mechanics:friction (1 Viewer)

[shazzam]

An engineer is to construct a banked road inclined at an angle alpha to the horizontal, where the radius of the curve is 100m

He bases his calculations on the fact that when a car of mass 1 tonne is travelling at a speed of 54km/h around a curve, the frictional force acting to keep it on the track is 514N

find alpha, the angle of inclination of the plane to the horizontal.

We could not resolve this. How can one find the direction of the friction? When we substituted quantities into the problem we found that the friction , exerted in either direction had a positive value...

 

[shazzam]

okok I would REALLY appreciate it if someone could offer me an answer (with explanation on this one), however even turtle seemed irresponsive. Am I not making my question clear enough? If anyone has any idea or perhaps suspect the conditions to be wrong please enlighten me

thank you thank you

 

[mervvyn]

this is just a stab in the dark -->

have you tried tan@ = (v^2)/rg (or whatever that formula is)
given that friction = u (or the greek symbol mu) and uN = tan@ , @ being the angle of inclination, N being the normal force.

That's just a guess, i don't even know if the maths is exactly right, but i think the theory is, you can work out the direction of the force with a vector diagram

Ok, now i'm confused. I think i'm on the right path, but i'll leave it for tomorrow or someone else. Sorry...

 

[shazzam]

yep mervvyn the answer to the question can be found by resolving forces. however I had a debate about the diretion of the friction with another student.
I thought that the friction force must be directed down the banked track slope, as it is the centripetal force that keeps the car on the banked track. I think I remeber briefly coming across this in physics in the Space topic. Hmm, does anyone disagree/agree/remember this from physics?

 

[Rorix]

The friction acts such to keep the particle in circular motion, so if the particle was too slow and thus inclined to fall into the circle, the friction would be outwards- and if the particle was too fast thus tending to escape the circle, the friction would be inwards

 

[Xayma]

this is just a stab in the dark -->

have you tried tan@ = (v^2)/rg (or whatever that formula is)
given that friction = u (or the greek symbol mu) and uN = tan@ , @ being the angle of inclination, N being the normal force.

That's just a guess, i don't even know if the maths is exactly right, but i think the theory is, you can work out the direction of the force with a vector diagram

Ok, now i'm confused. I think i'm on the right path, but i'll leave it for tomorrow or someone else. Sorry...

You use Friction=μ in maths ext 2?

Because μ is just the coefficient of friction for two surfaces.
F_f=μN

 

[shazzam]

The friction acts such to keep the particle in circular motion, so if the particle was too slow and thus inclined to fall into the circle, the friction would be outwards- and if the particle was too fast thus tending to escape the circle, the friction would be inwards

I realize this now, but I think that you can only use this assumption knowing the angle alpha at which the slope is inclined. However in this question alpha is the quantity which we are trying to find. Therefore I'm not at all sure how that (rorix's) generality could be used.

 

[mervvyn]

You use Friction=μ in maths ext 2?

Because μ is just the coefficient of friction for two surfaces.
F_f=μN

yeah i think so... i can't remember, which is scary, seeing as we only did it a few weeks ago, just before the trials. I think there is a maths way of looking at friction as well as the physics way you described - they are the same but different, i think.
More study required... *sigh*

 

[Xayma]

Yeah Im not suprised there is a different way, I was just suprised at the difference in notation.

I realize this now, but I think that you can only use this assumption knowing the angle alpha at which the slope is inclined. However in this question alpha is the quantity which we are trying to find. Therefore I'm not at all sure how that (rorix's) generality could be used.

No it is true. Friction always opposes the direction of motion, in this case if it is speeding out of the circle friction will be slowing it down/acting down the slope. If it is too slow to maintain it, it will still be slowed down but also the friction will be acting up the slope.

 

[shazzam]

hmmm yes. Sooo... if you wouldn't mind looking back at the original question, how can you know whether this particular car is spinning in or out?

 

[mojako]

weight of car = 1000*g Newtons
this will be the vertical component of Normal force.
so N = 1000g/cos@
horiz component of N = Nsin@ = 1000gtan@
taking inwards as positive,
1000gtan@ + Friction = mv^2/r, where Friction = F can be +ve or -ve at the moment
but you know mv^2/r = 10*(54km/h)^2 (convert that to m/s)
continuing, you'll get
tan@ = [ 10v^2 - F ] / 1000g
suppose F is +ve
if [10v^2 - F ] < 0 then F is negative
because [ 10v^2 - F ] must not < 0 for @ to be positive.

EDIT: if [ 10v^2 - F ] > 0 when we try it with +ve F,
F must be +ve and not -ve because of Rorix's "generality".
The "generality" says that F will be negative (i.e. outwards) for v up to a certain value. If v exceeds that value, F will be positive (inwards).
That "certain value" is when [10v^2 - F ] = 0.
(I think...)