# Banked Surface Problem

• Mar 31st 2010, 05:40 AM
craig
Banked Surface Problem
Quote:

Let $\displaystyle g = 9.8 ms^{-1}$

A car of mass $\displaystyle m = 1000kg$ travels on a circular circuit of radius $\displaystyle 200m$, (the circuit is rough). The circuit is banked at an angle of $\displaystyle \theta = \frac{\pi}{6}$ to the horizontal to try to reduce the chance of the car skidding outwards.

1. Find the coefficient of friction between the car and the circuit if the car starts to slip inwards when it's travelling at a speed of $\displaystyle 80 km h^{-1}$.

2. Find the magnitude of the normal reaction force on the car when it's travelling at $\displaystyle 80 km h^{-1}$.
I'm not too good on circular motion questions so please bare with me ;), here's my attempt.

Sort a few things out first:

$\displaystyle 80 km h^{-1} = \frac{200}{9}ms^{-1}$
In case we need it later, angular speed, $\displaystyle \omega = \frac{v}{r} = \frac{200}{9 \times 200} = \frac{1}{9}$

At $\displaystyle 80 km h^{-1}$ the car starts to slip inwards, so this means that the friction is acting up the slope, and that the friction is also equal to $\displaystyle \mu N$ where $\displaystyle N$ is the normal reaction of the slope on the car?

Resolving vertically I get $\displaystyle N = mg\cos{\theta}$

Resolving horizontally I get $\displaystyle F = mg\sin{\theta}$

We can rewrite the second equation as $\displaystyle \mu N = mg\sin{\theta}$, and substituting in our first equation we get $\displaystyle \mu mg\cos{\theta} = mg\sin{\theta}$, therefore $\displaystyle \mu = \tan{\theta} = \frac{\sqrt{3}}{3}$?

I have a feeling that Newton's $\displaystyle F = ma$ needs to be used but I'm not sure where?

Not sure how to approach the second question either, this has me a bit stumped.

Thanks in advance for the help
• Apr 1st 2010, 01:47 PM
xalk
Quote:

Originally Posted by craig
I'm not too good on circular motion questions so please bare with me ;), here's my attempt.

Sort a few things out first:

$\displaystyle 80 km h^{-1} = \frac{200}{9}ms^{-1}$
In case we need it later, angular speed, $\displaystyle \omega = \frac{v}{r} = \frac{200}{9 \times 200} = \frac{1}{9}$

At $\displaystyle 80 km h^{-1}$ the car starts to slip inwards, so this means that the friction is acting up the slope, and that the friction is also equal to $\displaystyle \mu N$ where $\displaystyle N$ is the normal reaction of the slope on the car?

Resolving vertically I get $\displaystyle N = mg\cos{\theta}$

Resolving horizontally I get $\displaystyle F = mg\sin{\theta}$

We can rewrite the second equation as $\displaystyle \mu N = mg\sin{\theta}$, and substituting in our first equation we get $\displaystyle \mu mg\cos{\theta} = mg\sin{\theta}$, therefore $\displaystyle \mu = \tan{\theta} = \frac{\sqrt{3}}{3}$?

I have a feeling that Newton's $\displaystyle F = ma$ needs to be used but I'm not sure where?

Not sure how to approach the second question either, this has me a bit stumped.

Thanks in advance for the help

How about the force acting on the car due to its cilcular motion??
• Apr 2nd 2010, 02:13 AM
craig
Quote:

Originally Posted by xalk
How about the force acting on the car due to its cilcular motion??

Hi thanks for the reply, how would I include this in my equations?

Thanks again
• Apr 2nd 2010, 05:34 AM
Haytham
vertical $\displaystyle N=mg\cos{\theta}+m\frac{v^{2}}{r}\sin{\theta}$ ------------------------------------------------------------------------------------------------------------------------------------ ----------- horizontal $\displaystyle F=mg\sin{\theta}-m\frac{v^{2}}{r}\cos{\theta}$
• Apr 2nd 2010, 06:52 AM
xalk
Yes my calculations agree with those of Haytham if by F he means friction
• Apr 2nd 2010, 07:29 AM
craig
Thanks for both replies, not got access to a pen & paper at the moment but will attempt the solution as soon as I do ;)
• Apr 2nd 2010, 08:08 AM
craig
Quote:

Originally Posted by Haytham
vertical
$\displaystyle N=mg\cos{\theta}+m\frac{v^{2}}{r}\sin{\theta}$

horizontal
$\displaystyle F=mg\sin{\theta}-m\frac{v^{2}}{r}\cos{\theta}$

So because the car starts to slip inwards, it's in limiting fraction, so am I right in saying that $\displaystyle F = \mu N$?

If so, then $\displaystyle \mu(mg\cos{\theta}+m\frac{v^{2}}{r}\sin{\theta}) = mg\sin{\theta}-m\frac{v^{2}}{r}\cos{\theta}$

So $\displaystyle \mu = \frac{g\sin{\theta}-\frac{v^{2}}{r}\cos{\theta}}{g\cos{\theta}+\frac{v ^{2}}{r}\sin{\theta}} = \frac{\frac{1}{2}g-\frac{200}{81}\cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}g+\frac{200} {81}\cdot \frac{1}{2}} \approx 0.284$?

Also for the second part of the question do I just calculate the value of the normal force?

Thanks again for all the help.
• Apr 2nd 2010, 08:12 AM
craig
Quote:

Originally Posted by Haytham
vertical $\displaystyle N=mg\cos{\theta}+m\frac{v^{2}}{r}\sin{\theta}$ ------------------------------------------------------------------------------------------------------------------------------------ ----------- horizontal $\displaystyle F=mg\sin{\theta}-m\frac{v^{2}}{r}\cos{\theta}$

Sorry could I just ask how you got the $\displaystyle +m\frac{v^{2}}{r}\sin{\theta}$ and the $\displaystyle -m\frac{v^{2}}{r}\cos{\theta}$?

I know that the $\displaystyle \frac{v^2}{r}$ is the acceleration, but how did you go about including these in your equations?

Thanks again for all the help.
• Apr 2nd 2010, 09:13 AM
Haytham
You have 3 forces, weight, friction and the reaction.

You chose this coordinate system
http://i270.photobucket.com/albums/j...Screenshot.png

As the movement is circular you know there is an inward net force represented in red (its the sum of the 3 forces) and its value is $\displaystyle F_{net}=m\frac{v^{2}}{r}$

now, you just decompose the forces in the axis

xx axis
$\displaystyle F_x=-\mu N$
$\displaystyle W_x=mg\sin{\theta}$
$\displaystyle F_{net/x}=m\frac{v^{2}}{r}\cos{\theta}$

yy axis
$\displaystyle N_y=N$
$\displaystyle W_y=-mg\cos{\theta}$
$\displaystyle F_{net/y}=m\frac{v^{2}}{r}\sin{\theta}$

and the sum of the 3 forces must be equal to the net force

xx axis
$\displaystyle m\frac{v^{2}}{r}\cos{\theta}=mg\sin{\theta}-\mu N$

yy axis
$\displaystyle m\frac{v^{2}}{r}\sin{\theta}=N-mg\cos{\theta}$

if you wish to, you could try to solve it again using a different set of coordinates. vertical (parallel to the W) and horizontal (parallel to the net force), i find it simpler.

Quote:

Also for the second part of the question do I just calculate the value of the normal force?
yes
• Apr 2nd 2010, 09:16 AM
craig
Thank you so much for that! Really helps, thankyou.