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!

Newton and friction <3

  1. Feb 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A block is shoved up a 22 deg slope with an Initial speed of 1.4 m/s the coefficient of Kinetic friction is .7...

    a) how far up the slope will hte block get?

    b) will the block come back down?


    note the bold is just a sign change, a diffrent possibility

    2. Relevant equations and The attempt at a solution

    x: f*g*sintheta+Friction of the surface = MAsubx

    y: n- Force of g * costheta = 0

    solve for Asubx and N

    put n into the Ax eqn

    Asubx = (m*g * sintheta - myuk * m * g * costheta) / m

    m cancels factor out g

    and use

    v^2 = v0^2 + 2a(x-x0)


    v^2 = v0^2 + (2 ( g (sin theta)- myuk *costheta)))*(x-x0)

    v = 0 m/s
    v0= 1.4 m/s
    g = 9.8 m/s^2
    theta = 22 degrees
    myuk = .7
    x0 = 0 m
    x = ? m

    i get .364 m for part a


    and i said no it will not return back down due to a small incline and small v0

    i THINK im right but im not 100% confident

    note: i have a free body diagram and such ill make on in PAINT if u want me to, lemme know, thx for your time!!!!!!!!!!!!!!!!!!!!!!
     
    Last edited: Feb 8, 2008
  2. jcsd
  3. Feb 8, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It looks fine to me. It won't come back down because the force of kinetic friction is greater than the gravitational force. What's the .1412m supposed to be?
     
  4. Feb 8, 2008 #3
    i deleted the bold parts i jsut rememberd that it is opposite so its negitive,
     
    Last edited: Feb 8, 2008
  5. Feb 8, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The force of friction and the gravitational force are acting in the same direction when the block is going up the ramp. If you are using g=+9.8 (and it looks like you are) then the plus sign was the right one. Both terms should have the same sign.
     
  6. Feb 8, 2008 #5
    so it was the .124 or so?
     
  7. Feb 8, 2008 #6
    v^2 = v0^2 + (2 ( g (sin theta)+ myuk *costheta)))*(x-x0)

    and yea i have g as positive and i noticed that after i calculated it



    i guess just filling the in ? will answer my question

    i belive its sintheta - myuk
     
    Last edited: Feb 8, 2008
  8. Feb 8, 2008 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    2*(g*sin(theta)+mu*g*cos(theta)), right?
     
  9. Feb 8, 2008 #8
    the part we are of interest right now we are trying to make sure is right is...

    Asubx = Fsubg*sintheta PLUS OR MINUS Myu * NormalForce


    WHERE the normal force IS Fsubg * costheta

    Fsubg =m * g

    this was pretty close to my freebody diagram not exact... i think my angle might be in the wrong spot lemme know

    [​IMG]

    can u make a FBD and compare to mine,... i actauly think mine is wrong now that i stare at it for a min lol
     
    Last edited: Feb 8, 2008
  10. Feb 8, 2008 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Add them. The frictional force points in the opposite direction to the direction of motion. The gravitational force is always down the ramp. When the block is going up, they point in the same direction. When they are going down then they would point in opposite directions, but remember that if the magnitude of the static frictional force (which is greater than the kinetic frictional force) is greater than the gravitational force, then it won't slide down at all.
     
  11. Feb 8, 2008 #10
    no static friction force was given.. but i made an assumption there is not way it is gonna return down, is .7 high for a kinetic force? what is the equivalent to, we just started work with friction this past week, this was my first of a few real problems

    Fsubg*sintheta PLUS Myu * NormalForce


    with that said

    v = 0 m/s
    v0= 1.4 m/s
    g = Minus 9.8 m/s^2
    theta = 22 degrees
    myuk = .7
    x0 = 0 m
    x = ? m

    this should be obviously negitive but i want to see what you haev to say about it

    Evaluate
    v^2 = v0^2 + (2 ( -g (sin theta + myuk *costheta)))*(x-x0)

    CORRECTION MY CALC WAS IN RADS LOL

    the answer is .097691 m
     
    Last edited: Feb 8, 2008
  12. Feb 8, 2008 #11
    alright this makes perfect sense if g is positive my answer is neg, so yea its soposed to be neg, if my math is right then that is the correct answer
     
  13. Feb 8, 2008 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    v=0, and v0=1.4m/sec. Yes, the 2*a*d term should be negative if you write the equation in that form. But I don't know how you are getting .141m. Just use your intuition about what direction things should point to adjust the signs. What do you get for g*sin(22)+0.7*g*cos(22)?
     
  14. Feb 8, 2008 #13
    -9.8 * (sin(22) + cos(22) * .7)
    -9.8 * (1.02364)
    -10.0316


    at that point in teh eqn it is

    0 = 1.96 - 2 ( 10.0316 ) *x

    -1.96 = -20.0633 * x

    x = .097691 m


    igonore the opening answer that is the real answer as we have moved through the problem i got in bold above
     
    Last edited: Feb 8, 2008
  15. Feb 8, 2008 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I agree with that.
     
  16. Feb 8, 2008 #15
    thanks i appreciate it, i had a feeling about that + or minus sign, in my conlcusion i talked about how that could have thrown my answer off... .3m did sound liek a big distance for a 1.4 m/s velocity and the small i was making x come out i knew i was on the right path

    take care have a good night =D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Newton and friction <3
  1. Newton 3 (Replies: 9)

  2. Newton's 3 Law of Motion (Replies: 13)

  3. Newton's 3 Laws Lab (Replies: 1)

Loading...