Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

ksaimouli

  • 2 years ago

A block is pulled by a string that makes an angle of 25º to the horizontal. If the mass of the block is 12.0 kg and the coefficient of friction is 0.25, what force would keep the block moving at a constant velocity?

  • This Question is Closed
  1. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1349633059599:dw|

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

    |dw:1349633142335:dw|

  3. ksaimouli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    what to do after this i am totally confused

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

    i know that ff=u*fn

  5. ksaimouli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we need to find fn

  6. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[F_n=mg-F_y=12*9.8-Fsin(25)\]

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

    i got this but how do u find the F

  8. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[F_f= \mu F_n= 0.25(117.6-Fsin(25))=F_x=Fcos(25)\]

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

    how did u get 117.6

  10. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    mg

  11. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    12*9.8

  12. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[29.4-0.25\sin(25)F=Fcos(25)\] \[\frac{29.4}{\cos(25)+0.25\sin(25)}=F\]

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

    thx

  14. ksaimouli
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    why =fcos(25)

  15. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1349634909926:dw|Basic trig

  16. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[F_f=F_x=Fcos(25)\]

  17. henpen
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    No acceleration => no net force in x-direction

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.