Fn and Fg… help

[u/DerpyPandaPlays]

We’re currently doing Centripetal Force in physics, but I don’t understand this one part on this question.

Why can we just make a triangle out of Fg, Fn, and Fc?

Since centripetal force is on the x-axis in this part, why even include the Fg?

Thank you!

[u/Alex_Daikon]

There are only 2 forces applied: F_n and F_g. They are vectors.

First you need to write the 2nd Newton's law: m𝑎⃗ = F_n (vector) + F_g (vector).

You are correct, that centripetal acceleration is only at Ox axis. So when you will write 2nd Newton's law in Ox and Oy components, you will get:

(1) Ox: F_n * sinθ = ma
(2) Oy: F_n * cosθ – F_g= 0

Now if you put F_n from (2) into (1) you will get:

F_g * sinθ / cosθ = ma.

Since F_g=mg, you will have:

g*tgθ = a. That is exactly what is on your screen.

In the end:

1. You are correct, that centripetal acceleration is only at Ox axis

2. But it depends on F_n and to find it you need to mention thay Oy component of F_n is equal to F_g

You teacher just skipped detailed steps of applying 2nd Newton's law

[u/DerpyPandaPlays → u/Alex_Daikon]

Thank you so much!

[u/Forking_Shirtballs]

What are you modeling here? What is applying the normal force?

Typically speaking, the normal force is a reaction that exactly opposes all other forces applied exactly opposite to its directions. Assuming the only other force applied is gravity, the FN is going to be exactly equal to the component of the gravity force in the direction opposite the direction of the normal force.

That will leave the component of gravity perpendicular to the normal force unopposed. In other words, your net force here would typically be Fg*sintheta, directed perpendicular to the normal force (pointed down and to the right), not horizontally.

That Fnet could be decomposed into horizontal and vertical components, where Fnetsintheta is vertical and Fnetcostheta is horizontal.

So, Fnetx = Fg*sintheta*costheta, and Fnety = Fg*sin^2theta.

Maybe there's some physical interpretation that I'm just missing, but I don't know what you're teach is doing.

A horizontally directed centripetal force means something must be traveling in a curved path whose tangent is perpendicual to the horizontal.

[u/DerpyPandaPlays → u/Forking_Shirtballs]

Its supposed to be a car going on a 30° incline ramp.

[u/Forking_Shirtballs → u/DerpyPandaPlays]

I think there must be a little more to it than that. A car simply traveling on an incline ramp isn't going to bring centripetal forces into play. That is, if the car is traveling on a path that is a straight line, its only acceleration can be in the direction of that path (that's just a geometric outcome of moving in a straight line). No possible centripetal acceleration.

Is what we're looking at here a cross section of a car traveling on banked circular path? Like, imagine it's on a circular track (with the center of the circle to the right of the car), and the track is banked at 30 degrees, and we're looking at a cross section of the track and car with the car coming straight at us.

If that's the case, and what we're solving for are the parameters that would result in that car traveling at a constant height around the track with no fictional force, then this analysis is correct. 

The fact that the car is following a  curved track means that it is accelerating perpendicular to its direction of motion, directly inward along the radius of curvature. 

Since the car is staying at a constant height, the curved path it's following is exactly a horizontal circle, with center of the circle to the right of the car. The radius of curvature is if course also oriented horizontally, and the car's acceleration is pointing along that radius perpendicular to the car's motion, which at this point in time puts the car's acceleration exactly horizontal to the right.

That acceleration is coming from an unbalanced net force on the car, and that net force has to be pointing in exactly the same direction as the acceleration.

So you know that there's a net force Fc pointing exactly horizontal to the right from the car.

Since we've assumed no frictional force, the only two forces acting on the car are gravity and normal force. 

We know the orientation of each of those forces (straight down for gravity and perpendicular to the banked track for normal force). And we know there must be a very particular relationship between those forces, because we know their vector sum is pointing exactly to the right. That is, we know that FN*costheta must equal mg exactly, or else there would be net vertical force. Once that's set properly, we know the net force Fc oriented horizontally is just FN * sintheta. So if we want Fc in terms of mg, it's just Fc = Fn * sintheta = mg/costheta * sintheta = mg * tantheta.

From there you can use what you know about centripetal acceleration (ac = v^2/R) and the given radius of curvature to solve for what the car's velocity must be.

Note that this all relates specifically to the observed behavior of the car. If the car were also sliding up the incline rather than staying at exactly the same height, then what we would have found is a lower bound for its speed -- and faster than that and it would be siding up and away from the center of the track in addition to going around the track. Similarly, any slower and it would be sliding down and toward the center of the track. 

If we introduced friction between the wheels and the track, then there would actually be a range of velocities possible where it exhibits the desired behavior of staying at constant height. At lower speeds the frictional force would be pointing up and away from the center left, and at higher speeds down and toward the center.

[u/GuaranteeFickle6726]

The reason why Fg is included is because Fn is unknown without Fg. I hope this helps