1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mechanice solving problem

  1. Jan 1, 2005 #1
    i hope someone can understand french and help me to solve the problem:

    dans cette étude d'un saut de ski, on négligera le frottement de l'air sur le skieur bien qu'en réalité ce dernier utilise manifestement la résistance de l'air lors de la phase aérienne.
    Un sauteur à ski, de masse 75 kg, s'élance sur un tremplin dont la piste, de longeur 150 m, est située entre l'altitude 1540 m et 1440 m. Ce tremplin se termine par une partie horizontale.

    a) quelle est la valeur de la vitesse du sauteur quand il quitte le tremplin en O, sachant que les frottements de neige sur les skis sont équivalents à une force de valeur constante et égale à 400 N?
    on prendra pour valeur de g= 10 m.s^-2

    b) la piste d'atterrissage est plane et inclinée à 45° par rapport à l'horizontale. Elle passe par un point A situé sur la verticale en O, à 5 m en dessous de ce dernier. Déterminer à quelle distance du point A le skieur touche le sol.

    figure approximée:


    _(sommet de la piste)
    \
    \
    \
    \
    -. (point O)
    |
    |.(point A)
    \
    \
    \
    ____α_\_

    α = 45°


    solution proposée pour la première partie:

    d'après le théorème de l'énergie mécanique, il y a des forces de frottements,
    on considère le système (sol, skieur, terre); il n'est pas isolé, l'énergie mécanique n'est pas conservée.

    Em1: position au sommet de la piste
    Em2: position en O.

    on considère l'horizontale passant par O comme niveau de référence.
    Alors Epp=0 au point O.

    Em2 - Em1 = ΣWƒext

    Em1= Ec1 + Epp1 (Ec1=0 car V=0)
    Em1= Epp1 = +mgh (or h= 1540-1440= 100 m)
    Em1= 75*10*100= 75000 j.

    Em2= Ec2 + Epp2 (Epp2=0 car le skieur se trouve en O)
    Em2= Ec2= ½mv²
    Em2= 37.5v²

    ΣWƒext= Wƒ= -ƒ*x= -400*150= -60000 j.

    ALORS:

    Em2 - Em1 = ΣWƒext

    37.5v²- 75000 = -60000
    37.5v² = 75000-60000= 15000

    v² = 400
    v = 20 m/s.



    PS: je crois et positivement sur que cette partie est vraie mais la partie b, j'ai beaucoup essayer de la résoudre or j'ai trouvé beaucoup de difficultés.
    C'est lorsque le skieur quitte le sol en O et entre dans la phase aérienne, on est demandé de trouvé à quelle distance de A il touche le sol!)

    j'éspère que quelqu'un m'aide et c'est rapide car je dois terminer cet exercice.

    PS: i'm sorry i tried first to translate it to english, but it is somehow difficult especially key words, anyway i hope that i will get your help and reach for the result by finding the answer.

    thank you
    JOE
     
    Last edited: Jan 1, 2005
  2. jcsd
  3. Jan 1, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The first part is okay,,indeed.For the second,u know that the skijumper is in free fall,and starts from a point O which is situated at a certain height above the point of landing.U know that he has an initial velocity parallel with the Ox axis.But u don't know the height.
    The second law of dynamics for the skijumper
    [tex] m\vec{a}=\vec{G}=m\vec{g} [/tex]
    ,projected on an orthonormal system of coordinates with Ox containing the landing point and Oy the points A and 0 and integrated wrt to corresponding limits,gives:
    [tex] x(t)=v_{0}t [/tex](1)
    [tex]y(t)=h-\frac{1}{2}gt^{2} [/tex](2)
    ,where h is the height of O wrt to the Ox axis (and the landing point)
    The trajectory is a parabola of equation:
    [tex] y(x)=h-\frac{1}{2}g\frac{x^{2}}{v_{0}^{2}} [/tex] (3)

    Consider 'd' as the point on the Ox axis where the skijumper lands.It can be found from (3) from the condition y=0
    It yields
    [tex] 0=h-\frac{1}{2}g\frac{d^{2}}{v_{0}^{2}} \Rightarrow h=\frac{1}{2}g\frac{d^{2}}{v_{0}^{2}} [/tex](4)

    Now,u know that in the point of landing (of coordinates (d,0)) the slope of the trajectory must be '-1' (an angle of 135° wrt to the horizontal,or 45° with the negative sense of Ox).
    The slope of the parabola (3) in a point 'x' is
    [tex]y'(x)=-\frac{g}{v_{0}^{2}} x [/tex](5)
    ,so in the landing point (x=d) it must be "-1".Therefore u find "d"
    [tex] -1=-\frac{g}{v_{0}^{2}} d [/tex](6)
    [tex] d=\frac{v_{0}^{2}}{g} [/tex] (7)
    From (7) and (4) u find "h":
    [tex] h=\frac{v_{0}^{2}}{2g} [/tex]

    Compute numerically "h" (is 20m),then drop 5m to find the height of 'A' and from the rightangle triangle with sides "d" and height of 'A' u can find,by applying the Pythagora theorem the distance the problem is aking u (it should be [itex] 5\sqrt{73}m [/itex].

    Daniel.
     
