Is the normal force at the top of a rollercoaster loop always directed upwards?

[mancity]

Homework Statement

why is the normal force at the top of a rollercoaster loop the same direction as the force of gravity?

Relevant Equations

f_n+f_g=f_c

 

[Halc]

why is the normal force at the top of a rollercoaster loop the same direction as the force of gravity?

Coincidence that they are depicted nearly equal, but the gravity vector is a function only of mass (and distance from Earth if it's a really tall loop), whereas the normal force is a function of mass, loop radius, and the speed at which the loop is taken.

That the normal vector at top and bottom appear similar implies implausibly that the speed is the same at top and bottom.
That the sum of the two component vectors in opposite direction add up to a larger magnitude vector is just wrong.

I also have issues with a_felt since one does not 'feel' gravitational acceleration. The guys in the ISS feel weightless despite having a coordinate acceleration in the frame of Earth.

 

[jbriggs444]

View attachment 332452

The left hand side of that drawing seems misleading. The sum of two oppositely directed vectors should not have a larger magnitude than either.

 

[Lnewqban]

Homework Statement: why is the normal force at the top of a rollercoaster loop the same direction as the force of gravity?
Relevant Equations: f_n+f_g=f_c

Welcome, @mancity !

The direction of any normal force is always perpendicular to the surface of contact between two bodies.
Therefore, in the case of the shown circle, the direction of the normal force at any point should be pointing towards or away the center of the geometrical circle.

The diagram represents the external forces acting on the car; hence, the normal shows the force with which the track is pushing against the wheels of the car: pointing always to the center of the loop.

Please, see:
https://www.physicsclassroom.com/class/circles/Lesson-2/Amusement-Park-Physics

 

[Cutter Ketch]

I believe the idea they are trying to convey is that the coaster is moving so fast and the curvature of the top of the hill is so tight that it takes more than just the force of gravity to keep the train on the tracks. The required v^2/r is greater than g, so there must be another downward force.

In those cases the rider feels negative g’s. (This is one of the things love about modern rollercoasters, so it’s very common). You float up in your seat and the only thing that keeps you from flying out of the car is the shoulder harness holding you down. More properly from the point of view of physics, the force of gravity isn’t enough to keep you following the train around the tightly curved top of the hill and you need an extra force holding you down.

Well, in those moments when you need an extra downward force to stay in the car it must also be true that the car needs an extra downward force to stay on the track. How do they manage that? If you look at the wheels of a modern steel coaster, you’ll see that they don’t just have wheels on top of the track. The wheels are in a bracket that has wheels on the top, sides, and bottom of the tubular steel track. That means that the wheels can hold down as well as hold up. The riders can experience negative Gs, and in those cases the presence of the underside wheels allow the track to provide a downward normal force so the train stays on the tracks.

 

[haruspex]

Coincidence that they are depicted nearly equal

@mancity did not ask about magnitudes, only directions.

why is the normal force at the top of a rollercoaster loop the same direction as the force of gravity?

Two bodies in contact exert equal and opposite repulsive forces on each other. The components of these two forces which are normal to the plane of contact are called the normal forces.
At the top of the loop, the track is above the cart, so a repulsive force exerted by the track on the cart is necessarily downwards, making it the same direction as gravity.

Did you think that the normal force on a body always acts upwards? I have seen that misunderstanding before.