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Minimum force to make box slide down slope

  1. Feb 12, 2016 #1
    1. The problem statement, all variables and given/known data

    If m=5.0kg, φ=30◦ and μs =0.87, what is the minimum force needed to make the box slide down the slope?



    2. Relevant equations
    force of status friction is less than or equal to the normal force multiplied by the coefficient of static friction
    F,normal=mgcosφ

    The image is a slope of angle 30 degrees with box, mass 5kg on it. The force is being applied so that the box will slide downward when enough force is applied.

    3. The attempt at a solution

    Because the F of static friction is equal to coefficient of static friction multiplied by the normal force I got;

    F,sf=((9.8)(5)cos(30))(0.87)=36.9N Therefore, the force would have to be greater than this, say 37N

    The correct answer is apparently 12N I do not know why!
     
  2. jcsd
  3. Feb 12, 2016 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Hint: Is friction the only force acting on the box?
     
  4. Feb 12, 2016 #3
    Thank you! x

    Fn=|Fn|cosθ k`
    Fg= -mg k`
    Ff= |Fn|(us) i`
    Fpush=Ff-mgsinθ

    Therfore,

    us(mgcosθ)-mgsinθ

    (0.87(5x9.8(cos30)))-5x9.8xsin30
    =12N

    yes? ahaha :p
     
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