Resultant and normal force of car on banked curve

In summary, the conversation discussed the concepts of design speed and normal force in a car on a banked curve. The normal force was found using the equation n = mg / cos\theta and the design speed was defined as the speed at which a car no longer relies on friction to stay on the bank. The relationship between design speed, corner radius, and bank angle was also explored through the use of trigonometric functions. The equation tan\theta = v^2 / rg was derived by considering the vertical and horizontal force components on the car.
  • #1
diffusion
73
0

Homework Statement


A car is on a banked curve of bank angle 18 deg, radius 70m. It's mass is 1020kg. Draw a free body force diagram approximately to scale, when the car is at the 'design speed'. What is meant by design speed? Find the resultant force and the normal force n.


Homework Equations


ncos[tex]\theta[/tex] = mg
Design speed = Speed at which a car no longer relies on friction in order to keep from sliding down the bank.

The Attempt at a Solution


To find the normal force, I think I am supposed to use the equation ncos[tex]\theta[/tex] = mg, or solving for n: n = mg / cos[tex]\theta[/tex]. Plugging in the values I get n = 1020kg x 9.81/ cos18, thus n = 10521 N. Is this correct? I'm poor with trig functions, so I'm not sure how or why cos relates to this equation. If someone could explain this that would be great, as I'm truly trying to understand this material.

Would the resultant force be = 0? If the car is at a constant "design speed", it is not accelerating, thus no net force.
 
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  • #2
The normal force 'Fn' is [tex]F_n=mg \cos{\theta}[/tex]. The concept here is of centripetal force. What force does the car experience when going around in a circle? Also what force is acting against it?
 
  • #3
djeitnstine said:
The normal force 'Fn' is [tex]F_n=mg \cos{\theta}[/tex]. The concept here is of centripetal force. What force does the car experience when going around in a circle? Also what force is acting against it?

Oh, so even though the speed doesn't vary, there is still acceleration because v is changing direction.

Can you tell me where the cos comes from?
 
  • #4
Actually, the normal force is not simply mgcos(theta). That is one component of it. Your solution, with n=mg/cos(theta) is correct.

Now, as for the derivation of this, it is somewhat more complicated. The first thing to understand is what is meant by the design speed. Do you understand that part?
 
  • #5
cjl said:
Actually, the normal force is not simply mgcos(theta). That is one component of it. Your solution, with n=mg/cos(theta) is correct.

Now, as for the derivation of this, it is somewhat more complicated. The first thing to understand is what is meant by the design speed. Do you understand that part?

Oh so we are shifting the coordinate system. Sorry for my erroneous remark.
 
  • #6
cjl said:
Actually, the normal force is not simply mgcos(theta). That is one component of it. Your solution, with n=mg/cos(theta) is correct.

Now, as for the derivation of this, it is somewhat more complicated. The first thing to understand is what is meant by the design speed. Do you understand that part?

The design speed is the speed of the car where it no longer relies on friction to keep it on the bank, yes? Should be constant as long as [tex]\theta[/tex] remains unchanged.
 
  • #7
djeitnstine said:
Oh so we are shifting the coordinate system. Sorry for my erroneous remark.

Well, in every physics problem I've ever seen, normal force is designed as normal to the surface. Your equation is correct for the component of normal force from gravity, but it is incorrect for the total normal force, which has another component as well.
 
  • #8
diffusion said:
The design speed is the speed of the car where it no longer relies on friction to keep it on the bank, yes? Should be constant as long as [tex]\theta[/tex] remains unchanged.

Correct. Design speed is a function of [tex]\theta[/tex], as well as of corner radius (which is a constant for this problem)
From that deduction, what can you say about the relationship between V, R and theta?
 
  • #9
cjl said:
Correct. Design speed is a function of [tex]\theta[/tex], as well as of corner radius (which is a constant for this problem)
From that deduction, what can you say about the relationship between V, R and theta?

Well, I know that tan[tex]\theta[/tex] = v^2 / rg, I just don't know why or where the tan comes from. Not sure if this is what you are referring to.
 
