Person sitting on scale on roller coaster

Click For Summary
SUMMARY

The discussion revolves around the physics of a 120-pound person sitting on a scale during a roller coaster ride, specifically analyzing forces acting on the person as the coaster navigates a curve with a 30-foot radius. The scale indicates a force of 25 pounds, prompting questions about the gravitational force (Fg), normal force (Fn), and the acceleration of the person. Participants clarify that the centripetal force equation (a = mv²/r) is applicable, and emphasize that friction is negligible in this scenario. The conversation concludes with the need to determine the roller coaster's velocity using the known radius and forces.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with centripetal acceleration and forces
  • Ability to draw and interpret free body diagrams
  • Knowledge of basic physics concepts such as weight and normal force
NEXT STEPS
  • Learn how to calculate centripetal acceleration using the formula a = v²/r
  • Study the relationship between gravitational force and normal force in circular motion
  • Explore the concept of free body diagrams in detail
  • Investigate the effects of friction in rolling motion versus sliding motion
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the dynamics of forces in circular motion.

cdornz
Messages
40
Reaction score
0

Homework Statement



A 120 pound person is sitting on a scale during a roller coaster ride. As the roller coaster goes over the top of a curve whose radius is 30 feet, the scale indicates a force of 25 pounds. The average speed of the roller coaster doesn't change while over this curve.

A) What forces are acting on the person?
B) Specify the magnitude and direction of each of the forces
C) Draw a free body diagram of the person
D) Describe the acceleration of the person
E) How fast was the roller coaster moving at that time?

Homework Equations



I know that a=mv(squared)/r : This equation goes into the right side of the ƩF=ma.
I know how to draw a free body diagram, start with the object at the center and draw the forces outward.

The Attempt at a Solution



A) I believed the forces to be: Fg=weight of person; Fn=force perpendicular to interface; Ff=mv(squared)/r
B) Fg=120pd, down; Fn=up, don't know magnitude; Ff=(120)v(squared) and I believe this is pointing inward, but I'm kind of confused by that.
C) This I am able to draw if I know the correct direction of the forces.
D) I have no idea how to get the acceleration, but I do have the equation. a=v(squared/r). Problem is, I only have the radius so I don't know how to go about solving for acceleration.
E) This also means I don't understand where to start to figure out the velocity of the roller coaster.
 

Attachments

  • 111.JPG
    111.JPG
    57.9 KB · Views: 523
Physics news on Phys.org
welcome to pf!

hi cdornz! welcome to pf! :smile:

(try using the X2 button just above the Reply box :wink:)
cdornz said:
A) What forces are acting on the person?
B) Specify the magnitude and direction of each of the forces
C) Draw a free body diagram of the person
D) Describe the acceleration of the person
E) How fast was the roller coaster moving at that time?

I know that a=mv(squared)/r : This equation goes into the right side of the ƩF=ma.

A) I believed the forces to be: Fg=weight of person; Fn=force perpendicular to interface; Ff=mv(squared)/r

what is "Ff" supposed to be an abbreviation for? :confused:

mv2/r is not a force, it's m times acceleration, and as you say, it goes on the RHS of ∑F = ma
B) Fg=120pd, down; Fn=up, don't know magnitude; Ff=(120)v(squared) and I believe this is pointing inward, but I'm kind of confused by that.

yes, you do know the magnitude of Fn: read the question!

(where does Fn come from?)
D) I have no idea how to get the acceleration, but I do have the equation. a=v(squared/r). Problem is, I only have the radius so I don't know how to go about solving for acceleration.

i don't understand :confused: … r is the radius
 


tiny-tim said:
what is "Ff" supposed to be an abbreviation for? :confused:

mv2/r is not a force, it's m times acceleration, and as you say, it goes on the RHS of ∑F = ma

yes, you do know the magnitude of Fn: read the question!

(where does Fn come from?)


i don't understand :confused: … r is the radius

Ff is supposed to represent force of friction, which I thought there would be since this is technically steel on steel.

As for the magnitude of Fn, would it be 145 pounds? I would assume that is the amount, because that's the amount of force pushing down..I wasn't sure if Fg and Fn would be equal or not.

As for acceleration, radius isn't needed, but I wasn't sure with the correct equation if I could solve for it.
 
cdornz said:
Ff is supposed to represent force of friction, which I thought there would be since this is technically steel on steel.

it's rolling … this is a rollercoaster

there's no friction (with the ground or the rail) when something rolls!

(and anyway why would it equal mv2/r ?)
As for the magnitude of Fn, would it be 145 pounds? I would assume that is the amount, because that's the amount of force pushing down..I wasn't sure if Fg and Fn would be equal or not.

why would it be more than the weight? :confused:

anyway, what object is this reaction force coming from?
As for acceleration, radius isn't needed, but I wasn't sure with the correct equation if I could solve for it.

i honestly have no idea what you're talking about :redface:

what equation do you propose to use for the acceleration?​
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
1K