Circular motion of car on banked track, with friction

1. Dec 27, 2008

JonTot

I recently ran into something between two teachers. For the longest time one has been teaching that when a car is traveling in circular motion on a banked track with friction, only the horizontal component of the friction force is to be taken into account, by adding to the centripetal force; the friction force of course being parallel to the banked track. However, another teacher has told me that the vertical component is taken into account, and this significantly complicates things. Since the car is not moving up or down along the incline, all vertical forces must cancel out. So the vertical component of the normal force must equal (in magnitude) the force of gravity and the vertical component of friction. But friction depends on the normal force. I've attached a diagram of the situation and the solutions, (using mathtype) as images to make sure you understand what each teacher is suggesting. What I want is a second opinion; do you take the vertical component of friction into account or not?

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• circular motion on a banked cuve with friction.GIF
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2. Dec 28, 2008

Ranger Mike

absolutely...having run these cars since 1971 I can tell you for a fact. Round track racing is about one thing and one thing only..TIRE SLIP ANGLE. or in engineering terms..the relationship between slip angle, coefficient of friction and cornering force. Spring rates, shock dampers (absorbers in USA) down force et all are about keeping as much tire contact patch on the pavement as possible. Weight, mass load and load transfer all relate back to the tire.
in simple terms..a car going into a 33 degree baked corner will do a whole lot of weight transfer..from the left rear to the right front. In fact Dayton super speed way cars have +/- 6 inch shock travel..it is that much!

other key factors in figuring out the best combination are- linear acceleration, acceleration thrust, rotational inertia, linear deceleration, centrifugal acceleration, ratio of sprung to unsprung weight. power to weigh ration. center of gravity, mass centroid , roll center of suspension system ( my favorite) roll axis, roll moment , polar moment of inertia, static weight distribution, dynamic load transfer and finally aerodynamic load..

to keep it real simple. try this test. go get the spare tire out of the trunk. Set it up like it would be on the car. now slide it across the garage floor..slides pretty easy..right.. now go get your fat little brother ( sister if applicable) and have him sit on the tire..now try to push it across the floor..takes a lot more work,,right..

this is what happens when suspensions unload weight from the rear of the car and cross load it to the right front on a steep left hand turn..the more load the more "grip" to a point..once the mechanical grip goes away..so does "handleing"
it is one huge balancing act between a lot of variables..

you get a handle on the above and you will see a lot of checker in your future....

Last edited: Dec 28, 2008
3. Dec 28, 2008

Staff: Mentor

Absolutely. All forces must be taken into account, including the vertical component of friction. (The first solution is incorrect.)

4. Dec 28, 2008

Ranger Mike

longitudinal load transfer which occurs in the longitudinal plane under linear acceleration or deceleration = acceleration (g) x Weight x ch height ( center of gravity) / wheelbase

assume 1760 lb formula car with 100 inch wheelbase, 13 inch center of gravity
704 lb front weight , 1056 rear wheel weight
braking at 1.2 g

crank thur the formula and 275 lbs. is tansfered ..i.e. ft wheel weight is now 979 lbs, rear wheel weight is now 781 lbs.
but...there is also left to right side weight that occurs as well!

this is lateral load transfer (lb) = Lateral accelaeration (g) x weight x cg height / track width

the track width is distance from the center the right rear tire to the center of the left rear tire...just about all race cars have wider rear track width so you use the widest track width on the car

one more wrinkle is " jacking weight" which occurs too
but that is another post i think

Last edited: Dec 28, 2008
5. Dec 28, 2008

Staff: Mentor

Determining if you can ignore a force is a simple matter of counting equations and unknowns.

Here, if you write the 2nd law equation for the horizontal components you have two* unknowns: the magnitude of the normal force and the magnitude of the friction force. You cannot solve one equation in two unknowns so you know that you must use the vertical components in order to get your second equation.

*There are two unknowns is assuming that the centripetal acceleration or force is given. Otherwise you will need a third equation.

6. Dec 28, 2008

Ranger Mike

what?

7. Dec 28, 2008

Staff: Mentor

The OP's confusion was due to the fact that one teacher thought you could ignore the vertical component of the friction force. I was just explaining the general method for determining when you can ignore a force or a component and when you cannot.

8. Dec 28, 2008

Ranger Mike

both of um got him to think..so both did what they get paid fer ..right..
and he asked a very good question every crew cheif asks every race day,,
looks like we all got a checker on this one!

9. Dec 28, 2008

JonTot

Thanks everybody, i was pretty sure you couldn't just ignore a force.
I've found a much more elegant way of presenting the solution for the second teacher's method. It's the same up till the part where the expression derived for the normal force is substituted in, and instead of doing that, this time i took out a common factor of the normal force before substituting, and then did some trig work to get a more elegant looking solution. Thought I'd share.

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