Results 1 to 10 of 10

Math Help - Slider crank

  1. #1
    Newbie
    Joined
    Aug 2011
    Posts
    24

    Slider crank

    Is anyone able to point me in the right direction to answering these questions please.

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,536
    Thanks
    778

    Re: Slider crank

    Surely you can solve b). Point a) can be solved with a simple application of the Pythagoras theorem.

    In general, let P' be the projection of P on OR. If you know angle POR, then you an find OP' and PP' using trigonometric functions. Knowing PP', you can find P'Q using the Pythagoras theorem. Therefore, you can find OQ.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2011
    Posts
    24

    Re: Slider crank

    Thank you. I will try to review trignomometric functions.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Aug 2011
    Posts
    60

    Re: Slider crank

    Is the answer for (b) = 3.25 - 1.5 = 1.75
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Slider crank

    Quote Originally Posted by arangu1508 View Post
    Is the answer for (b) = 3.25 - 1.5 = 1.75
    1. What you have calculated is the distance |\overline{OQ}| if P is placed on the left end of the diameter of the circle.

    2. As far as I understand the question you are asked to determine the length of the way the point Q is traveling on the line OR if the point P describes a semicircle. And that must be a complete diameter, that means the answer is 1.5 m.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2011
    Posts
    24

    Re: Slider crank

    Thank you once again earboth, with your help I was able to understand the question. Hopefully I have answered them correctly (although you have already confirmed b)

    a) Initially Q is @ 3.2 m after P rotates 90 degress using the pythagoras theorem we find

    c^2=b^2-a^2
     sqrt(2.5^2-0.75^2)= 2.38 m

    Distance the end Q travelled = 3.2 m - 2.38 m = 0.82 m

    b) Distance the end Q travelled (half a turn) 2 x .75 = 1.5 m

    c) using the sine rule we find sin(30) c = 3.12 m
    using the sine rule again we find sin(170) c =1.76 m

    Distance the end Q travelled 1.36 m

    If I have made any mistakes please let me know. Thank you.
    Last edited by Dcoz; August 29th 2011 at 11:16 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,536
    Thanks
    778

    Re: Slider crank

    Quote Originally Posted by Dcoz View Post
    a) Initially Q is @ 3.2 m after P rotates 90 degress
    It's 3.25 m, so the traveled distance in a) is 0.87 m, not 0.82 m.

    Quote Originally Posted by Dcoz View Post
    c) using the sine rule we find sin(30) c = 3.12 m
    using the sine rule again we find sin(170) c =1.76 m

    Distance the end Q travelled 1.36 m
    I don't understand this. First, what is c? 3.12 / sin(30) is not the same as 1.76 / sin(170). Second, sin(30) determines the vertical position of P, not the horizontal position of Q.

    As I said, one way is to find PP' and OP' (where P' is the projection of P) and then use Pythagoras theorem to find P'Q; then OQ = OP' + P'Q.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123

    Re: Slider crank

    Quote Originally Posted by Dcoz View Post
    Thank you once again earboth, with your help I was able to understand the question. Hopefully I have answered them correctly (although you have already confirmed b)

    a) Initially Q is @ 3.2 m after P rotates 90 degress using the pythagoras theorem we find

    c^m2=b^2-a^2
     sqrt(2.5^2-0.75^2)= 2.38 m

    Distance the end Q travelled = 3.2 m - 2.38 m = 0.82 m

    b) Distance the end Q travelled (half a turn) 2 x .75 = 1.5 m

    c) using the sine rule we find sin(30) c = 3.12 m
    using the sine rule again we find sin(170) c =1.76 m

    Distance the end Q travelled 1.36 m

    If I have made any mistakes please let me know. Thank you.
    1. I'll take emakarov's suggestions. I only use some different labels. (see attachment. I've modified your image a little bit)

    2.
    • |\overline{OF}| = p = 0.75 \cdot \cos(\theta)
    • |\overline{PF}| =  0.75 \cdot \sin(\theta)
    • |\overline{FQ}| = q = \sqrt{2.5^2-(0.75 \cdot \sin(\theta))^2}


    The traveled distance w:

    w = 3.25 - (p+q)

    3. The advantage of emakarov's method is that the traveled distance w depends only on the angle of rotation \theta. So you only have to plug in the angle to get the distance from the start-point.
    Attached Thumbnails Attached Thumbnails Slider crank-slidercrank.png  
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Aug 2011
    Posts
    24

    Re: Slider crank

    Thank you both for your help on this one. I'm not 100% certain I have got the answer to c) right

    this is what I have

    ϴ = 30˚ p + q = 0.6495 + 2.848 = 3.1336
    ϴ = 170˚ p + q = -0.7386 + 2.4954 = 1.756
    3.1336 1.756 = 1.38 m
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Aug 2011
    Posts
    60

    Re: Slider crank

    Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Numerical analysis with Crank-Nicolson
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: June 9th 2011, 03:55 AM
  2. Replies: 0
    Last Post: April 8th 2010, 05:45 AM
  3. Velocity of a slider?
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: November 11th 2009, 03:40 AM
  4. crank-nicolson for 2D nonlinear Fokker-Planck
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: September 20th 2008, 12:53 PM
  5. Crank Nicolson Scheme For Heat Equation
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: August 23rd 2006, 05:26 AM

Search Tags


/mathhelpforum @mathhelpforum