Circular motion and gravitation

[Inertialforce]

Homework Statement

Traffic is expected to move around a curve of radius 200m at 90km/h. What should be the value of the banking angle if no dependence is to be placed on friction?

Homework Equations

ΣFc = mac

The Attempt at a Solution

For this question what I did first was draw a top view of the situation, then I drew a side view of the same situation but with vectors to show what forces were acting on the car as it was going through the bank (free body diagram). As it turns out (since it says that no dependence is placed on friction) the only forces acting on the car along the bank are the forces Fn and Fg.

My question is when I put it into the centripetal force equation, what force do I incorporate into the equation?

For example is it:
ΣFc = mac
Fnsin(theta) = mac

or is it:
ΣFnsin(theta) + mgsin(theta) = mac

 

[alphysicist]

Hi Inertialforce,

Homework Statement

Traffic is expected to move around a curve of radius 200m at 90km/h. What should be the value of the banking angle if no dependence is to be placed on friction?

Homework Equations

ΣFc = mac

The Attempt at a Solution

For this question what I did first was draw a top view of the situation, then I drew a side view of the same situation but with vectors to show what forces were acting on the car as it was going through the bank (free body diagram). As it turns out (since it says that no dependence is placed on friction) the only forces acting on the car along the bank are the forces Fn and Fg.

My question is when I put it into the centripetal force equation, what force do I incorporate into the equation?

For example is it:
ΣFc = mac
Fnsin(theta) = mac

or is it:
ΣFnsin(theta) + mgsin(theta) = mac

First determine what direction the centripetal acceleration a_c is in. Then, in your centripetal force equation, you find the components of all forces in that direction; those components go in the centripetal force equation. What do you get?

(That is, what is the component of the normal force in the direction a_c is pointing, and the component of the gravity force in the direction a_c is pointing?)

 

[Inertialforce]

Hi Inertialforce,



First determine what direction the centripetal acceleration a_c is in. Then, in your centripetal force equation, you find the components of all forces in that direction; those components go in the centripetal force equation. What do you get?

(That is, what is the component of the normal force in the direction a_c is pointing, and the component of the gravity force in the direction a_c is pointing?)

In my free body diagram I have Fnsin(theta) and mgsin(theta) pointing towards the direction of my centripetal acceleration. But what I do not understand is that, I know I am supposed to write down all the forces pointing in the direction of the centripetal acceleration in my equation which would make:

ΣFc = mac
Fnsin(theta) + mgsin(theta) (because they are both pointing left to the direction of the centripetal acceleration)

However, in my physics textbook they have a example that is almost identical but in their free body diagram they only showed or expanded upon the forces of Fn (the free body diagram only includes the vectors Fn, Fncos(theta), Fnsin(theta) , and mg. I was just wondering why they do not include the cos and sin vectors for mg as well and also why they do not include mgsin(theta) or mgcos(theta) in the banked curve examples/questions that they show.

 

[alphysicist]

In my free body diagram I have Fnsin(theta) and mgsin(theta) pointing towards the direction of my centripetal acceleration. But what I do not understand is that, I know I am supposed to write down all the forces pointing in the direction of the centripetal acceleration in my equation which would make:

ΣFc = mac
Fnsin(theta) + mgsin(theta) (because they are both pointing left to the direction of the centripetal acceleration)

Here is the important part: the centripetal acceleration is to the left, so it is a horizontal acceleration. But the gravity force has no component in the horizontal direction; it is a vertical force.


(What confuses many people is the similarity in the diagrams for these problems and for problems involving boxes sliding down inclined planes. But the problems are different in that for the boxes, the acceleration is down the incline, and gravity has a component parallel to the incline.

But for these banked curve problems, the acceleration is horizontal, and gravity has no horizontal component.)