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Math Help - I need some help

  1. #1
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    Exclamation I need some help

    I am currently working on some of these review problems in the section I have been able to solve some. I am having a problem setting up these three problems. On the attachment below you will find problem 2.148, 2.150 & 2.154 these are the ones which I am kind of stuck on.

    Answers are:
    2.148= 1110 & 29.7 deg
    2.150 = 313 & 140
    2.154 = 8.78 -6.59j-8.78k

    Can some help me set up the problems?

    Thanks
    Jay

    Problem 2.148 and 2.150

    Problem 2.154
    Attached Thumbnails Attached Thumbnails I need some help-first.jpg   I need some help-second.jpg  
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  2. #2
    Grand Panjandrum
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    Question: 2.148.

    Resolve all three forces in to horizontal and vertical components:

    B=1500(\cos (50^{\circ}),\sin   (50^{\circ}))
    A=X(-\cos (\alpha ), -\sin (\alpha))
    W=600(0,-1)

    Then as A+B+W=\vec 0, we have a pair of simultaneous equations:

    1500 \cos (50^{\circ}) - X\cos (\alpha )=0,
    1500 \sin (50^{\circ}) - X\sin (\alpha )-600=0,

    which can be solved fairly easily.

    RonL
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  3. #3
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    I am not getting the answer of 1110

    1500sin(50)-600= Xsin(a)
    1149-600=549sin(a)

    I can not wait to finish this class.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by jaycas21
    I am not getting the answer of 1110

    1500sin(50)-600= Xsin(a)
    1149-600=549sin(a)

    I can not wait to finish this class.
    X\cos(\alpha)=1500\cos(50^{\circ})\approx 964.18
    X\sin(\alpha)=1500\sin(50^{\circ})-600\approx 549.07

    Dividing the second by the first gives:

    \tan (\alpha)=\frac{549.07}{964.18},

    so:

    \alpha = 29.66^{\circ}

    Substituting \alpha back into either of the equations above gives:

    X=1109.55.

    RonL
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  5. #5
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    With question 2.150 would I basically set it up the same way.

    I was going to place coordinates based on the horizontal and vertical axis's then figure out the components and find the magitude for one and basically subtract from D+ E+F=0

    what do you think.

    That a lot for the help its been great
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by jaycas21
    With question 2.150 would I basically set it up the same way.

    I was going to place coordinates based on the horizontal and vertical axis's then figure out the components and find the magitude for one and basically subtract from D+ E+F=0

    what do you think.

    That a lot for the help its been great
    Its exactly the same method, except that you are not told the angle, but
    you can find it form the dimensions of the truss (in fact you don't need the
    angle itself but the cos and sin of the angle which can be found directly
    with the aide of Pythagoras's theorem).

    So resolve the forces into horizontal and vertical components, then the
    sum of the horizontal components is zero, as also is the sum of vertical
    components.

    This will give you two equations in the magnitudes of forces E and F.
    These equations should be easier to solve than those we did before.

    RonL
    Last edited by CaptainBlack; February 2nd 2006 at 10:18 AM.
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  7. #7
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    Thus far I got

    Sqrt of100^2+50^2= 112

    Cosa=26.7
    Sina= 26.5

    280sin(26.5)=124.9
    but the answer in the book is F=140
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  8. #8
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    Quote Originally Posted by jaycas21
    Thus far I got

    Sqrt of100^2+50^2= 112

    Cosa=26.7
    Sina= 26.5

    280sin(26.5)=124.9
    but the answer in the book is F=140
    I have \tan (\alpha)=50/100=0.5, so \alpha\approx26.57^{\circ}, and so \cos(\alpha ) \approx 0.8944
    and \sin(\alpha ) \approx 0.4473.

    Then:

    D=280(-1,0)
    F=X(0,1)
    E=Y(\cos (\alpha ), -\sin (\alpha )),

    so D+E+F=0 gives us:

    -280+\cos (\alpha )Y=0
    X-\sin (\alpha)Y=0.

    Rearranging:

    \cos (\alpha )Y=280
    \sin (\alpha ) Y =X

    Dividing the second by the first gives:

    \tan (\alpha ) =X/280,

    so X=140 pounds, and Y=X/\sin (\alpha) \approx 140/0.8944=156.5pounds

    Opps, that last one should be Y=X/\sin (\alpha) \approx 140/0.4473=313.0pounds


    RonL
    Last edited by CaptainBlack; February 3rd 2006 at 06:22 AM.
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  9. #9
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    Right! I was able to get X=140 as I was working on it last night, but it still doesn’t make sense to me because the if the D=280 and F= 140 then would E also equal 140 to make the equation zero. I was trying it out this morning and I was trying to figure why the give the answer of E as 313

    Again thank so much for the help I am really trying hard on this class but it is kicking my butt. I have dynamics next.
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