Resultant and normal force of car on banked curve

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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.
 

Answers and Replies

  • #2
djeitnstine
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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
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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
cjl
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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
djeitnstine
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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
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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
cjl
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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
cjl
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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
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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
cjl
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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
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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
Doc Al
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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
cjl
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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.
 

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