Results 1 to 10 of 10

Math Help - A typical compound archery bow requiers

  1. #1
    Senior Member
    Joined
    Oct 2009
    Posts
    290

    A typical compound archery bow requiers

    A typical compound archery bow requiers a force of 133 N to hold an arrow at "full drwa " ( pulled back 71 cm ). ssunimg that the bow obeys hooke's law . what is its spring constant ?

    did conver 71 to m
    did use k = W/x to find spring constant ?

    plese help me ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    Yes, that's right. Hooke's law says that the force on a spring is given by the spring constant times the distance the spring is extended: F= kx. So k= F/x or your W/x= 71 N/.71 m= 100 N/m.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Oct 2009
    Posts
    290
    Hi

    F =W/x= 133N/.71 m= 1.87N/m.

    why you make F =W/x= 71 N/.71 m = 100

    ??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Oct 2009
    Posts
    290
    plese I want your help .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,807
    Thanks
    116
    Quote Originally Posted by r-soy View Post
    Hi

    F =W/x= 133N/.71 m= 1.87N/m. <== this is obviously wrong (Your arrow has a length of 71 m!)

    why you make F =W/x= 71 N/.71 m = 100 <== this is obviously a typo

    ??
    Quote Originally Posted by r-soy View Post
    plese I want your help .
    The correct answer should be k = 187\ \frac Nm
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Oct 2009
    Posts
    290
    187 X 133 /71 = 350.2

    please help me

    this answer is correct answer ??

    and what is the uint ??

    and from where we got the 187 ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member RHandford's Avatar
    Joined
    Sep 2010
    From
    N.E. Lincolnshire
    Posts
    70
    Yes I agree that this is the correct answer. The unit is Newton/meter. 187 is the result that has been calculated i.e. 133/.71
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Oct 2009
    Posts
    290
    I don't now from where we got 187 ??

    on the book writen
    K = mg/x , where k = force constant or spring constant
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,807
    Thanks
    116
    Quote Originally Posted by r-soy View Post
    A typical compound archery bow requiers a force of 133 N to hold an arrow at "full drwa " ( pulled back 71 cm ). ssunimg that the bow obeys hooke's law . what is its spring constant ?
    That's the question which should be answered. Nothing more, right?

    Quote Originally Posted by HallsofIvy View Post
    ... Hooke's law says that the force on a spring is given by the spring constant times the distance the spring is extended: F= kx. ...
    In other words: You need a force F to extend the spring by a distance d. And you need a constant k which refers to the properties of the individual spring. Hooke's law states then:

    F=k \cdot d~\implies~k=\dfrac Fd

    With your values:
    F = 133 N
    d = 0.71 m

    you'll get: k=\dfrac{133\ N}{0.71\ m}\approx 187\ \dfrac Nm

    And now your question is answered!

    Quote Originally Posted by r-soy View Post
    187 X 133 /71 = 350.2

    please help me

    this answer is correct answer ??

    ...
    What exactly do you want to calculate by that?
    I'll include the units into this term and you'll get:

    187\ \frac NM \cdot \dfrac{133\ N}{71\ cm}

    The result has the "unit" \dfrac{N^2}{cm \cdot m}

    That has definitely nothing to do with a compound bow.

    Quote Originally Posted by r-soy View Post
    I don't now from where we got 187 ??

    on the book writen
    K = mg/x , where k = force constant or spring constant
    This formula is only valid if a weight pulls at the spring (perpendicularly to the surface of Earth) and extend it.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Oct 2009
    Posts
    290
    thank you

    I have anothor quetion now I drwa graph between



    How do we find k from graph

    Are we calculate the slope then finid k

    like = 51 - 19 / 0.2 - 0.1 = 32/0.1 but convert 32 to m then 0.032/0.1 = 0.32 but what id here the unit of solpe ?

    and now find k = 1/0.032 =3.125 and what is the uint here ?

    plese check mu answer is correct and what is the uint of slope and k ?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 6th 2011, 12:53 AM
  2. Archery probability
    Posted in the Statistics Forum
    Replies: 4
    Last Post: March 17th 2011, 04:59 PM
  3. Curve fitting when typical regression models fail
    Posted in the Business Math Forum
    Replies: 3
    Last Post: February 3rd 2011, 07:38 PM
  4. [SOLVED] Planes &amp; Archery
    Posted in the Geometry Forum
    Replies: 2
    Last Post: April 1st 2009, 11:40 AM
  5. Typical Integration..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 7th 2008, 09:45 AM

Search Tags


/mathhelpforum @mathhelpforum