Centripetal Force Banked Curves

  • #1
nick76342
7
0

Homework Statement


This is a conceptual issue which I am trying to understand:

When you are describing a vehicle traveling on a banked curve, force parallel (the force which is found to be parallel to the surface of the road pointing down the bank) is omitted from the FBD and the equations that follow. Yet, force perpendicular still exists. I am not sure why force parallel is omitted since gravity is still acting on the car and should apply some force pulling the car down the hill. I can see where the velocity of the vehicle as it travels along the curve would counter this force, yet I do not understand why force parallel would not be included. In a standard incline plane problem, force parallel dictates an object's motion down the incline. Shouldn't the same apply to a banked curve regardless of the speed of the car, or whether friction is acting on the car?\

Any help would be much appreciated.

Homework Equations


a = v^2 /r

The Attempt at a Solution


Perhaps force parallel is somehow already included in the centripetal force?
 
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  • #2
Hello Nick, :welcome:
nick76342 said:
force parallel is omitted from the FBD
Is it really? Or is it for a specific case ? Do you have an example ?
 
  • #3
BvU said:
Hello Nick, :welcome:
Is it really? Or is it for a specific case ? Do you have an example ?
One example would be:

For a car traveling with speed v around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required.

In this case we are looking to derive the design angle. But how do we account for force parallel. Or is force parallel under the name of the x component of Fn?
 
  • #4
No friction means net force is centripetal force. Gravity and normal force are the only ones that appear in FBD and they add up to centripetal. No need for a parallel component from friction for that particular combination of speed/radius/banking angle
 
  • #5
nick76342 said:
One example would be:

For a car traveling with speed v around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required.

In this case we are looking to derive the design angle. But how do we account for force parallel. Or is force parallel under the name of the x component of Fn?
The only applied force that is purely in the downslope direction would be friction, but there is no friction in this scenario. Since we do not need to find the normal force, I would expect forces in the downslope direction to be considered, and (in the lab frame) these to consist only of the downslope component of gravity. This would equate to the downslope component of the centripetal force.
(In the car's frame, there wouid be the downslope component of centrifugal force, with a net force of zero.)
 
  • #6
BvU said:
No friction means net force is centripetal force. Gravity and normal force are the only ones that appear in FBD and they add up to centripetal. No need for a parallel component from friction for that particular combination of speed/radius/banking angle

I am looking for where gravity is pulling the car down the bank (not that the car is moving down the bank or not). So does that mean that the x component of the normal force is really what I am looking for?
 
  • #7
haruspex said:
The only applied force that is purely in the downslope direction would be friction, but there is no friction in this scenario. Since we do not need to find the normal force, I would expect forces in the downslope direction to be considered, and (in the lab frame) these to consist only of the downslope component of gravity. This would equate to the downslope component of the centripetal force.
(In the car's frame, there wouid be the downslope component of centrifugal force, with a net force of zero.)

So the down slope component is simply a component of the centripetal acceleration?
 
  • #8
nick76342 said:
So the down slope component is simply a component of the centripetal acceleration?
The downslope component of the acceleration is simply a component of the centripetal acceleration; the force that produces it is (in this instance) just the downslope component of gravity.

But I am still unclear about the method you refer to in your original post. Please post the whole method.
 
  • #9
I think I understand this all now. Thank you everyone!
 
  • #10
nick76342 said:
I am looking for where gravity is pulling the car down the bank
nick76342 said:
For a car traveling with speed v around a curve of radius r, determine a formula for the angle at which a road should be banked so that no friction is required.
Could you be clear and complete instead of bringing in pieces of inconsistent problems/examples ?
 
  • #11
For the purposes of clarification:
I did not initially give an example since the question I had did not pertain to the root of any question I had been given in particular, but was my own question in drawing FBDs and wondering where the force parallel had gone which I remembered from working with incline planes. After doing some more problems in addition to your input, I now have a better perception of the relationships of these forces. I cannot further elaborate on any particular question since no question asked for this, just my own curiosity of how forces interact in these types of situations. Sorry for the confusion.
 
  • #12
OK. Thing to remember is normal force is always perpendicular to the surface and gravity is always down :smile:
Friction is sideways and opposes motion, so it can be either direction (up or down) parallel to slope.
 
  • #13
nick76342 said:
For the purposes of clarification:
I did not initially give an example since the question I had did not pertain to the root of any question I had been given in particular, but was my own question in drawing FBDs and wondering where the force parallel had gone which I remembered from working with incline planes. After doing some more problems in addition to your input, I now have a better perception of the relationships of these forces. I cannot further elaborate on any particular question since no question asked for this, just my own curiosity of how forces interact in these types of situations. Sorry for the confusion.
In case its not obvious from earlier replies... You can't always ignore forces parallel with the slope.

Typically a banked curve is designed for one particular speed. At that speed no friction up or down the bank is required. However at other speeds friction is required.

A problem might state the bend is designed for 60mph and ask you to calculate the minimum speed below which you slide down, and the Max speed above which you slide up the bank.
 

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