  4. Jan 1, 2005 #3
    thanks in deed u've been so helpful

    sincerly,
    JOE
     
  5. Jan 1, 2005 #4
    sorry i was discussing with my classmate.. i still have one question
    first what did u mean by "wrt"
    second, how did u obtain "h" in the equation of Y(t)

    thanks again
     
  6. Jan 1, 2005 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    wrt est une abréviation pour "with respect to", ce qui veux dire "par rapport à".

    Moi j'ai une question pour toi... comment fais-tu pour écrire, par exemple, des lettres grecques, ou des f "stylisés" sans LaTeX? Connais-tu un site qui explique les commandes pour écrire ainsi?

    Merci.
     
    Last edited: Jan 1, 2005
  7. Jan 1, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Je ne parles pas Français.Je suis très desolée.
    To the first question u've been given an answer.
    I) Initial conditions plus integration:
    [tex] \frac{d^{2}y(t)}{dt^2}=-g [/tex](1)

    Make the substitution:[tex] \frac{dy(t)}{dt}=v_{y}(t) [/tex] (2).
    Then (1) becomes
    [tex] \frac{dv_{y}(t)}{dt}=-g [/tex] (3)
    .Separate variables and integrate knowing that at t=0,[itex]v_{y}(0)=0 [/itex]
    U'll get:
    [tex] v_{y}(t)=-gt [/tex] (4)
    Use the substitution (2) to put (4) in the form:
    [tex] \frac{dy(t)}{dt}=-gt [/tex] (5)
    .Separate variables and integrate knowing that at t=0,[itex]y(0)=h [/itex]
    U'll get:
    [tex] y(t)=h-\frac{1}{2}gt^{2} [/tex] (6)
    ,which is just the relation i used.

    II) Make use that the gravitational acceleration "g" is constant,so the body of the skijumper will have an uniform accelerated movement along the "y" axis.Then use the law of movement with constant accleration which is
    [tex]y(t)=y_{0}+v_{y,0}t +\frac{1}{2}at^{2} [/tex]
    ,plus the assumptions:
    [tex] a=-g;v_{y,0}=0;y_{0}=h [/tex]
    ,to get the same thing.

    Daniel.
     
    Last edited: Jan 1, 2005
  8. Jan 1, 2005 #7
    Google m'a aidé: voir ici. Le f "stylisé" (ƒ) est ALT+159.
     
  9. Jan 2, 2005 #8

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Cool! I had never noticed this but the "ALT commands" are actually listed in Windows' Character Map... There we can see that all the greek letters can be written, but not all of them have an associated ALT+...

    It seems only phi and mu does. For the others we have to copy/paste them :yuck:.

    ΩΣδφΔζ
     
  10. Jan 2, 2005 #9
    some functions

    α = alt + 224
    ß = alt + 225
    Γ = alt + 226
    π = alt + 227 (pi)
    Σ = alt + 228
    σ = alt + 229
    µ = alt + 230 (mu)
    τ = alt + 231 (tau)
    Θ = alt +233
    Ω = alt +234
    ∞ = alt +236
    ε = alt+238
    ∩ = alt+239
    ° = alt+248
    √ = alt+251 (squared root)
    ⁿ = alt+252 (exponant n)
    ² = alt+253
    ± = alt+241
    ≥ =alt+242
    ≤ =alt+243
    ┴ =alt+193

    i hope it's good enough :)
     
  11. Jan 2, 2005 #10

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    :eek:

    Thanks A I, you rock! :biggrin:

    But where did you find those codes?
     
  12. Jan 3, 2005 #11
    On dirai qu'il y en a un bon nombre de personnes qui parlent la langue Fraicaise dans ce forum!! :surprised
     
  13. Jan 3, 2005 #12
    Oui...je suis agréablement surpris. Mais de l'autre part, ça semble raisonnable puisqu'il y a presque seize mille membres.
     
  14. Jan 4, 2005 #13
    let's return to english please! :)
    i guess it's more modern
    if we want to share langage knowledge
    a suggestion can be made to open a langage section
    by the way, i also speak spanish and arabic other than english and french and some words in german
    but anyway it is a PHYSICSFORUM talking about different fields of SCIENCE
    let's just stay here!
    i guess it would be more than enough :)
     
  15. Jan 11, 2005 #14

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I tought I'd use this thread where people speak french to ask this question I haven't been able to find an answer to: What is the french translation for "Manifold"??

    Merci.
     
  16. Jan 11, 2005 #15
    "Tubulure" :)
     
  17. Jan 11, 2005 #16

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Where did you get that, a translation site?

    Google gives no result on "Tubulure Riemanienne" or even on "Tubulure+mathématique" so that can't be it.

    That would have been funny though. :biggrin:
     
  18. Jan 12, 2005 #17
    hey guys,
    we finally got our grades..
    the graded homework consisted of 3 exercises
    the first two were excellent.. i did them by myself..
    the third which i asked for help:
    it appeared that there is a mistake

    the slope of the trajectory is not -1 because the slope of the line is -1
    and not of the parabola
    therefore we can not use -1 as the slope
    i took 18/20 :)
    lol
    its good anyway
     
  19. Jan 14, 2005 #18
    Hi quasar...

    Je crois que la meilleure traduction de "manifold" c'est "varieté".
     
    Last edited: Jan 14, 2005
  20. Jan 14, 2005 #19

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ok, thanks clive!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Mechanice solving problem
  1. Problem to solve (Replies: 1)

  2. Problem Solving (Replies: 1)

  3. Problem solving? (Replies: 2)

Loading...