This part will let you solve for the frictional force and the normal force. From those you can calculate the necessary coefficient. Larger coefficients will keep the car from skidding. Smaller ones will not.
hollystella
hollystella is offline
#5
Oct12-06, 05:50 AM
P: 6
Thanks so much for your help. This is what I calculated it out to:

I found the sum of the forces in the vertical direction: ncos(angle) + (-mg) = 0, so ncos(angle) = mg, and in terms of mg, n = (mg)/(cos[angle]).

To find out what that angle was, I used the sum of the forces in the horizontal direction: nsin(angle) = m[(v^2)/(R)], and then replaced n with what I found above to make it sin(angle)/cos(angle) = m[(v^2)/(gR)], so tan(angle) = (v^2)/(g*R) = (20^2)/(9.8 * 120) = 0.34. So tan of 0.34 = 18.8 degrees.

Then I added the static friction in the x-direction:
nsin(angle) + f[s] = m[(v^2)/(R)]
After replacing n and doing all that I did above, I got tan(angle) = (v^2)/(g*R) - (f[s]). I replaced the angle with 18.8, the v with 30 m/s, and g = 9.8 m/s^2, and R = 120 m, solved and got 900 for the right part of the equation, and got f[s] = 500.16.

To find the normal force, I again took nsin(angle) = 900 - 500.16 (from the f[s]), and got n = 1240.7. I replaced that in f[s] = mu[s]*n, and used the 500.16, and got 0.40 as my coefficient of static friction.

Does that seem like a reasonable answer? I looked at a table of coefficients of static friction, and couldn't determine whether it was large enough to keep the car from skidding.

Thank you soooo much! :-)
OlderDan
OlderDan is offline
#6
Oct12-06, 06:33 AM
Sci Advisor
HW Helper
P: 3,033
Quote Quote by hollystella
Then I added the static friction in the x-direction:
nsin(angle) + f[s] = m[(v^2)/(R)]
The logic for finding the angle is perfect. I'll assume your computations are correct. I think you have a problem with friction. Friction acts parallel to the track. You have to find horizontal and vertical components of friction, just as you found horizontal and vertical components of the normal force.
Thomas_
Thomas_ is offline
#7
Oct7-07, 10:36 AM
P: 21
Hello,

I'm sorry to bring up this old thread again, but I'm having the exact same problem. However, I decomposed friction into horizontal and vertical components.

I ended up with these equations:

X: Nsin(A) + fsin(A) = mV^2/R
Y: Ncos(A) - fcos(A) = mg

After dividing I get:
(N+f)/(N-f) tan(A) = V^2/Rg
= N(1 + u)/N(1-u) tan(A) = V^2/Rg
= (1 + u)/(1-u) = V^2/Rg

After solving this for the coefficient (u), I get ~ 0.38. However, the answer is supposed to be 0.34.

g = 9.8
A = 18.785
V = 30


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