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Math Help - Problem I have been struggling with

  1. #1
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    Problem I have been struggling with

    I am new to this forum and I really could use some help if any one has the answer please let me know how you got it

    The magnitudes FA=40 N and FB=50 N FC=40 N
    the angles alpha=50 deg and beta = 80 degrees
    Determine the magnitude of FA+FB+FC

    The book has the answer as 83 N and I have not clue on how they got this answer.

    Can anyone help
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jaycas21
    I am new to this forum and I really could use some help if any one has the answer please let me know how you got it

    The magnitudes FA=40 N and FB=50 N FC=40 N
    the angles alpha=50 deg and beta = 80 degrees
    Determine the magnitude of FA+FB+FC

    The book has the answer as 83 N and I have not clue on how they got this answer.

    Can anyone help
    I think you will have to provide a bit more information.

    1. What does the N denote (Newtons)?

    2. The angles alpha and beta, what are they the angles between?

    RonL
    Last edited by CaptainBlack; January 15th 2006 at 11:37 AM. Reason: spelling
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  3. #3
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    I got your question in my PM.
    But you need to draw a triangle and explain better.
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  4. #4
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    OK here is the clearification for the question posted

    OK attached to the link is the question I have 2.3 and 2.6 . I am having a hard time setting up the problem. Please click on the question link
    Question
    Attached Thumbnails Attached Thumbnails Problem I have been struggling with-scan0002.jpg  
    Last edited by MathGuru; January 18th 2006 at 05:32 PM.
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  5. #5
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    Remember that if vectors \vec{V_1},\vec{V_2} have magnitudes a,b respectively, then,
    ||\vec{V_1}+\vec{V_2}||^2=a^2+b^2-2ab\cos \alpha where \alpha is one of the angles. (this is the law of cosines).
    Last edited by ThePerfectHacker; January 18th 2006 at 08:01 PM.
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  6. #6
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    Quote Originally Posted by jaycas21
    I am new to this forum and I really could use some help if any one has the answer please let me know how you got it

    The magnitudes FA=40 N and FB=50 N FC=40 N
    the angles alpha=50 deg and beta = 80 degrees
    Determine the magnitude of FA+FB+FC

    The book has the answer as 83 N and I have not clue on how they got this answer.

    Can anyone help
    That is suppopsed to be your problem/question, which is 2.4 per your attached figure below, but in that posting below, you said your problems are 2.3 and 2.6.
    So, which or what is really the problem you want solved?
    2.4, or 2.3 and 2.6 ?

    --------------------------------
    Let us say it is 2.4.

    It is mentioned there that you solve it graphically, meaning, using drawing instruments and scales, plot those data on paper, then:
    ---get the resultant of FA+FB using the parallelogram method.
    ---then get the resultant of [(FA+FB) resultant] +FC, using the parallelogram method again.
    This second resultant is (FA+FB+FC). Using the scale, it should be about 83N according to the book.

    Analytically, or using Trigonometry, we can get the exact (FA+FB+FC).
    Let us that resultant R. Or, R = (FA+FB+FC), in vectors.

    In the figure, FA looks horizontal, so I assume it to be horizontal.
    We put the origin (0,0) of the x,y axes setup at the ring of the eye-screw or at the common point of the 3 vectors. So if FA is on the x-axis, then FB is 50 degrees counterclockwise from the x-axis, and FC is (50+80=) 130 degrees counterclockwise from the x-axis.
    Or, FA is on the x-axis, FB is in the 1st quadrant, and FC is in the 2nd quadrant.

    ---Horizontal component of R = Sum of horizontal components of FA,FB,FC.
    Rh = (FA)h +(FB)h +(FC)h
    where h is read "sub h" and it signifies the horizontal component of the vector.
    Rh = FA +(FB)cos(50deg) -(FC)cos(50deg)
    FC is 50 degrees clockwise from the negative side of the x-axis.
    Rh = 40N +(50N)cos(50deg) -(40N)cos(50deg)
    The author of this problem hates 50.
    Using a calculator,
    Rh = 40N +32.1394N -25.7115N
    Rh = 46.4279N ---------------------***

    ---Vertical component of R = Sum of vertical components of FA,FB,FC.
    Rv = (FA)v +(FB)v +(FC)v
    Rv = (FB)sin(50deg) +(FC)sin(50deg)
    Since FA is horizontal then it has no vertical component.
    Rv = (50N)sin(50deg) +(40N)sin(50deg)
    Rv = (90N)sin(50deg)
    Rv = 68.944N --------------***

    Then, by Pythagorean Theorem,
    R^2 = (Rh)^2 +(Rv)^2
    R = sqrt[(46.4279N)^2 +(68.944N)^2]
    R = 83.12N ------------------------------answer.
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