Circular motion - banked angle?

[paperdoll]

Homework Statement

a 1250 kg car follows a circular path around a roundabout of radius 18m at a constant speed of 24 kmh-1

at what angle would the road need to be banked for there to be no need to rely on friction to maintain the circular path?

Homework Equations

SOH CAH TOA?
F=ma
Fc=mv^2/r

The Attempt at a Solution

v=6.6666 ms-1
r=18
m=1250

f=mv^2/r
f=3085.8 N

I can't figure out how to do the question from here :\

 

[Doc Al]

Start by drawing yourself a free body diagram of the car. What forces act on it? Then apply Newton's 2nd law.

 

[paperdoll]

Start by drawing yourself a free body diagram of the car. What forces act on it? Then apply Newton's 2nd law.

I must be drawing my diagram wrong because I get 14.6 degrees as my answer but the answer in the back of the book is 15.3

let me see if I can take a photo

 

[PeterO]

I must be drawing my diagram wrong because I get 14.6 degrees as my answer but the answer in the back of the book is 15.3

let me see if I can take a photo

I trust you were using the tangent function for this moving body, and not the sine function you would use for a stationary body (when calculating necessary friction)

 

[paperdoll]

I trust you were using the tangent function for this moving body, and not the sine function you would use for a stationary body (when calculating necessary friction)

Hmm...I was using the sin function. I haven't used tangent before, is this specially for circular motion?

here is a picture of what I was trying to do:
http://img513.imageshack.us/img513/5274/photoon20120121at2143.jpg

Uploaded with ImageShack.us

if tan function is to be used, does that mean gravity is not counted? I am confused

 

[Doc Al]

here is a picture of what I was trying to do:
http://imageshack.us/photo/my-images/841/photoon20120121at2143.jpg/

In your diagram, what forces act on the car? I can see gravity, but what other force acts?

if tan function is to be used, does that mean gravity is not counted?

Gravity is definitely counted.

 

[paperdoll]

In your diagram, what forces act on the car? I can see gravity, but what other force acts?

Gravity is definitely counted.

the centripetal force is the one which is 3085.8 N and I think it's acting on the diagonal to the car? (on diagram)

since gravity is on the hypotenuse, I'm not sure how to derive a tangent function

 

[Doc Al]

the centripetal force is the one which is 3085.8 N and I think it's acting on the diagonal to the car? (on diagram)

Ah... Some problems:
(1) 'centripetal force' is just the name we give to whatever net force give rise to the centripetal acceleration. It should not appear on a free body diagram. Only real forces should appear on the diagram. So... What force besides gravity acts on the car?
(2) What's the direction of the centripetal acceleration? It's not along the diagonal!

 

[PeterO]

Hmm...I was using the sin function. I haven't used tangent before, is this specially for circular motion?

here is a picture of what I was trying to do:
http://img513.imageshack.us/img513/5274/photoon20120121at2143.jpg

Uploaded with ImageShack.us

if tan function is to be used, does that mean gravity is not counted? I am confused

The free body diagram gives the answer.
The two forces acting in the car are:
1. gravity [vertically down] and
2. the Normal reaction Force [perpendicular to the surface].
The net force is the centripetal force [horizontally towards the centre of the circle].

When you draw those three in a triangle, the Normal reaction force [which you don't know the size of yet**] is the hypotenuse.
The Weight Force and the Centripetal Force are the adjacent and opposite sides - thus the tangent function.

** the reason you don't know the size of the Normal Reaction force yet is that as a reaction Force, it becomes "as big as is necessary". Only after a full analysis will you know how big that is.

Example: what is the normal reaction force on a billard ball dropped onto a surface?

If you place a billiard ball on a block of putty, you will create a small indentation - indicating that there was a certain force needed to support the ball.
If you DROP a billiard ball onto a block of putty, you will create a bigger indentation - indicating that a larger force needed to STOP the ball.

 

[paperdoll]

Ah... Some problems:
(1) 'centripetal force' is just the name we give to whatever net force give rise to the centripetal acceleration. It should not appear on a free body diagram. Only real forces should appear on the diagram. So... What force besides gravity acts on the car?
(2) What's the direction of the centripetal acceleration? It's not along the diagonal!

okay, I just went to google and did some research and the centripetal motion should be on the horizontal...but I thought they shouldn't appear on a free body diagram?

the only other force I can think of is the "normal force" which would be 12250 N acting upwards from the car

 

[Doc Al]

okay, I just went to google and did some research and the centripetal motion should be on the horizontal...

Right. The acceleration is horizontal (towards the center of the circle the car is making).

but I thought they shouldn't appear on a free body diagram?

The centripetal acceleration will appear when you apply Newton's 2nd law to analyze the forces.

the only other force I can think of is the "normal force"

Good!

which would be 12250 N acting upwards from the car

No. Since the car is accelerating and on an angled road, you cannot assume that the normal force equals the weight of the car. Just call it "N", an unknown. You won't need to solve for it to find the angle.

You have the two forces: Gravity, which acts downward, and the normal force. Now apply Newton's 2nd law to the vertical and horizontal force components. You'll get two equations.

 

[PeterO]

the only other force I can think of is the "normal force" which would be 12250 N acting upwards from the car

That is the wrong size for the Normal force, and the "Normal" part of the name "Normal Force" is a reference to the force being perpendicular to the surface.

Since a banked track is not horizontal; the Normal Force will not be vertical. [the weight force will of course be vertical].

Re-read my earlier post and you may see how/why it is Tan that you use, in conjunction with weight and centripetal Force, to get the angle of banking.