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Homework Help: Ball in a wedge Equillibrium

  1. Dec 4, 2008 #1
    1. The problem statement, all variables and given/known dataA solid sphere of radius 1 m and mass 5 kg
    is placed in a wedge with θ = 29 ◦
    The inner surfaces of the wedge
    are frictionless.
    The acceleration of gravity is 9.8 m/s2 .
    Determine the force exerted by the wedge
    on the sphere at point B. Answer in units
    Figure, ASCII style!
    | /
    |O/ <-b
    v

    point a is on the flat side towards the ball.

    Edit: I realize that the ascii diagram may suck, but.
    2. Relevant equations
    Fnet=0


    3. The attempt at a solution

    Alright, so I tried this.
    Fx=a+bcos(119)
    Fy=bsin(119)-5*9.8.


    This shouldn't be as annoying as it is. I mean, by my method, B is easily given, but the answer i get is wrong. Which means I'm not seeing something. Ideas?
     
  2. jcsd
  3. Dec 4, 2008 #2
    If you draw the triangle of forces acting on the ball, there the the horizontal reaction A (towards the right), joined to the reaction B upwards and to the left, and the vertical force representing the weight of 5 kg. (5*9.8)

    |<\
    |**\B
    |***\
    v---->
    ...A

    The * are just fillers to make a nicer triangle, ignore them.
    The angle between A and B is 29 degrees.
    So if you try to solve this right triangle, you will get the values of A and B.
     
  4. Dec 4, 2008 #3
    Just tried the triangle, just gave me the same answer as the vector resolution. :/

    5*9.8/sin61 = 56.024349 Which apparently is not the answer.

    This is why it's annoying! The intuitive answers are always wrong!
     
  5. Dec 4, 2008 #4
    Sin([tex]\theta[/tex]) = opposite / hypotenuse
    Cos([tex]\theta[/tex]) = adjacent / hypotenuse
     
  6. Dec 4, 2008 #5
    Well I hope I know that xP But yeah. From your triangle, the 29 degree shoots up to the top of the triangle, and the 61 becomes the most useful for a sin substitution. hence 5*9.8/b=sin61, so 5*9.8/sin61=b. :/ Not sure if i'm screwing up somewhere.
     
  7. Dec 4, 2008 #6
    The (lousy) triangle I drew is a triangle of forces which has to close for equilibrium.
    Force A is the normal (90degrees) to the vertical side, so it is horizontal.
    Force B is the normal to the slanting side, so it makes 29 degrees with the horizontal.
    Well, the vertical component is mg=5*9.8, which I am sure you know.
    Want to give it another try, if you are solving for B?
     
  8. Dec 4, 2008 #7
    Hah, I fail epiclly at Trig. xD Sin29 worked like a charm, on the last try, too. Thanks, man.
     
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