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Banked curve involving friction 
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#1
Apr105, 11:37 PM

P: 171

Hi could some of you guys please help me with question, its question 77 in Understand Physics By Cummings.
For those that don't have access to the book its a question dealing with a banked curve with radius R and a angle alpha. Now there is also a friction force stoping the car from sliding of the banked curve, now my question is how do you derive the equation to explain the situation. I know for a normal banked curve without friction the answer is just, Nsin(angle) = m(v^2)/r and Ncos(angle) = mg So therefore tan(angle) = (v^2)/rg Now how do you derive the equation is friction is also a force helping the car sliding down, i know the answer is suppose to be vmax = (rgtan(angle + tan^1(coefficient of f))) 


#2
Apr205, 05:54 AM

P: 527

If the speed of the car is any greater than the speed you calculated without friction, then the car will need an additional friction force to keep it moving in a circle. This force is parallel to and pointing down the slope. You can decompose this force into its horizontal and vertical components. The horizontal component adds to the component of normal force in the horizontal direction, and the vertical component adds to the force of gravity.



#3
Apr205, 06:37 PM

P: 171

(Vmax)² = Rg(sinx + (coeff)cosx)/(cosx  (coeff)sinx) Where coeff = coefficient of friction for the particular surface. How do i change this equation into the answer which is: (Vmax)² = Rg(tan(x + arctan(coeff))) 


#4
Apr205, 10:47 PM

P: 171

Banked curve involving friction
Back to the question, i was wondering if you can use the identity
tanx = coeffcient of friction Its just that friction is acting down the plane for this circumstance...if you can then i know that arctan(coeff) will just be the the angle of the bank. That way it could simplify down to tan2x but this doesn't yield the same answer as tan(x + arctan(coeff)). Any suggestions guys? 


#5
Apr205, 11:15 PM

P: 527

I don't think the statement [itex]\tan x = \mu[/itex] is true. Your equation [itex]gr\left(\frac{\sin x + \mu\cos x}{\cos x  \mu\sin x}\right)[/itex] is mathematically equivalent to [itex]gr\tan\left(x+\tan^{1}\mu \right)[/itex].



#6
Apr305, 03:28 AM

P: 171




#7
Apr305, 03:51 AM

P: 527

You begin with
[itex]gr\left(\frac{\sin x + \mu\cos x}{\cos x  \mu\sin x}\right)[/itex]. Divide the numerator and the denominator by [itex]\cos x[/itex] [itex]gr\left(\frac{\tan x + \mu}{1  \mu\tan x}\right)[/itex]. Using the identity [itex]\tan\left(\alpha + \beta\right) = \frac{\tan\alpha + \tan\beta}{1  \tan\alpha\tan\beta}[/itex] you get [itex]\alpha = x[/itex] and [itex]\tan\beta = \mu \Rightarrow \beta = \tan^{1}\mu[/itex]. The result is [itex]gr\left(\frac{\sin x + \mu\cos x}{\cos x  \mu\sin x}\right) \equiv gr\tan\left(x + \tan^{1}\mu \right)[/itex]. 


#8
Apr305, 03:59 AM

HW Helper
Thanks
P: 10,524

You know the addition law for the tangent function: [tex] \tan(x+y) = \frac{\tan(x) + \tan(y)}{1\tan(x) \tan(y)}[/tex] Let [tex] y = \arctan(\mu) \rightarrow \mu = \tan(\arctan(\mu)) [/tex] Replace for mu in your formula and you get the desired form. ehild 


#9
Apr305, 04:55 AM

Sci Advisor
HW Helper
P: 6,672

The horizontal forces are: (1)[tex]F_{xfriction} = \mu_sF_Ncos\theta[/tex] (2)[tex]F_{Nx} = F_Nsin\theta[/tex] Therefore: (3)[tex]F_Nsin\theta + \mu_sF_Ncos\theta = mv^2/r[/tex] One has to look at the vertical components of the forces to find the normal force. These have to sum to zero (since there is no vertical acceleration). The friction force has a downward vertical component and this, together with gravity, equals the vertical component of the normal force: [tex]mg + \mu_sF_Nsin\theta = F_Ncos\theta[/tex] So: (4)[tex]F_N = mg/(cos\theta  \mu_ssin\theta)[/tex] So substituting into (3): [tex]\frac{mg}{(cos\theta  \mu_ssin\theta)}(sin\theta + \mu_scos\theta) = mv^2/r[/tex] (5)[tex]v = \sqrt{\frac{rg(sin\theta + \mu_scos\theta)}{(cos\theta  \mu_ssin\theta)}[/tex] AM 


#10
Apr305, 06:27 AM

P: 171

Thanks to all for helping out, especially to ehlid and Jd for providing the proof for the trig identity.



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