  • #10
diffusion said:
Well, I know that tan[tex]\theta[/tex] = v^2 / rg, I just don't know why or where the tan comes from. Not sure if this is what you are referring to.
That is what I am referring to, but you do have to understand where the equation comes from. What are the two main forces that occur in this problem? How are they related? Once you understand this relationship, the tangent becomes obvious. If you can't seem to get it, draw a free body diagram of the car.

Note: Also, for this work, think of trig functions as the triangle side definitions. I.E., think of sin(x) as the opposite side of the right triangle divided by the hypotenuse.
 
  • #11
cjl said:
That is what I am referring to, but you do have to understand where the equation comes from. What are the two main forces that occur in this problem? How are they related? Once you understand this relationship, the tangent becomes obvious. If you can't seem to get it, draw a free body diagram of the car.

Note: Also, for this work, think of trig functions as the triangle side definitions. I.E., think of sin(x) as the opposite side of the right triangle divided by the hypotenuse.

I think I see what you're saying. Forgive me for my difficulty, I'm in a college physics course which assumes experience with high school physics; I have none. :(

So I drew the following two diagrams. Radius was easy to locate, direction of velocity would be tangent to the curve where the car is located, so there's my opposite and adjacent. Seems to make sense.

car2.jpg


What confuses me now is where does the g come from in the equation tan[tex]\theta[/tex] = v^2 / rg. If I didn't already know the equation, by looking at the diagrams here I would have put: tan[tex]\theta[/tex] = v / r
 
  • #12
diffusion said:
What confuses me now is where does the g come from in the equation tan[tex]\theta[/tex] = v^2 / rg. If I didn't already know the equation, by looking at the diagrams here I would have put: tan[tex]\theta[/tex] = v / r
Derive the equation by considering vertical and horizontal force components. Write Newton's 2nd law for each direction, then combine the two equations.
 
  • #13
diffusion said:
I think I see what you're saying. Forgive me for my difficulty, I'm in a college physics course which assumes experience with high school physics; I have none. :(

So I drew the following two diagrams. Radius was easy to locate, direction of velocity would be tangent to the curve where the car is located, so there's my opposite and adjacent. Seems to make sense.

car2.jpg


What confuses me now is where does the g come from in the equation tan[tex]\theta[/tex] = v^2 / rg. If I didn't already know the equation, by looking at the diagrams here I would have put: tan[tex]\theta[/tex] = v / r

You have the right idea, but are applying it to the wrong triangle.

As Doc Al said, consider vertical (which you already have, with the mg arrow in your diagram) and horizontal forces acting on the car.
 

1. What is a banked curve?

A banked curve is a type of curved road or track that is designed with a slope or banked angle to assist vehicles in making turns at high speeds. This slope helps to counteract the centrifugal force acting on the vehicle and allows it to make the turn safely.

2. What is the resultant force on a car on a banked curve?

The resultant force on a car on a banked curve is the combination of two forces: the normal force and the centripetal force. The normal force is the force exerted by the road surface on the car, perpendicular to the surface. The centripetal force is the force that keeps the car moving in a circular path.

3. How does the banked angle affect the resultant force on a car?

The banked angle of a curve affects the resultant force on a car by changing the balance between the normal force and the centripetal force. If the banked angle is too steep, the centripetal force may exceed the normal force, causing the car to slip or skid. If the banked angle is too shallow, the normal force may exceed the centripetal force, causing the car to roll outward.

4. What factors influence the magnitude of the normal force on a car on a banked curve?

The magnitude of the normal force on a car on a banked curve is influenced by several factors such as the speed of the car, the mass of the car, the banked angle of the curve, and the coefficient of friction between the tires and the road surface. A higher speed or a heavier car will result in a larger normal force, while a steeper banked angle or a lower coefficient of friction will decrease the normal force.

5. What happens to the resultant force on a car if the banked angle is increased?

If the banked angle of a curve is increased, the resultant force on a car will also increase. This is because the steeper banked angle will provide more support to the car and increase the normal force, allowing the car to make the turn at a higher speed. However, if the banked angle is too steep, the car may experience a larger centripetal force and become unstable, causing it to slip or skid.

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