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Basic Mechanics Incline Question

  1. Mar 14, 2007 #1
    Hello - My geometry is rusty and I'm having some difficulty visualizing a question. Picture a box sitting on an inclined plane with an incline angle of 30 degrees from the horizontal (floor). The box is at rest. There is a 10N normal force (perpendicular to horizontal) and a force, F, perpendicular to the plane itself.

    I'm looking for the setup. Is the angle formed between the normal force (10N) vector and the force (F) perpendicular to the incline the same as the angle at the angle of incline (30 degrees)?

    What are the rules for this in geometry - i.e. how can I tell where angles are congruent with relative ease?

    Thanks in advance.
     
  2. jcsd
  3. Mar 14, 2007 #2

    cristo

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    Are you sure the force F is not perpendicular to the horizontal, and the normal force perpendicular to the plane?
     
  4. Mar 14, 2007 #3

    Mentz114

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    Apart from right angles, your set up only has angles of 30 and 60 deg. So it's one or the other. If you make a diagram it will be obvious.
     
  5. Mar 14, 2007 #4
    Sorry if I was vague

    Here's a diagram if it's viewable... a bit small, but I had to scale it...:biggrin:

    It's strange, but I have a real problem discerning X and Y components. I know I need to skew the X and Y axis but then I have a hard time with the free body setup with regard to SIN, COS etc... I know it's really simple...

    thanks again
     

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    Last edited: Mar 14, 2007
  6. Mar 14, 2007 #5

    Mentz114

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    Here it is. As you can see, the angle between the normal and the vertical is 30.
     

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  7. Mar 14, 2007 #6
    Thanks for your reply. I'm with you. So, what's the easiest way to draw the free body accurately each and every time if I'm not given those angles (N and mg) and there's a box sitting on an inclined plane or an object on a hill that forms an angle, theta, with the ground or horizontal? I've got it in a book here but it doesn't seem very intuitive -unless I'm making something very simple more difficult. What approach do you take?
     
    Last edited: Mar 14, 2007
  8. Mar 15, 2007 #7

    Mentz114

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    You should be able to see from the diagram that the angle between the vertical and the normal to the plane is always the same as the plane angle with the horizontal.
    There's not actually an approach, it's simple geometry. Your diagram was no good because you hadn't drawn the angles. Bone up on elementary geometry.
     
  9. Mar 15, 2007 #8
    Thanks

    Will do, thanks.
     
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