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Crate on a frictionless ramp, constant speed.

  1. Jan 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A 130 kg crate is pushed at constant speed up the frictionless 23° ramp shown in the figure. What horizontal force F is required?

    2. Relevant equations

    F=ma
    F(parallel)=mgsin(theta)
    F(perpendicular)=mgcos(theta)

    3. The attempt at a solution

    I know that because it is at a constant speed the net force equals zero, therefore the force caused by gravity equals the applied force. So first I solved the parallel force caused by gravity, which equals 498.2994 N. From there to create a net force of zero I flipped the sign, -498.2994 N. Solved for the horizontal component of that force: -498.2994cos23 =-458.687 N.

    But obviously that was not right. I think I might have to include the normal force by gravity in that equation too though. Any thoughts?
     
  2. jcsd
  3. Jan 24, 2012 #2

    Doc Al

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    Staff: Mentor

    Since the applied force is horizontal, you have found the component of F parallel to the incline. Now you need to find the full force F. Express that mathematically so you can solve for F.
     
  4. Jan 24, 2012 #3
    I am afraid that I am a little lost at the moment. I need to express the full force of the equation?
     
  5. Jan 24, 2012 #4

    Doc Al

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    If the applied force F is horizontal, what would be its component parallel to the incline?
     
  6. Jan 24, 2012 #5
    Force applied(parallel) = Force applied(sin(theta))

    So: 498.2994/sin(23) = 1275.3 N [force applied]
     
  7. Jan 24, 2012 #6

    Doc Al

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    Not exactly. Realize that the applied force is horizontal, not vertical (like gravity).
     
  8. Jan 24, 2012 #7
    Okay so the applied force in the direction of parallel, would that be an equal but opposite force to the force parallel caused by the force of gravity?
     
  9. Jan 24, 2012 #8

    Doc Al

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    Yes, if I understand what you mean. The net force in any direction must be zero. Thus the forces parallel to the ramp must be zero. Thus the parallel component of gravity and the parallel component of the applied force must be equal and opposite.
     
  10. Jan 24, 2012 #9
    Okay.

    If I take the horizontal component of the Normal force and the horizontal component of the applied force and add them together, will that give me a total horizontal force.

    (498.299cos23)=458.6866 N
    (1380sin23)=539.208 N

    So, 997.8946 N?
     
  11. Jan 24, 2012 #10

    Doc Al

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    Since the crate is in equilibrium, the net horizontal force will be zero. Some problems with this approach:
    - The applied force is totally horizontal, so the horizontal component of the applied force is the applied force
    - The normal force does not simply equal mgcosθ

    You'll have a much easier time if you stick to the force components parallel to the ramp.
     
  12. Jan 24, 2012 #11
    Okay.

    So.

    Fcos(theta)-mgsin(theta)=ma
    Fcos(23)-(130kg)(9.81 m/s^2)sin(23)=0 (it equals zero because of the zero net force)

    Solve for F.

    F= 541.332 N

    Sorry for being a pain, I have been fine with similar questions, just this one was confusing. You were a huge help, thank you. That would be the answer if I am not mistaken.
     
  13. Jan 25, 2012 #12

    Doc Al

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    Good!
     
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