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Newtons second law/friction problem

  1. Feb 16, 2009 #1
    1. The problem statement, all variables and given/known data
    A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15 degrees above the horizontal.

    (a) If the coefficient of static friction is .50, what minimum force magnitude is required from the rope to start the crate moving?
    (b) If [tex]\mu[/tex]k=0.35, what is the magnitude of the initial acceleration of the crate?
    2. Relevant equations
    F=ma


    3. The attempt at a solution
    part a:

    I calculated the weight (9.8)(68 kg)=666N

    then I got static friction: (666N)(.50)=333N

    I substituted all the forces into newtons second law and got:
    ma=Normal+Gravity+Static+Applied force

    after I plugged in the variables I got 0=333N + (force applied)(cosx)
    (It's 0 because the box isnt moving so a=0)

    when i solve for force i get 344.7N but the answer in the book is 300N
     
  2. jcsd
  3. Feb 16, 2009 #2

    LowlyPion

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    Homework Helper

    The Force of the rope has 2 components.

    The Y component is lightening the load of the 666 N by F*Sinθ . And it's F*Cosθ that is the force applied in the X direction to move it.

    Try looking at what happens then if you account for that with θ = 15.
     
  4. Feb 16, 2009 #3
    The force is unknown though. How can I find the components of the force if the magnitude is an unknown?
     
  5. Feb 16, 2009 #4

    LowlyPion

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    Homework Helper

    But F is your only unknown.

    Write out your equation and solve for F.

    You know the components of F from θ = 15.
     
  6. Feb 16, 2009 #5
    the original equation I had was 333N=Fcosx

    would it be 333N=Fcosx + Fsinx?

    sorry I'm very confused.
     
  7. Feb 16, 2009 #6
    wait I see it now. thanks
     
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