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Normal Forces on Tyres while cornering 
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#1
Mar2412, 05:04 PM

P: 5

I'm trying to derive an expression for the normal forces on a cars tyres travelling around a circular banked track. Initially I was going to assume the centre of gravity was located longitudinally and laterally centre between all four tyres (so front and rear had the same forces on them). Also the car will be travelling at constant angular velocity.
However even with an unbanked track, I seem to be missing something, I know that the normal force must be velocity dependent (since it is going to decrease on the inside tyres and increase on the outside tyres as velocity increases). In the image uploaded the axis of rotation is to the left. Taking moments about the centre of gravity you get... [tex] \sqrt{\frac{x^{2}}{4}+h^{2}}·(N_2sinθN_1sinθF_1cosθF_2sinθ)=0 [/tex] and I know F_{1}+F_{2}=mrω^{2} as well as the coefficient of friction being the same for both tyres. I'm struggling to combine these two to give a velocity dependent expression for N_{1} & N_{2}. Confusingly when I take moments about the base of each tyre I get the same value for N_{1} & N_{2} after rearranging... [tex] \sqrt{\frac{x^{2}}{4}+h^{2}}·\frac{mgsin\phi }{x} [/tex] which is not velocity dependent or correct. Cheers for any help. EDITSorry, I can't figure out how to move this post over to the homework/courseworktype questions forum or how to delete it. 


#2
Mar2412, 10:51 PM

P: 264

I'm not sure of this but I'll give it a go.
When you go around a turn, choose the frame of the car. The CG is pushed out due to a centrifugal force and that's [itex]mv^{2}/r[/itex]. Now, we balance the torques using say, the outer wheels as our axis. The torque from the centrifugal force would be [itex]mv^{2}h/r[/itex]. The torque equation reads [itex] mgx/2N_{1}mv^{2}h/r=0[/itex]. Do the same analysis with the inner wheels as your axis to get [itex] mgx/2N_{2}+mv^{2}h/r=0[/itex] I think the force of friction doesn't matter since those go through our axis of rotation anyway. At first glance it looks correct because it is velocity dependent and the normal force for the outer wheels (wheel 2) is indeed larger. 


#3
Mar2512, 06:43 AM

P: 5

I was trying to keep it in an inertial reference frame ideally (sorry, should have said). Although I agree with your working.
Surely it's possible to do this in an inertial reference frame; since all those equations are literally just Normal ± the torque created by centripetal force. In this scenario is centripetal force not just the friction forces? and isn't the torque created by friction not acting through the centre of gravity? Sorry, realised I was being an idiot above. It should read: if you take moments about the centre of gravity you get: [tex] \frac{x}{2}(N_2N_1)  h(F_1+F_2)=0 [/tex] and if you take moments about either tyre you get: [tex] \frac{x}{2}mgN_2=0 \,\,\,\,\,\, or \,\,\,\,\,\, \frac{x}{2}mgN_1=0 [/tex] although this is the same problem, so I'm still obviously missing something. 


#4
Mar2512, 05:46 PM

P: 264

Normal Forces on Tyres while cornering
Okay your first equation is identical to what I posted earlier because [itex]F_{1}+F_{2}=mv^{2}/r[/itex]. By the way, small mistake in my earlier post, it should be [itex] mgx/2N_{1}xmv^{2}h/r=0[/itex] and [itex] mgx/2N_{2}x+mv^{2}h/r=0[/itex]. So that's settled.
For your second set of equations, I can't quite explain why the [itex]mv^{2}/r[/itex] term needs to be there but it definitely does. Hopefully someone else can explain it. 


#5
Mar2512, 10:26 PM

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I see your first equation
but on moments around individual tyres, pardon me if i'm dumb. it's been a long time. you wrote : ##\frac{x}{2}mgN_2=0 \,\,\,\,\,\, or \,\,\,\,\,\, \frac{x}{2}mgN_1=0## If i sum moments about bottom of right hand tire taking CCW positive i see: ##mg\frac{x}{2}N_1x  \frac{mv^2}{r}h=0 ## exactly what McLaren said. 


