Forces, friction and car turn radius

[osker246]

Homework Statement

A curve of radius 151 m is banked at an angle of 11°. An 798-kg car negotiates the curve at 94 km/h without skidding. Neglect the effects of air drag and rolling friction. Find the following.

(a) the normal force exerted by the pavement on the tires
(b) the frictional force exerted by the pavement on the tires
(c) the minimum coefficient of static friction between the pavement and the tires

Homework Equations

a=v^2/r

The Attempt at a Solution

Well for some reason which I cannot understand, I simply cannot figure out this problem. I first proceeded by making a freebody diagram, giving me...

F_net= F_N+F_g+F_f=ma

I broke up the diagram into x and y componets...

X-Component

F_{net x}= F_Nx+F_gx+F_fx=ma
F_{net x}= 0 +F_gx+F_fx=ma

Since, a=$$\frac{v^{2}}{r}$$

-mgSin($$\Theta$$)-F_fx=$$\frac{mV^{2}}{r}$$

F_fx= -$$\frac{mV^2}{r}$$ - mgSin($$\Theta$$)

So from here I can have the Friction force, where I can easily fund the friction coefficiant. Now I need to find normal force which I believe is...

F_n=mgcos($$\Theta$$)

Now using these equations to answer the questions above, I for some reason receive wrong answers. I can't figure out where I am going wrong either...any help is appreciated, Thanks.

The answers are

A) 8.37 kN
B)2.03 kN
C)0.244

 

[Alibeg]

It seems to me, that you are using inconsistent coordinate system.

If you say:

a = a_x = v^2/r, that means that you have chosen coordinate system in a such way, that radial inward acceleration is in the same plane as the road (ignore the effect of banking, but try to look at the road as a circle arc - there is only one plane that contains that arc).

In other words, your x-axis is horizontal an y is in the same direction as gravity force (assuming Cartesian coordinate system).

But at the same time, the normal reaction force has no x-component which means that y-axis has to be tilted at 11 degrees (again, assuming Cartesian coordinate system).

So there is a contradiction.


To solve the problem, you should try to position yourself right behind the car so you can see it from behind...
What you see now is: a) cross section of the road and b) taillights of your car.
Now, you can also see the slope of 11 degrees.


Approach I:
Take x-axis to be parallel to the road slope (11 degrees) so y-axis will then coincide with the normal force, BUT acceleration (v^2/r) will have 2 components - then use trigonometry to solve the rest of the task. Also weight will have two components, but friction will have only x-component.
(summary: 2 forces - friction, normal reaction force - that coincide with chosen coordinate system's axes and 2 - weight, acceleration - that have to be divided into components).

Approach II:
Take x-axis to be horizontal and y-axis vertical. This way, acceleration will be completely in x-direction and weight of the car in y. Problem is that normal reaction force and friction will have to be divided into components.
(summary: again 2 forces vs 2 forces)


Either approach should get you to the same result.
Don't forget that after you find the friction force and divide it with normal reaction force, you'll get coefficient of friction (Ffric = Fnorm * coefficient mu).

EDIT: If you are unsure what is the direction of the friction force, just imagine that there is no friction for a moment - and try to analyze in which side the car would drift. If it tended to drift outwards then friction force is directed from the outer side of the road to the inner - to oppose such drift. Or, if car drifted to the inside of the road, the friction force would have opposite direction.

EDIT: One question:

The answers are

A) 8.37 kN
B)2.03 kN
C)0.244

are your or official results?


If you get stuck again, post back.
I hope this helped at least a bit. :D

 

[osker246]

Hi Alibeg,

First off thank you for the lengthy reply! The answers I listed down below are the correct answers to my homework, the answers I am calculating are different.

I am using approach number one as you described. But I did not set acceleration in the y component equal V^2/r, I set it equal to 0. But from what you are saying, then...

Normal force = (mv^2)/r + mgcos($$\Theta$$)
=(798 kg*(26.11 m/s)^2)/151m +(798kg*9.81m/s^2*cos(11)=11279.51 N= 11.27 kN

As you can see this does not match with correct value of 8.37 kN. So I am not sure where I am going wrong...

 

[osker246]

Okay, so I am thinking my FBD is incorrect. Is what I had drawn correct or incorrect?

 

[Alibeg]

Hi. :) Correct.

You need to be careful:
The correct way to solve the problem would be (with your choice of coordinate system) to set these equations:

x axis: sum of all forces in x direction = m * a_x
y axis: sum of all forces in y direction = m * a_y

So:

x axis:
x-component of Fn + x-component of Ffr + x-component of G = m a_x
y axis:
y-component of Fn + y-component of Ffr + y-component of G = m a_y

(Why did I put it this way? Well, acceleration v^2/r corresponds to radial acceleration of a body that has curved path. Since curved path is completely in horizontal plane, so is acceleration) - one note: Road is not entirely in horizontal plane, but car's path really is.

On your picture, acceleration vector would be from car's position to the left. (Don't add the force vector - there is no force, but there is acceleration)

Again we write:
x-axis:
x-component of Fn + x-component of Ffr + x-component of G = m a_x
0 - Ffr - Fg sin11 = m (- v^2 / r) cos11

(Why those minuses? Well, it is because you have oriented the x-axis that way. There is no problem with that, but you must use it that way to the end of the task)

As you can see: a_x = (- v^2 / r) cos11
Because: a = v^2 / r
If you sketch it, you will see what am I talking about.

Multiply the equation with (-1)
Ffr + Fg sin11 = mv^2 cos 11 / r---> (we would get the same if we oriented the x-axis the other way - this is one way to show that there is no need to worry if you orient your axes in a different way from official solution)

Now:
y-axis:
y-component of Fn + y-component of Ffr + y-component of G = m a_y
Fn + 0 - Fg cos11 = m (v^2 / r) sin11
Fn - Fg cos 11 m v^2 sin11 / r

We combine these equations:
Ffr + Fg sin11 = m v^2 cos11 / r (eq.1)
Fn - Fg cos 11 m v^2 sin11 / r (eq.2)
_____________________________
We must add one more equation:
Ffr = N * mu (coefficient of friction) ---> We have to add this equation. We cannot read it from the picture, but it is something that we know, and it is independent of the task (in most cases - sometimes task can explicitly say that Ffr = N * ... * ... and those situations are quite specific - don't worry).

We have three equations with three unknowns - I think you can solve it. If you encounter problems again, feel free to post again. :D

 

[osker246]

Alright! I got it, I guess the hardest part of the problem was actually understanding how acceleration could be broken up into components. But it makes perfect sense now. Thanks a bunch! I appreciate the help.