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Particle force question

  1. Apr 18, 2008 #1
    The question is, "A particle of mass 10kg is at rest on a rough plane inclined at 30° to the horizontal. A horizontal force of magnitude 10N acts on the particle. a) Find the magnitude of the friction force on the particle."

    URL TO IMAGE (I added the green to help you understand why I am confused): i209.photobucket.com/albums/bb180/newguyjb/forcesdiagram.jpg

    The book provides the answer with the working, which is:
    F+10cos30°=98sin30°
    F=98sin30°-10cos30°
    =40.3N

    However, I don't understand how they have decided that it is 10cos30°. On the diagram I provided (which is the link) in green the area which is not in the diagram in the book, but which relates to why I am confused as to why it is 10cos30°. Could someone explain why it is 10cos30°.

    Thanks for any help.
     
  2. jcsd
  3. Apr 18, 2008 #2

    alphysicist

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    Hi King,

    That equation is the components along the incline of all the forces. What is the angle between the 10N force and the incline? The component of the 10N force that you want in that equation would be the side adjacent to that angle.
     
  4. Apr 18, 2008 #3

    Pythagorean

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    Hrm... I get a different answer too. I may be making an error myself, but here's how I do it:

    If you rotate it so that the surface is flat and below the particle, you can define a new "horizontal" that has forces to the right that are a component of the applied horizontal force and gravity. The magnitude of our friction is now acting completely to the right in this frame (positive x in the new system).

    The total force in this new "horizontal" frame is zero and the applied force (10 N) pulls off at an angle of 30 degrees downwards and to the right and gravity pulls downward and to the left at and angle of 60 degrees from the new "horizontal".

    The normal force can be ignored since it acts "vertically" in our new system and our friction force acts "horizontally" in our new system.

    So you have:
    (firction + applied force both acting "right") = (gravity acting "left")
    F+10Ncos30=98Ncos(60)
    F = [98cos(60)-10cos(30)]N = 47.5N

    So I get the same 10Ncos30 term they do, but my gravity term is off :/
     
    Last edited: Apr 18, 2008
  5. Apr 18, 2008 #4
    You have your force diagram wrong;

    pic1.jpeg

    or whilst that's waiting to get validated

    stopouts.com/pic1.jpeg

    Since frictional forces act parallel to the plane and the object's at rest, the force pulling it down the plane must be equal to the frictional force keeping it in place + the component of the 10N force acting parallel to the plane.
     
    Last edited: Apr 18, 2008
  6. Apr 18, 2008 #5

    alphysicist

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    Hi Pythagorean,

    Your formula looks correct. cos(60)=sin(30)


    I think you might have made a numerical error; I'm not getting 47.5 from your expression.
     
    Last edited: Apr 18, 2008
  7. Apr 18, 2008 #6

    Pythagorean

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    ah, of course:

    cos(60) = sin(60+90) = sin(150) = sin(30)

    I was thrown off because our end calculations still don't match.
     
  8. Apr 18, 2008 #7
    Hi all,
    Thank you for all the replies :)
    I understand how to resolve forces, and from the question I understand that it is F+????-98sin30°, just the ???? part confuses me. If you take the slope it is 180°, the forces (one going right and the other going down) are 90°, thus 90° remains as the sum of the missing angles to create 180°. Also, one can calculate that the right-angled triangle on the left has 90°+30°=120° and so 60° remains. Thus, if we go back to where the 10N force is, the angle adjacent to the horizontal must be 30°. And oh my God I just solved my own question. lol!

    Thanks for everyone's help.
     
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