Unbanked and banked Curves problem

  • Thread starter Thread starter dvela
  • Start date Start date
  • Tags Tags
    Curves
dvela
Messages
4
Reaction score
0
1. A roller coaster of mass 320 kg (including passengers) travels around a horizontal curve of radius 35 m. Its speed is 16 m/s. What is the magnitude and direction of the total force exerted on the car by the track?



Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
thats the entire problem?
 
yeah
 
as always this deals with centripetal acceleration covered by the formula Fc = (mv/r^2)

plug in the values and you get Fc= (320*16*16)/(35)

which solves for 2340.6N

Have a great day.. at least show some effort.
 
Last edited:
i tried that but the answer is supposed to be 3900 N at 53 degrees above the horizontal, but thanks anyways
 
if there's a bank you didn't give me the full question. I cannot see an angle anywhere in the eq'n.. maybe some is cut off or elsewhere on the page..
 
Two forces acting on the roller coaster. Normal force , weight, frictional force.
 
Last edited:
Welcome to PF!

Hi dvela! Welcome to PF! :wink:

joshmdmd's formula for centripetal acceleration is wrong, it should be v2/r (not v/r2).

Try again. :smile:

(and joshmdmd, please don't give full answers on this forum)

joshmdmd said:
as always this deals with centripetal acceleration covered by the formula Fc = (mv/r^2)

plug in the values and you get Fc= (320*16*16)/(35)

which solves for 2340.6N

Have a great day.. at least show some effort.
 
inky said:
Two forces acting on the roller coaster. Normal force , weight, frictional force.

There is no Friction in the question and there is normal force, gravitational force and centripetal force. Basically you are looking for magnitude and direction for normal force.. (got an infraction for solving the eq'n completely so i guess I am not giving out more xD)
 
  • #10


tiny-tim said:
Hi dvela! Welcome to PF! :wink:

joshmdmd's formula for centripetal acceleration is wrong, it should be mv2/r (not mv/r2).

Try again. :smile:

(and joshmdmd, please don't give full answers on this forum)

I already edited that out xD
 
  • #11
joshmdmd said:
There is no Friction in the question and there is normal force, gravitational force and centripetal force. Basically you are looking for magnitude and direction for normal force.. (got an infraction for solving the eq'n completely so i guess I am not giving out more xD)

They were warnings, not infractions. And as explained, it's fine to give hints and ask probing questions, but it's not okay to solve homework problems for students here.
 

Similar threads

Replies
9
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
3K