# Find speed: Banked Curve w/ coeff static fric given, but fric parallel slope must be0

1. Jul 17, 2012

### smsport

Ʃ1. The problem statement, all variables and given/known data

A car of mass 2,000 kg is moving round a curve on a
banked track (see diagram-We were actually only given the top picture "a" and not the bottom one, but both are relevant) at a constant speed. The
coefficient of static friction between the car's tires and
the track is μs = 0.140. The radius of curvature of the
car's path is r = 200m, and the angle of bank of the track
is θ = 10.0o . Showing all your work, find the speed
that the car must travel such that the force of static
friction parallel to the slope is zero.

2. Relevant equations

Ʃ Fx= Fnx= Fnsin10=Fc = mv^2/r (No Ffrictionx because it should =0) Equation #1
Ʃ Fy= Fny -Ffrictiony-mg=0
Ffrictiony= 0.140 sin 10
Fny= Fncos10
ƩFy= Fncos10-0.140Fnsin10=mg Equation #2

3. The attempt at a solution

Divide equation 1 by equation 2:
(mv^2/r =Fnsin10)/(mg=Fn(cos10-0.140sin10):

m's cancel and Fn's cancel then solve for v:

v=√(rg(sin10/cos10-0.140sin10)
v=18.8 m/s

My question is this: I understand that the so-called design speed is the speed a car can get around the track without friction and that this is expressed as v=√(rgtan theta) because there is no element of friction to use. For the problem above, I initially solved it using this simpler "design speed" formula and got an answer very close to my answer above. However, close is not necessarily right! I looked at the question again and figured that since the coefficient of static friction was given that I needed to include the Ffriction in the y direction. Am I right on this?

I am finding banked curve free body diagrams with friction to be confusing. There seem to be so many elements and it's hard to keep track. I also have some difficulty understanding the X and Y directions of friction as for what they really mean. I get that the x direction in this problem involves the Fnx as the only force acting as the centripetal force only because the problem states that the x direction static friction is zero. If it didn't and was perhaps asking for the min speed it could go around curve w/o slipping then I would have Fstaticfriction in the opposite direction as Fnx (ie: static friction away from center)? I am having a hard time wrapping my brain around these banked tracks problems.

Am I at least on the correct path with how I worked the problem above?

Thanks!

#### Attached Files:

• ###### Banked Track.pdf
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2. Jul 17, 2012

### PeterO

Re: Find speed: Banked Curve w/ coeff static fric given, but fric parallel slope must

Just checking:

Are your directions x and y parallel and perpendicular to the slope, or horizontal and vertical.

If parallel and perpendicular, then there is no friction in either direction, so that 0.14 sin... factor will disappear and you are back to what you called the design speed - which I believe is what you want.

Also - the centripetal force is not one of the acting forces.

The actual forces here are weight [force due to gravity], which is vertically down, and the Normal reaction Force, which is perpendicular to the slope [that is what Normal means]. The vector sum of those two will give the resultant force - which in this case is the required Centripetal Force.

3. Jul 18, 2012

### smsport

Re: Find speed: Banked Curve w/ coeff static fric given, but fric parallel slope must

I do consider the normal force perpendicular to the slope.

When you note that the only actual forces are mg and normal reaction force, there is no friction force vertically down along with the weight force? For some reason I thought that was the case as I've seen other banked curve problems where it seems there is a friction force vertically down. Perhaps I am misunderstanding something about the vertical in these banked track problems. My initial assumption (which is not what I did above where I included a y direction friction force) for the Force in the y direction is that it is just FN=mg x cos theta plus the weight force and that this should equal zero because the car is neither levitating nor falling through the earth? And you seem to be saying that the resultant of the weight and the normal force perpendicular is the required centripetal force and so that means there is no required centripetal force because these two are zero? Am I following you on that correctly?

Also, because friction is zero then that implies there is no force whatsover in the direction parallel the slope?

At any rate, then for this problem, it appears you are saying my inital thoughts were correct in that this question is basically asking for the design speed, or r times g times tantheta=v^2. Then the coefficient of static friction given, just like the mass of the car, has no use in finding the speed the car must travel such that the force of static friction parallel to the slope is zero for this problem?

