Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Edutopia

On a carnival ride the static friction between a person and the walls of the ride keep them from falling. The normal force is perpendicular to the vertical. Given that the engineers of the ride do not want the normal force to be more than 1.5 times any persons weight, what is the minimum coefficient of static friction needed to keep riders from falling? parameters below...

  • one year ago
  • one year ago

  • This Question is Closed
  1. Edutopia
    Best Response
    You've already chosen the best response.
    Medals 0

    radius: 6.7m

    • one year ago
  2. Edutopia
    Best Response
    You've already chosen the best response.
    Medals 0

    my thinking is that bounding the normal force to 1.5 times a persons weight means that the centripetal acceleration must not exceed 1.5 (as F=ma). If this is so then, it is my reasoning, that the static friction coefficient would have to be something of the order of 300%

    • one year ago
  3. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 1

    Well, the downward force of gravity on the person is \[ F_g = -mg \] On the other hand, the force due to friction is \[ F_f = \mu N \] In balance, \( F_f + F_g = 0 \). You have an upper bound on N. Use that this equation to solve for \( \mu \).

    • one year ago
  4. Edutopia
    Best Response
    You've already chosen the best response.
    Medals 0

    u(1.5)-9.81=0, u(1.5)=9.81, u=6.54, (but u cant be greater than one!?)

    • one year ago
  5. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 1

    Careful. You know that \( N \leq 1.5mg \). Hence as \(-F_g = F_f \), we have that \[ mg = \mu N \leq 1.5 \mu mg \] Now you take it from there.

    • one year ago
  6. Edutopia
    Best Response
    You've already chosen the best response.
    Medals 0

    why is N < 1.5mg, why is gravity involved in the horizontal direction

    • one year ago
  7. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 1

    You are given that "the engineers of the ride do not want the normal force to be more than 1.5 times any persons weight"

    • one year ago
  8. Edutopia
    Best Response
    You've already chosen the best response.
    Medals 0

    AH!, you know what i did, i was thinking 1.5 times a persons mass!, and to me that ment that acceleration had to be 1.5, and i was thinking well duh this is impossible. Weight is a force!!! duh, lol

    • one year ago
  9. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 1

    Weight is mg, yes

    • one year ago
  10. JamesJ
    Best Response
    You've already chosen the best response.
    Medals 1

    what is important here however is not the direction of the weight vector, but the magnitude of N relative to the number mg.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.