Centripetal acceleration speeding car problem

[Dumbledore211]

Homework Statement

A car is traveling round a bend which is banked at an angle of 30 to the horizontal. The bend is assumed to be in the shape of an arc of a circle of radius 80m. the surface of the road is rough and the coefficient of friction between the tyres of the and the surface of the road is 0.3. Find the greatest speed and the least speed without slipping occurring

Homework Equations

Cp= v^2/r

The Attempt at a Solution

So, here the bend is assumed to be an arc of a circle which is why the angle θ=30 is assumed to be formed at the center of the circle. The centripetal acceleration is also caused by the maximum frictional force which has a coefficient of 0.3. The equation that best describes the motion of the car is as follows frictional force - mgcosθ= mv^2/r
or, fs-mgcos=mv^2/r
or, μR- mgcosθ=mv^2/r
What befuddles me about this problem is the value of R which is supposed to be the normal reaction force. How do I go about solving for R?? Can any of you guys drop a hint as to where I am going wrong.??

 

[mfb]

All your equations look strange. Did you mix horizontal and vertical forces there?

How did you define R?

 

[haruspex]

So, here the bend is assumed to be an arc of a circle which is why the angle θ=30 is assumed to be formed at the center of the circle.

No, the arc is of unknown length. The arc is in the horizontal plane. The 30 degrees is the angle that the road is banked laterally, at all points along the arc.

The centripetal acceleration is also caused by the maximum frictional force which has a coefficient of 0.3.

That would be true on an unbanked road, but here the normal force also supplies some centripetal force. Correspondingly, only the horizontal component of the frictional force contributes to the centripetal force.
Write out the ∑F=ma equation for both vertical and horizontal. Don't forget that the frictional force can act up the slope, not just down the slope.

 

[Dumbledore211]

The 30 degrees is actually the angle created by the arc with the horizontal plane. So, the vertical force in this case is the normal force i.e R=mgcos30 and the horizontal component of the forces in this case are as follows mgsin30-fs=mv^2/r
or, mgsin30-μR=mv^2/r
or, mgsin30-0.3mgcos30=mv^2/r
or, gsin30-0.3gcos30=v^2/r
Are my set of equations correct?

 

[haruspex]

The 30 degrees is actually the angle created by the arc with the horizontal plane.

No, I believe you are still misreading it. The arc is horizontal, but the road surface slopes from one side to the other at an angle of 30 degrees. That's what 'banked' means.

 

[Andrew Mason]

I noticed recently that people seem to be using R for the normal force instead of the usual N. I don't know where this new convention started. First of all R is also the gas constant, so it can be confusing in some cases. Second, if R is intended to refer to the "Reaction" force to the body's weight and/or an outward force, it introduces a new level of confusion to Newton's third law. It is really just the perpendicular component of a stress in the surface that is equal and opposite to the stress in the body that is in contact with the surface.

AM

 

[Dumbledore211]

@Andrew mason It is just not me. Almost all my textbooks use R to denote the normal reaction force

 

[Andrew Mason]

@Andrew mason It is just not me. Almost all my textbooks use R to denote the normal reaction force

By calling it a "reaction" force they are referring to another force (ie. its Third Law pair). It can be very complicated to determine what, exactly, it is a "reaction" to. It is not a third law pair to the weight of the body, for example.

All forces are third law "reaction" forces so singling out this as R "the Normal reaction force" is misleading and confusing at best.

AM

 

[Andrew Mason]

The 30 degrees is actually the angle created by the arc with the horizontal plane. So, the vertical force in this case is the normal force i.e R=mgcos30 and the horizontal component of the forces in this case are as follows mgsin30-fs=mv^2/r
or, mgsin30-μR=mv^2/r
or, mgsin30-0.3mgcos30=mv^2/r
or, gsin30-0.3gcos30=v^2/r
Are my set of equations correct?

What does the sum of all forces (friction, normal force and gravity) have to equal?

AM