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2-D kinematics/Forces (weight up an incline)

  1. Sep 29, 2006 #1
    The Question:
    A 123 kg chesterfield is pushed up a frictionless ramp at a constant speed by a delivery person.
    If the ramp is inclined at 26.9° to the horizontal, what horizontal force must the delivery person apply to the chesterfield?


    The way I am thinking about the problem is that the answer should be the horizonal component of the F(g) (weight) of the chesterfield (C). Since balancing that force would produce a net force of 0 in the horizontal dimension.

    Drawing it out, the y' of w is wcosθ and x' is wsinθ. So the horizonal component would be (wsinθ)cosθ. The number I get for that is 486.85N (using g=9.81). This is apparently incorrect.

    So my question is: How am I supposed to think about this problem? What am I doing wrong? Also, the question seems a little ambiguous also, but how would any of you interpret it?

    Thank you for your time and consideration.
  2. jcsd
  3. Sep 29, 2006 #2
    To find the magnitude normal force use the following equation:
    Fg = Fn cos0 (because the chesterfield is on a tilt)
    Fn = Fg/cos0

    Then I found the magnitude of the normal force in the X direction. That should be the force of the push.
  4. Sep 29, 2006 #3

    Chi Meson

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    Stop thinking in terms of x an y axes. Rotate you axes so that they are parallel and perpendicular to the surface. You now have to find the componants of weight and the applied force that are "parallel" and "perpendicular."

    Note: the normal force will balance all componants perpendicular to the
    surface. So all that is left are the componants of applied force and weight parallel to surface.
  5. Sep 29, 2006 #4
    Sorry, my x' was 'parallel' to the ramp and y' was perpendicular.

    What I have are:
    F(normal)=wcosθ and


    But that is incorrect...
  6. Sep 29, 2006 #5


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    Constant velocity implies equilibrium. So, the sum of the three forces, the normal force N (reaction from the ground to the chair), the weight of the chair W and the pushing force F, has to vanish, i.e. [tex]\vec{N}+\vec{W}+\vec{F}=\vec{0}[/tex]. So, what applies to vectors, applies to their components, too. Place a coordinate system and off you go.
  7. Sep 29, 2006 #6

    Chi Meson

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    Since the pushing force is not paralle to the surface, then only the parallel componant of the push will be 546 N. What is the full magnitude of the push if this is one of the componants?
  8. Sep 29, 2006 #7
    That was my first post, I believe.
  9. Sep 29, 2006 #8

    Chi Meson

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    If 545.9 N is a componant of hte pushing force, then the pushing force itself must be a larger magnitude. Go back and examine what you did with the cos theta.

    The answer is 612 N
  10. Sep 29, 2006 #9
    ah yes, I see now.

    I was making some incorrect equalities with my angles.

    Thank you for your help.
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