Inclined Plane, Centripetal motion: Find the Velocity that suits the Turn

[Jaccobtw]

Homework Statement

You have a weekend job selecting speed limit signs to put at road curves. The speed limit is determined by the radius of the curve and the bank angle. For a turn of radius 400 m and a 7.0 degree bank angle, what speed limit should you post so that a car traveling at that speed negotiates the turn successfully even when the road is wet and slick?

Relevant Equations

F = ma
a = v^2/r

So here is what is going on in my mind:

We have a turn that is 400m away from the center of the turn. The faster the car goes, the harder it is for it to maintain its radius.

We have a component of the normal force that points towards the center, and static friction which does the same.

I can't identify a force that goes in the opposite direction.

As the car drives too fast, it overcomes the static force.

I need help at this point. Thank you

 

[haruspex]

Relevant Equations: F = ma
a = v^2/r

I can't identify a force that goes in the opposite direction.

Quite so. So it will accelerate, according to your two equations.
You are not given a coefficient of friction. Instead, it just says "wet and slick", as though you are supposed to assume no friction. That's a little unreasonable, but we can leave that for now and just go with that assumption.
Find the horizontal force.

 

[Jaccobtw]

I have the normal force = mgcos(theta). But the key says its mg/cos(theta). How can this be correct?

I have the horizontal force equal to mgcos(theta)sin(theta) = mV^2/ 400

I got a velocity of about 22 m/s

But once again, I'm not sure if I expressed the normal force correctly

The magnitude of the normal force should be less than the gravitational force, making mg the hypotenuse.

 

[haruspex]

I have the normal force = mgcos(theta)

In what direction is the net force if there is to be no slipping?

 

[Jaccobtw]

In what direction is the net force if there is to be no slipping?

The net force is to the left, based on my drawing.

 

[Chestermiller]

Let’s see your free body diagram showing the forces N and mg acting on the car.

 

[Jaccobtw]

Let’s see your free body diagram showing the forces N and mg acting on the car.

The only two forces are N which is perpendicular to the ramp, and mg which is perpendicular to the ground

 

[Chestermiller]

The only two forces are N which is perpendicular to the ramp, and mg which is perpendicular to the ground
View attachment 245819

Good. Now please next express the force balance in the vertical direction.

 

[Jaccobtw]

Σ F(y) = Ncos(theta) - mg = 0

The vertical component of the normal force minus mg = 0

 

[Jaccobtw]

why doesn't using trig work on the bottom triangle, where mg is the hypotenuse, to find the normal force?

 

[TSny]

why doesn't using trig work on the bottom triangle where mg is the hypotenuse to find the normal force?

For the typical problem where a block slides up or down an inclined plane, the acceleration is along the plane. So, there is no component of acceleration perpendicular to the plane. So, the components of the forces that are perpendicular to the plane add to zero. This leads to N = mgcosθ if N and mg are the only forces.

But in this problem, the acceleration direction is not parallel to the plane. So, there is a nonzero component of acceleration perpendicular to the plane. The components of the forces that are perpendicular to the plane do not add to zero. You no longer get N = mgcosθ. The force components that are vertical add to zero, as in your post #9.

 

[Jaccobtw]

But in this problem, the acceleration direction is not parallel to the plane. So, there is a nonzero component of acceleration perpendicular to the plane. The components of the forces that are perpendicular to the plane do not add to zero. You no longer get N = mgcosθ. The force components that are vertical add to zero, as in your post #9.

So the normal force is actually greater than gravity...When usually its the other way around... Is this correct?

 

[TSny]

So the normal force is actually greater than gravity...When usually its the other way around... Is this correct?

Yes. Avoid carrying over a formula that is valid in a specific situation to a different situation. Use your free-body diagram along with what you know about the direction of the acceleration to derive what you need.

 

[Jaccobtw]

Yes. Avoid carrying over a formula that is valid in a specific situation to a different situation. Use your free-body diagram along with what you know about the direction of the acceleration to derive what you need.

Could we talk about why the normal force is greater?

Here is my perspective that may need more light:

The ramp pushes back on the car with a force that increases as the car’s velocity increases. If the car goes too fast, it skids off the road. If it goes too slow, it falls inward (down) toward the “center”.

I’m still confused as to how exactly to force can be derived from velocity. Anyways I don’t really know how accurate this is. I’d deeply appreciate what you guys think

 

[haruspex]

why doesn't using trig work on the bottom triangle, where mg is the hypotenuse, to find the normal force?
...
Could we talk about why the normal force is greater?

Because the resulting acceleration has a component normal to the plane. If you resolve forces in that direction you get $N-mg\cos(\theta)=ma\sin(\theta)$. That was why I asked about the direction of a in post #4.
...
Since the net force is to be horizontal, the normal force must not only balance gravity by itself but also provide the centripetal force.

If the car goes too fast, it skids off the road. If it goes too slow, it falls inward (down) toward the “center”.

Yes, that's what was bothering me in post #2. It doesn’t make practical sense to set the max speed so that it can make the turn with zero friction. A more logical approach is to find the speed at which the friction coefficient required to stop it sliding up the slope is the same as that to stop it sliding down when stationary.