Check on a basic kinematic problem (FBD of a cabin on a Ferris wheel)

[greg_rack]

Homework Statement

Write and explain the free-body-diagram of the forces acting on a cabin of a Ferris wheel, in a uniform circular motion at the instant in which it's at the same height as the wheel's pivot point.

Relevant Equations

Standard motion equations

Hi guys,
given the "blacker" to be the cabin under consideration,

I firstly wrote its weight force; then, my confusion started when drawing the force applied on the cabin by the structure($F_{r}$).
I concluded it must have been both counter-acting the weight, and acting as a centripetal force... but I'm quite uncertain about it and I'd appreciate knowing if I'm correct or no.

The path I'm taking, for now, is that vertical component $F_{ry}$ must be equal to the weight, while the horizontal $F_{rx}$ to the centripetal force, resulting thus in an inclined force with respect to the standard vertical direction, pointing in the second quadrant of an $xy$ plane.

What do you think about it? Have I made any mistakes or imprecisions?

 

[Steve4Physics]

Homework Statement:: Write and explain the free-body-diagram of the forces acting on a cabin of a Ferris wheel, in a uniform circular motion at the instant in which it's at the same height as the wheel's pivot point.
Relevant Equations:: Standard motion equations

Hi guys,
View attachment 277031given the "blacker" to be the cabin under consideration,

I firstly wrote its weight force; then, my confusion started when drawing the force applied on the cabin by the structure($F_{r}$).
I concluded it must have been both counter-acting the weight, and acting as a centripetal force... but I'm quite uncertain about it and I'd appreciate knowing if I'm correct or no.

The path I'm taking, for now, is that vertical component $F_{ry}$ must be equal to the weight, while the horizontal $F_{rx}$ to the centripetal force, resulting thus in an inclined force with respect to the standard vertical direction, pointing in the second quadrant of an $xy$ plane.

What do you think about it? Have I made any mistakes or imprecisions?

That's good. As far as your wording goes, I'd say $F_{rx}$ IS the centripetal force.

When you draw $F_r$ on your diagram, make sure you get its length/direction reasonably accurate. If drawn correctly, then 'by eye' (using the parallelogram method) you should see the vector-sum of $F_r$ and weight acts horizontally.

[EDIT: typo' corrected.]

 

[kuruman]

And if you were asked to draw the FBD for a passenger in the cabin, you would draw an arrow parallel to the one you described above with a magnitude scaled by the ratio of the passenger's mass to the cabin's mass.

And if the passenger held a box of popcorn, $\dots$ you get the picture.