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Statics Friction problem -- A 500lb box is being pushed up a ramp

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    A 500lb box is being pushed up a ramp which is sloped at 10 deg. The center of mass of the box is at the geometric center of the shown area. If the static coefficient of friction is 0.35 and the kinetic coefficient of friction is 0.28, what is the force P that will cause the box to tip? What is the force P that will cause the box to slide?

    2. Relevant equations
    *I resolved the weight into x and y components as my online homework allows me to set the axis to whatever I want, so I set it to move 10 degrees.
    Wx=Weight*sin(10)
    Wy=Weight*cos(10)

    ΣFx:0=P-Slipping Force-Wx
    ΣFy:0=N-Wy
    Slipping force= μ(kinetic)N
    θ

    3. The attempt at a solution
    So I solved for the force P that causes the box to slide by the sum of the forces in the x direction.
    P=Slipping Force+Wx
    P=μ(kinetic)N+Weight*sin(10)
    P=.28*Weight*cos(10)+Weight*sin(10)
    P=.28*500*cos(10)+(500*sin(10))=224.7lb

    That is correct, but I keep getting the wrong answer for the force that causes the box to tip. I am using a moment equation about the right bottom corner of the box(it is being pushed from the left, top side) and setting it equal to zero. Is a moment equation the right way to go about this or am I using the wrong approach?
     
  2. jcsd
  3. Nov 12, 2014 #2
    A moment equation about the point where the box will tip is the correct approach. How far off the center of mass does the force P push?

    Note that it would be impossible to tip the box if the static friction doesn't give enough counter force. So in other words the box would slip before it would tip.
     
  4. Nov 12, 2014 #3
    P is pushing 15in above the center of the box. The box's dimensions (probably should've included) are
    Height=70in
    width=30in
    The distance from the bottom to force P is 50 in
     
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