Possible friction force needed for a car on a banked curve

[coneheadceo]

Homework Statement

A 1200kg car rounds a dry curve (μ= .6)with a radius of 67 m banked at an angle of 12°. If the car is traveling @ 95 km/hr (26.4 m/s), will a friction force be required? If so how much and in what direction?

Homework Equations

Fn = mg cos 12°
ƩFn sin 12° = m (v^2/r)
Ffrmax = μFn

The Attempt at a Solution

Seen many different versions of this question asked so here is what I have attempted...

Vertical Force on the car at that angle

Fn = 1200 * 9.8 * cos 12°
= 1.2 x10^4 N

Horizontal force

ƩFn sin 12° = m (v^2/r)
ƩFn sin 12° = 1200 ( (26.4)^2/67)
= 5.8 x 10^4 N

max total static friction force of the
Ffrmax = μFn
= (.6) (1.2x10^4)
= 7200 N

Based on my answers I would say additional force would be needed (5.8 x 10^4 N-7200 N=5.1x10^4) perpendicular to the curve.

Would you guys agree or where have I gone wrong?

 

[PeterO]

Homework Statement

A 1200kg car rounds a dry curve (μ= .6)with a radius of 67 m banked at an angle of 12°. If the car is traveling @ 95 km/hr (26.4 m/s), will a friction force be required? If so how much and in what direction?

Homework Equations

Fn = mg cos 12°
ƩFn sin 12° = m (v^2/r)
Ffrmax = μFn

The Attempt at a Solution

Seen many different versions of this question asked so here is what I have attempted...

Vertical Force on the car at that angle

Fn = 1200 * 9.8 * cos 12°
= 1.2 x10^4 N

Horizontal force

ƩFn sin 12° = m (v^2/r)
ƩFn sin 12° = 1200 ( (26.4)^2/67)
= 5.8 x 10^4 N

max total static friction force of the
Ffrmax = μFn
= (.6) (1.2x10^4)
= 7200 N

Based on my answers I would say additional force would be needed (5.8 x 10^4 N-7200 N=5.1x10^4) perpendicular to the curve.

Would you guys agree or where have I gone wrong?

The Normal force here is what you get if the car was parked on the banking. When driving in the banking, the normal force is bigger.

It is the same effect when comparing parking at the bottom of a dip to driving through that dip.

 

[CAF123]

Banking and turning the corner means you are under the influence of a centripetal force. It is this centripetal force that contributes to the normal force and that is why it is 'bigger'.

 

[PeterO]

Banking and turning the corner means you are under the influence of a centripetal force. It is this centripetal force that contributes to the normal force and that is why it is 'bigger'.

Not quite the way I would word it, but generally yes.

 

[CAF123]

Just out of interest, how would you describe it?

 

[coneheadceo]

So what changes do I need to make to my equations so I address the problem correctly?

 

[CAF123]

This is what I did:
if you set up the coordinate system such that the positive $x$ direction is pointing radially inward and positive $y$ upwards, then I get that;

$\sum F_x = \frac{mv^2}{r},$ which gives $\small Nsinθ = \frac{mv^2}{r}.$
Similarly, $\sum F_y = 0,$ which gives $\small Ncosθ = mg$

The centripetal force required is $F_c = \frac{mv^2}{r},$ so inputting numbers yields $F_c ≈1.2$ x $10^{4} N$.
Now in this coordinate system, with the component of normal force contributing, we see that it provides $≈ 2.5$ x $10^{3} N$, so the rest must come from friction, so that the car can successfully navigate the turn.

 

[coneheadceo]

I somewhat follow the coordinate system rewrite but am lost on where you got
"the component of normal force contributing, we see that it provides ≈2.5 x 10 3 N"

and where do I go from here?

 

[CAF123]

I solved for $N$ first to give $$N = \frac{mg}{cosθ}$$ using the given values. Multiplying this by $sinθ$ gives the contribution of the normal force to the centripetal force.

This means the contribution from friction must be ≈ 1 x 10⁴ N, but according to your calculations the maximum friction force falls short of the required amount. Something is not adding up here.
Do you have an answer from a book and/or does the question say anything sbout the car not making the corner?

 

[coneheadceo]

based on how the question is worded I would assume if the car is going to stay on the road is additional force needed, if it is needed how much and in what direction?

 

[PeterO]

Just out of interest, how would you describe it?

Since the car is on a banked turn it is undergoing centripetal acceleration, so the net force acting is horizontal.
The weight force, mg, is unaltered, so the Normal force has to be bigger to give a net horizontal force.
Since the angle between mg and N is fixed at 12^o there may need to be a friction force parallel to the slope to make the vector sum correct [horizontal and the right size].

It is that friction force [if any] you are after in this problem.

 

[coneheadceo]

and this is one of those where the example in the book is for a flat surface and it does not have a solution in the back.

 

[coneheadceo]

Okay so from all of the posts and help I just need some help making this unconfusing. I'm not looking for the answer outright but if you guys could reorder the correct equations so I know how to work through it, I know I can walk through it on my own. I'm feeling like I have a brain fart that is caught and won't get out over this.

 

[coneheadceo]

Sometimes it takes a day away to clear the head and get things lined up right. Thanks guys for the help.. Got it straigthened out now.

 

[CAF123]

How did you get on in the end?
Was a friction force required?

I ask because I fear my method above may be incomplete - i may have not properly accounted for the friction or something.
How did you solve it?

 

[azizlwl]

Please correct me if it is wrong.

F_NCosθ=mg +F_fSinθ
F_NSinθ +F_fCosθ=mv²/r

 

[CAF123]

Surely what you have written depends on what coordinate system you choose?
How have you set yours up?

 

[CAF123]

I think I realized your coordinate system.
I agree with the first, but for the second, the Nsinθ and F_fcosθ components point rightwards, and we want a centripetal acceleration directed leftwards. I am not entirely sure if this changes things. Maybe another user could clarify things?