Uniform Circular Motion and Centripetal Force

[AJayS]

Hey guys, first post here! Hoping to get a little help.

Homework Statement

You are a traffic safety engineer in charge of determining safe speeds for roads. A particular banked curve has a radius of 11.0 meters and is banked at an angle of 8.00°. The coefficient of static friction between common tires and this road is 0.870. What is the maximum speed that a car can drive this curve? Use both the bank of the curve and the friction on the tires in determining your answer.

Homework Equations

f=μn
Fc=mv^2/r

The Attempt at a Solution

So what I did was split the force of gravity, mgcos8 in the direction perpendicular to the ramp and mgsin8 parallel. Also, the force of friction towards the center of rotation and set all forces towards the center of rotation to mv^2/r.

Essentially, I had mgsin8 + (μ X mg X cos8)=mv^2/r.

Common factor of m cancels and I solve for V as everything else is given. I receive an answer of 10.4 m/s. The correct answer is apparently 11.1 m/s, so close, but not close enough for a rounding issue I believe. My only other guess is that somehow I've split the vector wrong. Any help's much appreciated, thanks very much in advance.

 

[Simon Bridge]

Welcome to PF;
Notice that gravity acts straight down - so there is no component of the gravity force acting directly towards the center of the turn.

Did you draw a free-body diagram?
You need the vector sum of the forces to point horizontally towards the center - the forces are gravity, friction and the normal force from the road and they all act in different directions.

 

[AJayS]

Hi Simon, thanks for the response.

Now I see it a little clearer. I did draw a free body diagram but it was incorrect. I was treating the parallel surface of the bank as straight horizontal. I'm still getting an incorrect answer however.

As I have it, I have the sum of the horizontal forces as the normal force in the x direction (Fnsin8). I am confused as to where friction fits into all of this. If I'm correct, doesn't friction always oppose the direction of motion? In other words, would the ∑Fx= Fnx - friction in the x direction?

 

[Simon Bridge]

If I'm correct, doesn't friction always oppose the direction of motion?

Nope - a car accelerating forward in a straight line has a net friction force acting on it pointing in the same direction as the acceleration.

In this case, this is static friction. Static friction acts one stationary objects too - more accurately: between surfaces that are instantaneously stationary wrt each other.

It opposes the direction the surfaces would move if there were no friction.

Consider:
http://t0.gstatic.com/images?q=tbn:ANd9GcSfjwcftYv3-4OgKxziSLn1cHg850F5nLClXo9lngoNk7IXNoqNvQ

 

[HalfLostAndSearching]

Hi there! As a fellow scientist, I would like to provide some feedback on your solution.

Firstly, your approach is correct in considering the forces acting on the car on the banked curve. However, there are a few areas that need to be addressed in your solution.

1. The forces acting on the car should be split into components parallel and perpendicular to the banked curve, not the ramp. This is because the banked curve is the direction in which the car is turning, not the direction of gravity.

2. The force of friction should be in the opposite direction of the car's motion, towards the center of the curve. This is because friction acts to prevent the car from sliding outwards and helps maintain its circular motion.

3. When solving for the maximum speed, you should consider the maximum value of the coefficient of static friction, which is 0.870 in this case. This means that the car can only drive at a speed where the force of friction is equal to the maximum value, or else it will start to slide outwards.

Taking these factors into account, the correct equation to use would be:

mgsin8 + (μmax x mgcos8) = mv^2/r

Solving for v, we get v = √(rgtan8 x μmax)

Substituting in the given values, we get v = √(11 x 9.8 x tan8 x 0.870) = 11.1 m/s

I hope this helps clarify your solution. Keep up the good work in your studies!