Results 1 to 6 of 6

Thread: statics

  1. #1
    Senior Member
    Joined
    May 2016
    From
    england
    Posts
    377
    Thanks
    3

    statics

    a particle of weight w is attached to two light inextensible strings each of length 40cm. the other ends of the strings are attached to two points on the same horizontal ceiling 64 cm apart. find the tension of the strings in terms of w

    so i am trying to solve this resolving forces vertical and horizontal and then equaling them to each other. but its not working out the answer is 5/6 w. i can post my method is people need it
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,337
    Thanks
    2859

    Re: statics

    You have an isosceles triangle with sides of length 40, 40, and 64. Dropping a perpendicular divides that into two right triangles with hypotenuse of length 20 and horizontal leg of length 32 so the vertical leg has length \sqrt{32^2- 20^2}= \sqrt{624}= 4\sqrt{39}. Taking the tension in each string to be 'T', the vertical and horizontal components are \frac[4\sqrt{39}}{40}T= \frac{\sqrt{39}}{10}T and \frac{32}{40}T= \frac{4}{5}T. The total of the two vertical components is \frac{\sqrt{39}}{5}T and that must be equal to the weight, w. That is, T= \frac{5}{\sqrt{39}}w , not 5/6 w, though it is pretty close, 0.801w as compared to 5/6w= 0.833w.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    5,768
    Thanks
    2417

    Re: statics

    Quote Originally Posted by HallsofIvy View Post
    You have an isosceles triangle with sides of length 40, 40, and 64. Dropping a perpendicular divides that into two right triangles with hypotenuse of length 20 and horizontal leg of length 32 so the vertical leg has length \sqrt{32^2- 20^2}= \sqrt{624}= 4\sqrt{39}. Taking the tension in each string to be 'T', the vertical and horizontal components are \frac[4\sqrt{39}}{40}T= \frac{\sqrt{39}}{10}T and \frac{32}{40}T= \frac{4}{5}T. The total of the two vertical components is \frac{\sqrt{39}}{5}T and that must be equal to the weight, w. That is, T= \frac{5}{\sqrt{39}}w , not 5/6 w, though it is pretty close, 0.801w as compared to 5/6w= 0.833w.
    I'm don't think this is correct.

    You do have a 40,40,64 isosceles triangle but splitting that in two gets you a 32,24,40 triangle
    Last edited by romsek; Apr 25th 2017 at 10:52 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,831
    Thanks
    1083

    Re: statics

    I get the vertical leg has length $\sqrt{40^2-32^2} = \sqrt{576}=24$. Thus the vertical components are $\dfrac{24}{40}T = \dfrac{3}{5}T$. Thus, $\dfrac{6}{5}T = w \Longrightarrow T = \dfrac{5}{6}w$. I think the book's solution is correct.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,106
    Thanks
    3647

    Re: statics

    $2T\sin{\theta} = W$

    $2T \cdot \dfrac{3}{5} = W$

    $T = \dfrac{5W}{6}$
    Attached Thumbnails Attached Thumbnails statics-equilibrium_f.jpg  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    May 2016
    From
    england
    Posts
    377
    Thanks
    3

    Re: statics

    Thanks i was doing it wrong
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Statics
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: Mar 8th 2017, 03:08 AM
  2. Statics
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Aug 21st 2009, 05:14 AM
  3. Statics
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Mar 20th 2009, 03:24 AM
  4. STatics Help!!
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Feb 20th 2009, 02:55 AM
  5. 3-D Statics
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Dec 15th 2007, 07:16 AM

/mathhelpforum @mathhelpforum