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Math Help - Resultant force question

  1. #1
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    Resultant force question

    hello, im just working through a structures past exam paper just wondering if somebody could check my work, its the holidays and i have no work this is next years work that im just getting a head start on so answers would not be classed as cheating in this case . so if im wrong please correct me, no ones taught me this so far im just working from any examples i could find online, thank you

    bobby

    ive attached what i have done in a word document.
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  2. #2
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    Re: Resultant force question

    If you are using the standard convention, that positive x is to the right, then it should be obvious that the x-component and y-component of F1 are negative, not positive. (The angle you need to use is the angle with the positive x-axis which, for F1, is 180+ 40= 220 degrees, not just 40.)
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  3. #3
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    Re: Resultant force question

    in kN ...

    \sum{F_x} = 50\cos(30)-100\cos(60)-25\cos(40) < 0

    \sum{F_y} = -50\cos(30)+100\cos(60)-25\cos(40) > 0

    your magnitude is fine, however, note the direction of the resultant force is in quad II

    \theta = \arctan\left(\frac{F_y}{F_x}\right) + 180
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    Re: Resultant force question

    ok so what your saying is that my table is correct up to the totals? also the f1 coloum is defintly postive not negative because im comparing that to a previous example. also if my resultant force is wrong how do i calculate it. i understand you have to do total of vertical squared + total of vertical squared , then square root answer, but my problem is this dosent work with a negative value, so what i have done is minused the totals as well as squaring, is this correct? i understand what is being said about the angle however when i wored it out useing your numbers the angle seems far to high compared to other examples i have. also does the direction of the arrows have any influence such as making the vertical or horizontal components negative or postive?

    thank you
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  5. #5
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    Re: Resultant force question




    note that a vector that has direction in quad I has positive x and y components

    quad II ... negative x and positive y components (that would be the 100kN source)

    quad III ... both x and y components are negative (the 25 kN force)

    quad IV ... x component is positive, y component is negative (the 50 kN force)

    finally, the resultant angle is normally referenced to the positive x-axis.


    recommend you view this series of videos as a tutorial on dealing with vectors ...

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