Banking a Curve without Friction

[logan3]

Homework Statement

A highway curve of radius 80 m is banked at 45°. Suppose that an ice storm hits, and the curve
is effectively frictionless. What is the speed with which to take the curve without tending to
slide either up or down the surface of the road?

r = 80m
\theta = 45^o
g = 9.8m/s^2

Homework Equations

F_c = F_N sin \theta = mv^2/r
F_N = mgcos \theta

The Attempt at a Solution

F_N sin \theta = \frac {mv^2}{r} \Rightarrow (mg cos \theta)sin \theta = \frac {mv^2}{r} \Rightarrow v = \sqrt{rg cos \theta sin \theta}
\Rightarrow \sqrt{(80m)(9.8m/s^2)cos 45^o sin 45^o} = \sqrt{392m^2/s^2} = 19.8m/s

I know the answer's supposed to be 28m/s and I'm missing a tangent somewhere to make it v = \sqrt{rgtan \theta} instead, but I can't figure out where.

Thank-you

 

[ehild]

Homework Statement

A highway curve of radius 80 m is banked at 45°. Suppose that an ice storm hits, and the curve
is effectively frictionless. What is the speed with which to take the curve without tending to
slide either up or down the surface of the road?

r = 80m
\theta = 45^o
g = 9.8m/s^2

Homework Equations

F_c = F_N sin \theta = mv^2/r

F_N = mgcos \theta

The last line is wrong. The normal force here it is not the same as the normal force when sliding on a slope.
In that case, the normal force canceled the normal component of gravity. Here the normal force has to provide the horizontal centripetal force, so the vertical force components have to cancel. Make a drawing! The resultant of the normal force and gravity must be horizontal.

 

[logan3]

I thought the normal force always acted perpendicular to a surface?

 

[ehild]

I thought the normal force always acted perpendicular to a surface?

Yes, it is normal to the surface. But it is not mgcos(theta) always.

 

[logan3]

If the forces have to cancel out to keep the object stationary, then the normal force needs to be equal to the sum of the centripetal force and mg, right? Or is it that since the centripetal force acts inwards, then it must be counteracted by the sum of mg and normal force?

 

[ehild]

If the forces have to cancel out to keep the object stationary, then the normal force needs to be equal to the sum of the centripetal force and mg, right? Or is it that since the centripetal force acts inwards, then it must be counteracted by the sum of mg and normal force?

The car is not in equilibrium, but travels along a horizontal circle. Circular motion requires that the forces applied on the car supply the centripetal force. Draw the free-body diagram for the car.

 

[logan3]

Sorry, this problem just doesn't make sense to me. BTW, I've drawn the FBD many times and even have a reference picture, even before I made this thread. Thanks anyways.

 

[ehild]

Sorry, this problem just doesn't make sense to me. BTW, I've drawn the FBD many times and even have a reference picture, even before I made this thread. Thanks anyways.

can you show your picture?

 

[vela]

If the forces have to cancel out to keep the object stationary, then the normal force needs to be equal to the sum of the centripetal force and mg, right? Or is it that since the centripetal force acts inwards, then it must be counteracted by the sum of mg and normal force?

It's best not to think of the object as being stationary. After all, it's going around in a circle. It's also better to think in terms of centripetal acceleration instead of a centripetal force, which is not a third force acting on the car. There are only two forces on the car, the normal force and its weight. They show up on the lefthand side of $\sum \vec{F} = m\vec{a}$ while the centripetal acceleration appears on the righthand side. Can you write down the equations for the horizontal and vertical directions now?

 

[NTW]

I get the same result as logan3...

The side forces, parallel to the road, are m * g * sin θ and m * v²/r * cos θ

For equilibrium, they must be equal => m * g * sin θ = m * v²/r * cos θ

Solving for v => v = 19,8 m/s

 

[logan3]

I had to step away and think about it for awhile. Actually, I pretty much thought about it non-stop for the past two hours. It helped thinking about how the normal force changed in other problems, such as a box on a horizontal flat surface and a box sliding down an incline.

The components of the normal force are the horizontal and vertical forces -- the horizontal is the centripetal force and the vertical is the force acting opposite the force of weight (mg). Since the only net force is the centripetal force, then the vertical normal force component must cancel out with the force of weight acting down.
F_N cos \theta = mg \Rightarrow F_N = \frac{mg}{cos \theta}

 

[ehild]

I had to step away from everything and think about it in my head for awhile. Actually, I pretty much thought about it non-stop for the past two hours. What helped was thinking about how the normal force changed in other problems, such as a box on a horizontal flat surface and a box sliding down an incline.

The components of the normal force are the horizontal and vertical forces -- the horizontal is the centripetal force and the vertical is the force acting opposite the force of weight (mg). Since the only net force is the centripetal force, then the vertical normal force component must cancel out with the mg acting down, therefore:
F_N cos \theta = mg \Rightarrow F_N = \frac{mg}{cos \theta}

You got it at last!
Here is the free-body diagram:

From the yellow triangle you see that the vector sum of N and mg is equal to the centripetal force Fcp.

 

[NTW]

The weight mg has two components, one normal to the surface, m * g * cosθ, and the other one parallel to the surface, m * g* sinθ...

 

[ehild]

The weight mg has two components, one normal to the surface, m * g * cosθ, and the other one parallel to the surface, m * g* sinθ...

Yes, but you do not want the car sliding downward, but moving horizontally around a circle.

 

[NTW]

Yes, but you do not want the car sliding downward, but moving horizontally around a circle.

Well, I'm contemplating an instantaneous situation, 'a frame of the movie', and the forces parallel to the road are, for the component of the weight, m * g * sin θ, and the component of the centrifugal, 'inertial force', parallel to the road, that may be fictitious, but is valid from my frame of reference is m * (v²/r) * cosθ.

Equating the two, and solving for v, it results v = 19,8 m/s...

 

[vela]

You're making an algebra mistake solving for $v$. If you move the $\cos \theta$ over, you get
$$\left(\frac{mg}{\cos\theta}\right)\sin\theta = m\frac{v^2}{r}$$ which is exactly the same equation logan3 finally got.

 

[NTW]

You're making an algebra mistake solving for $v$. If you move the $\cos \theta$ over, you get
$$\left(\frac{mg}{\cos\theta}\right)\sin\theta = m\frac{v^2}{r}$$ which is exactly the same equation logan3 finally got.

Yes, sorry, you are right. My mistake... Now everything is clear...