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Force on Frictionless Surface

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    In the figure, a crate of mass m = 111 kg is pushed at a constant speed up a frictionless ramp (θ = 30°) by a horizontal force F . The positive direction of an x axis is up the ramp, and the positive direction of a y axis is perpendicular to the ramp. (a) What is the magnitude of F ? (b) What is the magnitude of the normal force on the crate?


    2. Relevant equations



    3. The attempt at a solution
    For part A, I got 628.67 N which is correct.

    I do not have part B, but I do have some other information that I found, which is correct.

    Gravitational Force along x-axis is -544.455 N

    Fx component of the applied force is 544.45 N

    Fy component of the applied force is -314.341 N

    Y component of the gravitational force is -943.023 N

    Acceleration components are zero.

    And the normal force of the crate is in the positive direction of the y-axis.
     
  2. jcsd
  3. Sep 24, 2009 #2

    kuruman

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    Draw a free body diagram, put in all the forces and find what the normal force is. In this case it is not mg*cosθ, so don't use that. Remember that the acceleration in a direction perpendicular to the incline is zero.
     
  4. Sep 24, 2009 #3
    would it just be 0? F=M*A...A=0, so F is equal to zero?
     
  5. Sep 25, 2009 #4

    kuruman

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    If by F you mean "the net force", in other words the sum of all the forces, then it is zero. That's because the problem is telling you that the velocity is constant. Constant velocity means zero acceleration which means zero net force according to Newton's Second Law.
     
  6. Sep 25, 2009 #5
    ok so since the sum of all the forces is zero, could it be -628.67?
     
  7. Sep 25, 2009 #6
    or should I find the magnitude of the gravitational force. And set F(g) + F + F(n) = 0?
     
  8. Sep 25, 2009 #7

    kuruman

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    In what direction is the normal force? The sum of all the components of forces in that direction must be zero. You know all the components of forces in that equation except for the normal force.
     
  9. Sep 25, 2009 #8
    Does 1257.36 seem right?
    I did (544.45i - 314.34j) + (-544.45i - 943.02j) + F(N) = 0

    I get 0i + (-1257.36j) = 0

    So F(N) is the Square Root of 1257.36^2, which is just 1257.36

    I want to be completely sure of my answer, cause I only have one try left!
     
  10. Sep 25, 2009 #9

    kuruman

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    That looks about right if your components along the y-direction are correct - I didn't check them - but if you say they are, I believe you.
     
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