Results 1 to 7 of 7

Math Help - Multi - Balancing An Object

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    221

    Multi - Balancing An Object

    Suppose we have a landmass of an island that needs to be balanced by an elephant using its tusk. We need to find the right place to place the elephant's tusk to balance it. The center of the universe is some sdistance away from this island (in the same plane as this island). If we measure from the center of this universe, the island is bounded by the following functions:

    y = x^2 + 4 and y = x + 6

    If the density of the island at point (x,y) is the inverse of the square (reciprocal) of the distance from the x-axis, where should the tusk be placed in order to balance the island?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Look below. I will not solve, I will set up.
    Attached Thumbnails Attached Thumbnails Multi - Balancing An Object-picture5.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by ThePerfectHacker View Post
    Look below. I will not solve, I will set up.
    Hello, and thanks in advance for your time

    I'm familiar with the formula

    int(int_R f(x,y)))dA which is what you have but "D" .. same thing.

    Now, I don't know why you have an extra "x" and "y"..only time I've seen this is when finding the volume using polar coordinates, where the dA then becomes r*dr*d(theta) .. a variation of fubini's theorem.

    Then, I see how you got the region by:

    x^2 + 4 <= y <= x + 6.. but don't see how you got the -2 <= x <= 3 ..

    And further, I have no idea what (delta, I think it is)*x = k (curvature?)/x^2 is ... and then you have:

    x_bar = M_y/M and y_bar = M_x/M (this reminds me of the chinese remainder thm, but obviously this isn't number thry!). I suspect it's "mass" . This part got me lost.. the final step in your work is the one that kind of makes sense:

    M = int(int_R f(x,y)))dA, but what is our f(x,y) .. and limits of integration? That'd be the region you described? We'd use fubini's theorem to compute this, such that:

    int (from c to d) of int (from a to b) of f(x,y) dy dx

    OR, obviously, int (from a to b) of int (from c to do) of f(x,y) dx dy

    Thanks for the help, and I apologize for not understanding your work.

    EDIT: Ohh!! I kind of see now how you got the M_x, M_y..

    I remember this being tricky..

    M_x = mass of lamina ... the x-coordinate of the center of mass.. = M_y/m .. and then similarly for M_y = M_x/m..

    Now it makes sense why you had the extra x, y.. but I still don't get what limits correspond to which integral and what f(x,y) is.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2006
    Posts
    221
    Take 2:

    So now that it makes a little more sense to me, would we have the following:

    M_x/m (where m is the total mass, and M_x is the moment with respect to the x-axis) = int*int_R y*f(x,y) dA

    And then for M_y/m = int*int_R x*f(x,y) dA

    But first we want to find what m is, which I believe is just

    int*int_R f(x,y) dA ..

    So this should be simple enough I think after I know what f(x,y) is.. and how you got the limits of integration.

    Going back to the word prob., what exactly does it mean by:

    "If the density of the island at point (x,y) is the inverse of the square (reciprocal) of the distance from the x-axis..." isn't that just f(x,y)? Is that how you figure out f(x,y)?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Ideasman View Post

    x^2 + 4 <= y <= x + 6.. but don't see how you got the -2 <= x <= 3 ..
    I think I might have made a mistake.
    Attached Thumbnails Attached Thumbnails Multi - Balancing An Object-picture9.gif  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2006
    Posts
    221
    Ahh okay, so it is -1 <= x <= 2. Thanks for confirming it !

    So now the only barrier stopping me from doing the various integrals and finding M_y/M and M_x/M is finding what f(x,y) is .. you have the delta*x = k/x^2 and I have no idea what that is.. (I hope your "k" doesn't mean curvature, or then you totally lost me).

    that "inverse of the square" wording is very confusing, and I have no idea how to come up with a function based on that information. Once I find this out the rest will follow from there, since I will have the func, and then finding x_bar and y_bar are easy, along with mass which is all the info. I need for the problem.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Ideasman View Post
    Ahh okay, so it is -1 <= x <= 2. Thanks for confirming it !

    So now the only barrier stopping me from doing the various integrals and finding M_y/M and M_x/M is finding what f(x,y) is .. you have the delta*x = k/x^2 and I have no idea what that is.. (I hope your "k" doesn't mean curvature, or then you totally lost me).
    "k" is just a constant function (positive). Like 2 or 3 or pi.
    Just treat it like a number. In the end it shall cancel.

    that "inverse of the square" wording is very confusing, and I have no idea how to come up with a function based on that information. Once I find this out the rest will follow from there, since I will have the func, and then finding x_bar and y_bar are easy, along with mass which is all the info. I need for the problem.
    Inverse - as in "inversely proportional".
    Thus, an inversely proportional square is 1/x^2

    But you need to include "k" in front because there is also something called: the konstant of proportion.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Pole Balancing
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: May 7th 2011, 06:01 PM
  2. linear balancing
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 3rd 2011, 12:48 PM
  3. [SOLVED] Balancing an Equation
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 18th 2010, 01:24 PM
  4. Balancing Coins
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: September 24th 2010, 01:28 PM
  5. Replies: 0
    Last Post: April 25th 2010, 08:23 AM

Search Tags


/mathhelpforum @mathhelpforum