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Statics - Mechanics

  1. Jan 24, 2007 #1
    This isnt a homework problem, I just have a general problem with solving AS level (M1) statics questions in mechanics that are similar to this example (our teacher have moved on to the next module now but i'm still stuggling with these):

    example
    A particle of mass 3kg rests in limiting equilibrium on a rough plane inclined at 30degrees to the horizontal. the coeffiecient of friction between the particle and the plane is 1/3. A horizontal force of magnitude XN is applied to the particle so that the equilibrium will be brocken by the particle moving upwards. Find the Value of X.

    :confused: I always seem to get these wrong and i dont understand why. Here is how i wuld go about it, if anyone can tell me where and why my method is wrong please do! I really need to understand this soon. :frown:

    attempt

    Fr=1/3R
    Fr=3g/sin30=58.8N
    therefore R=58.8*1/3=19.6N

    and there, this is where i get stuck and have absolutely no idea where to go from now. :cry:
     
    Last edited: Jan 24, 2007
  2. jcsd
  3. Jan 24, 2007 #2

    Hootenanny

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    Welcome to the forums Beautiful Mess,

    Well, to begin with you've found the normal force and frictional force incorrectly. Have you drawn a free body diagram?
     
  4. Jan 24, 2007 #3
    Draw a free-body diagram and re-calculate the component of g along the incline.
     
  5. Jan 24, 2007 #4
    how do you calculate the component of g along an incline? (i feel really stupid)
    our teacher never told us, she just expects us to work everything out ourselves.
    I have a free body diagram, somehow it doesnt seem to help me at all. Ok i'm trying again:

    does R=3gcos30 ?

    and if so Fr would = (3gcos30)/3
     
  6. Jan 24, 2007 #5

    Hootenanny

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    That is correct! :approve: So now, this frictional force acts parallel to the incline (30o) to the horizontal. So what force is required to make this particle move with a constant velocity?
     
  7. Jan 24, 2007 #6
    yay i understood a little! :D

    (ive also just noticed that the force X is applied at 30degrees to the plane)

    so
    Xcos30-Fr=R
    Xcos30=3gcos30+(3gcos30/3)
    X=all above/cos30

    erm that looks wrong, i'll keep trying on paper
     
  8. Jan 24, 2007 #7
    oooh it doesnt = R it equals 0

    and then perdendicular to plane

    Xsin30 - R =0

    so, no that must be wrong now i can get 2 different values for X...
     
  9. Jan 24, 2007 #8

    HallsofIvy

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    This is more a physics problem than math. I'm moving it to "homework-introductory physics"
     
  10. Jan 24, 2007 #9
    Your first post says the force is horizontal, this means that it will contribute to the normal force as well as the friction force
     
  11. Jan 25, 2007 #10
    HallsofIvy - I can see why you might class this as Physics but in England this is classed as a Module (Statics-Mechanics) of a Maths A level Course (pre-university qualification) and so we (British) would expect to find such questions under Mathematics (or Applied Mathematics) rather than physics.
     
  12. Jan 28, 2007 #11
    Yea thats why I put it there, sorry.

    Also I know my first post said horizontal but i mis-read it (i mixed it up with the question above). I seem to be understanding this better now though, so thank you for everyone's help :smile:
     
  13. Feb 21, 2007 #12
    I am officially a genious! not, but its amazing what a bag of marshmallows can help you learn! I get it now, its damn easy, I was such a fool not to see the way before! (ignore the way i'm speaking it sometimes happens when i have too much sugar...)
     
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