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

Mathhelp346 Group Title

please HELP!! Does adding weight affect the coefficient of friction?

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. HELLSGUARDIAN Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    yes it does..as by adding weight the amount of normal reaction acting increases and hence frictional force also increases..as \[F =\mu N\] and N is Directly proportional to weight..

    • 2 years ago
  2. HELLSGUARDIAN Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1350009988789:dw|

    • 2 years ago
  3. Mathhelp346 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    whats n

    • 2 years ago
  4. HELLSGUARDIAN Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Normal reaction..the reaction which is acting on a body having weight..

    • 2 years ago
  5. Mathhelp346 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok thank you

    • 2 years ago
  6. HELLSGUARDIAN Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    your welcome..

    • 2 years ago
  7. SWdrafter Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    The increase of weight only results in the increase of the frictional force not the coef. of friction for this case. Note that the u is a constant.

    • 2 years ago
  8. Mathhelp346 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    does it not increase because the coefficient of friction is a ratio? that's the definition i have for coefficient of friction

    • 2 years ago
  9. SWdrafter Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    The ratio is derived from the observed frictional force and normal force. For normal dry friction this ratio remains constant.

    • 2 years ago
  10. Mathhelp346 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    so since its constant it does not change

    • 2 years ago
  11. SWdrafter Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    that is correct

    • 2 years ago
  12. Mathhelp346 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    and what do you mean by normal dry friction?

    • 2 years ago
  13. SWdrafter Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    This is between to surfaces with no lubricant in between them.

    • 2 years ago
  14. Mathhelp346 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok thank you

    • 2 years ago
  15. SWdrafter Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Thanks for the medal.

    • 2 years ago
  16. SWdrafter Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    For more info, check out: http://en.wikipedia.org/wiki/Friction

    • 2 years 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.