Last edited: Jul 18, 2012
4. Jul 18, 2012

### PeterO

Re: Find speed: Banked Curve w/ coeff static fric given, but fric parallel slope must

Commenting on the five bits I highlighted in red, in order.

1: Friction, if present, is always parallel to the surface, since it is opposing any tendency to slip across the surface. The only way to have friction acting vertically down would be to have an object slip vertically up on a surface - like if your leaned you hand on a wall and forced it to slide up that wall. You would then even feel that friction force!

2: This body is accelerating! It is travelling in a circle, not a straight line, so the net force will not be zero. also FN=mg x cos theta is what applies when a body is placed on a sloping ramp. When travelling around a banked track this does not apply.
When a car is parked on a road, FN = mg [but in the opposite direction of course].
When that same car is driven through a dip in the road FN > mg [but in the opposite direction of course].
Similarly, when a car is parked on a banked track FN = mg.cosθ [in the direction perpendicular to the slope, not opposite to mg].
When the car is driver around that banked track FN > mg.cosθ [in the direction perpendicular to the slope, not opposite to mg].
It can be shown that no lateral friction is necessary if FN = mg.tanθ [in the direction perpendicular to the slope, not opposite to mg].
If we have friction, it is possible for FN to be smaller or greater than mg.tanθ.
NOTE: that since FN is a reaction force, we have to change the physical conditions to alter it.
for FN = mg.tanθ you have to travel at just the right speed.
Any slower then FN < mg.tanθ
Any faster then FN > mg.tanθ
In those second two cases, we need some friction to enable the car to proceed without sliding down / up the slope.
If you have ever watched NASCAR, you will see the effect when a car that was travelling around the banked curves, with the tyres supplying the required amount of friction, if that car should come across a section of track where the friction force changes to a smaller, insufficient force [perhaps oil or some other debris].

3: Your reference to "normal force perpendicular" is slightly worrying. Don't forget that it is perpendicular to the surface, not perpendicular to the weight force !!!

4: Neither of the forces [Weight and Normal Reaction Force] is zero, and the vector sum of the two is definitely not zero. Only a pair of forces in opposite directions could possibly sum to zero; weight acts vertically down, Normal Reaction Force acts obliquely up in a direction perpendicular to the slope.

5: "Also, because friction is zero then that implies there is no force whatsover in the direction parallel the slope?"
That is true - meaning neither of the acting forces is parallel to the slope. Of course you could resolve the acting forces to get a component parallel to the slope, but it is not necessary to do so.

5. Jul 18, 2012

### smsport

Re: Find speed: Banked Curve w/ coeff static fric given, but fric parallel slope must

Yes, I do have/know that Normal force is perpendicular to the slope and not the weight force. I know that one very well. I can start the free body diagram similarly to a box on an incline. I must have seen something somewhere with friction force vertically that threw me off, but the way you explain helps me to see that, of course, this situation does not have friction vertically. I am sure I saw something somewhere that made it appear there was some sort of y component of friction that led me astray. Apologies for mistaken thoughts, there.

I did not initially believe there was no net force because the car is moving so apologies for such a inane question. Just trying to wrap my brain around exactly what is going on here.

Does anyone have a link to a good tutorial (video or otherwise) going through a fbd of a car on a banked track?

I thought I had it figured out when I convinced myself that using r x's g x's tan theta= v^2 for this problem is what was necessary, but it seems now I'm going in circles, no pun intended, and would love a link to help me see it in front of me better. My prof went over minimum/max and design speed in about a minute and my book doesn't go over this, my notes don't seem to help me understand what my prof was doing. It seems there are some videos and tutorials online but if a wise physics helper here can point me to a really good one I'd be ever so grateful. I'm an excellent student, but with physics it takes me a bit more time and I need more step by step and visual instruction on certain concepts and then it will click and I'll have it and wonder how the heck I didn't see it at first.

Thanks!

6. Jul 18, 2012

### PeterO

Re: Find speed: Banked Curve w/ coeff static fric given, but fric parallel slope must

Just sent you a message about a power point display.

Peter

And a second message as there was an error in the first!!

Last edited: Jul 18, 2012
7. Jul 18, 2012

### smsport

Re: Find speed: Banked Curve w/ coeff static fric given, but fric parallel slope must

Got it! Thanks again. I also think I figured out what the issue was, but I will be ever so grateful to look at your file also.