1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetic Friction not so easy

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    The block is pushed to the right at a constant velocity. If the coefficient of kinetic friction is 0.5, what is the magnitude of the pushing force?


    2. Relevant equations
    I dont know...


    3. The attempt at a solution
     
  2. jcsd
  3. Sep 20, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    Did they give the mass of the block? What's your first idea? Give it a shot...
     
  4. Sep 20, 2007 #3
    no they didnt give the mass of the block, that is why im so confused my teacher said something about sigma Fx=0 and sigma Fy=0 but i dont know what to do with those numbers, all i know is velocity is constant which makes acceleration 0 which makes force of x and y 0, but after that im lost
     
  5. Sep 20, 2007 #4

    learningphysics

    User Avatar
    Homework Helper

    Yes, net force in the x direction is 0. Net force in the y-direction is 0.

    What are the forces acting in the y-direction?

    What are the forces acting in the x-direction?
     
  6. Sep 20, 2007 #5
    In the x direction is the pushing force and force of friction and in the y direction is force of gravity on a box whos mass is not given and the normal force, i think the lack of given numbers is confuing me
     
  7. Sep 20, 2007 #6

    learningphysics

    User Avatar
    Homework Helper

    Yeah, just use the variables... let mass = m.

    Write the [tex]\Sigma\vec{F} = ma[/tex] equation for the x-direction... then for the y-direction... just use the variables...
     
  8. Sep 20, 2007 #7
    in x direction it would be (mass)(0) so resultant force would be 0, because acceleration would be 0 due to constant velocity and in y direction (mass)(9.8) and then i dont really know what you could do with that...
     
  9. Sep 20, 2007 #8

    learningphysics

    User Avatar
    Homework Helper

    Write the equations using Fpushing, friction, mg, Fnormal... how do they add or subtract?
     
  10. Sep 20, 2007 #9
    mg-Fnormal=0

    this is where i get confused with x because they are not in equilibrium so they wouldnt equal 0
    Fpushing-friction=?? idk
     
  11. Sep 20, 2007 #10

    learningphysics

    User Avatar
    Homework Helper

    It equals 0, because acceleration is 0 (constant velocity). Now, you also know that friction = [tex]\mu * F_{normal}[/tex]

    Solve for Fpushing using your two equations, in terms of mass, [tex]\mu[/tex] and g.
     
  12. Sep 20, 2007 #11
    ok, i understand why it equals zero now but when you use that equation would it be
    friction=(0.5)(9.8m) and which two equations are you referring to?
     
  13. Sep 20, 2007 #12

    learningphysics

    User Avatar
    Homework Helper

    Yes, that's right...

    The two equations I meant were:

    Fpushing - friction = 0
    Fnormal - mg = 0

    you actually used the second equation to get friction = 0.5*9.8*m

    So what does Fpushing come out to...
     
  14. Sep 20, 2007 #13
    Fpushing-(0.5)(9.8m)=0
    Fpushing-4.9m=0
    Fpushing=4.9m??
     
  15. Sep 20, 2007 #14
    Seems good to me.
     
  16. Sep 20, 2007 #15

    learningphysics

    User Avatar
    Homework Helper

    yup. looks good to me too.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Kinetic Friction not so easy
  1. Kinetic Friction (Replies: 1)

  2. Kinetic friction (Replies: 3)

Loading...