#6
Mar2512, 11:38 PM

P: 264

Jim, I think the OP wants to do it in an inertial frame of reference. Even I'm not sure where the [itex]mv^{2}h/r[/itex] term comes from when this is analyzed in an inertial frame. I have a feeling that I am missing something obvious because it is a pretty simple problem but it still evades me.
Could you explain again why you have that term? When we see it going around the turn from an inertial frame, the centripetal force is provided by the friction so this has no torque right? 


#7
Mar2612, 09:24 AM

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I don't quite understand inertial frame  so my bad .
the centripetal force indeed is developed by the tires and goes through my summing point, ergo zero torque . But the centrifugal force acts on CG at distance h above my summing point and is coupled through the automobile suspension so produces torque, ergo nonzero moment. I should know what you mean by inertial frame but i don't. I seem to remember something called "Center of Mass Coordinates" ... My bad, i'll look it up. Thanks !! old jim 


#8
Mar2612, 10:18 AM

P: 5

By "inertial reference frame" I meant an observer watching the car with constant velocity. From the reference frame of the car (i.e. considering the car as stationary) you have to introduce pseudoforces (such as centrifugal forces) to explain why the car isn't moving in that reference frame.
However, after ALOT of searching I stumbled across this site on car rollover. It says that an acceptable thing to do, even in an inertial reference frame, is to consider the friction forces acting radially outwards from the centre of gravity since it creates the same torque that the friction creates from the base of the wheels. So according to that website both your calculations are correct (in terms of inertial and noninertial reference frames, I guess ). The paragraph in question: 


#9
Mar2612, 10:56 AM

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from Wikipedia which of course makes tires pull other direction aka centripetal force have i achieved thought equivalent of inertial reference? They've assigned names to a lot of things we 'just did' when i was in school and sliderules were king. 


#10
Mar2612, 11:17 AM

P: 5

The thing is that centrifugal force doesn't actually exist, it's a 'pseudoforce' that emerges when you build Newton's laws of motion in an accelerating reference frame. You shouldn't need it to explain the turning forces experienced by some objects in circular motion when being watched by an observer whose acceleration is zero (i.e. in an inertial reference frame).
see: http://en.wikipedia.org/wiki/Fictiti...ircular_motion I'm not referring to definition of centrifugal force as the reaction force to a centripetal force by the way (which in this instance would be the friction expierienced by the track as supposed to that expierienced by the tyres) 


#11
Mar2612, 11:47 AM

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Thanks for that link !
I think i've got it. It's one of those 'fine points'. Road pushes car sideways ##\sum{F} = ma## and that's a dynamics problem and in my school days only difference between statics and dynamics was, in statics ##\sum{F} = 0## So there'd be a statics problem also to test whether car successfully turns or rolls over and that's the one where we'd include ##\frac{mv^2}{r} ## Thanks for your patience . 


#12
Mar2612, 01:10 PM

P: 5

Cheers guys, think I've cracked it now.
Now onto the joys of sprung, unsprung weight and lateral load transfer. 


#13
Mar2612, 07:05 PM

P: 264

HB, I'm glad you cracked it. This thread might be helpful because it's a simplified version of your problem with an excellent analogy from A.T. in post 11.
The key point that I missed earlier is that in an inertial frame, the net change in angular momentum about either tyre is not zero. The car is accelerating towards the center of its circular path and this is a change in angular momentum whose magnitude is exactly the missing [itex]mv^{2}h/r[/itex] In uniform circular motion, there is an acceleration towards the center but there is never any velocity along the radial direction. Similarly, in our problem, there is a torque (and a change in angular momentum) about the axis at the tyre but the axis moves such that there is never any angular momentum with respect to it at any time. 


#14
Mar2612, 07:19 PM

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What does a car have to do with a city in Lebanon?



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