A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

How do you find the acceleration of two objects if the coefficient of kinetic friction between the 7kg object and the plane is 0.250? Given: m1=7kg, m2=12kg, and there is an angle of 37.0 degrees.

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I need more information. What does the set-up look like?

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

    Okay. There is a triangle, with a 37 degree angle in the left bottom corner, holding up a block that weighs 7kg. This block is attached to a levy on the right upper corner of the triangle which is holding up a 12 kg block hanging downward on the right.

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    First draw a free body diagram. For the 7kg object you should get the two equations: Fx=T-Ff-mgsin37=m1a and Fy=N-mgcos37=0. So N=mgcos37=54.79. The friction force(Ff)=.25N=13.7. Substitute this into Fx and you will get Fx=T-54.98=m1a. This is the same as T=m1a+54.98. Then from the 12kg object, you should get the equation: Fy=m2g-T=m2a. Which is the same as T=m2g-m2a. Substitute this in for the T in the Fx equation and solve for a. I got 2.76, but I may have done some math wrong somewhere, so you should probably check everything :)

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so much!

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'll try it our right now.

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No problem :)

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    How did you get 13.7 from the friction force?

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The normal force is mgcos37 which is 54.79. Friction force is normal force time the coeffecient of friction. So it is 54.79*.25= 13.7

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh. lol Thanks.

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got up to the last step. m2g-m2a= m1a + 55.0N 118N-12kg(a)= 7kg+55.0N How would I solve this? :?

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, I got it. Thank you again!

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.