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Math Help - Help with Force Problem.

  1. #1
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    Help with Force Problem.

    Two traffic signals are temporary suspended from a cable as shown. Knowing that the signal B weighs 300 Newtons, determine the weight of signal C.

    (I know how to compute vectors and how to find them. All I need to know is which force vectors are used. And what is the equation. Just write out the vectors used.)
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by ThePerfectHacker
    Two traffic signals are temporary suspended from a cable as shown. Knowing that the signal B weighs 300 Newtons, determine the weight of signal C.

    (I know how to compute vectors and how to find them. All I need to know is which force vectors are used. And what is the equation. Just write out the vectors used.)
    I assume that all the strings are tight and the whole system is stationary i.e. the balls B and C are in equilibrium.
    Let the tension in the part AB of the string be T_{AB}(I don't know how to make overhead arrow, just assume it a vector). Its direction is from A to B. T_{BA} represents the same tension with direstion reversed.
    Equations are
    T_{BA}+T_{BC}+W_B=0
    T_{CB}+T_{CD}+W_C=0
    WEIGHTS ARE DIRECTED DOWNWARDS

    Malay
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  3. #3
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    Good that I understand.
    But how do I solve them?
    There are 5 variables over here and 2 equations.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    Good that I understand.
    But how do I solve them?
    There are 5 variables over here and 2 equations.
    Write them out in full, you have two 2-vector equations, which
    makes 4 scalar equations, and the unknowns are the absolute values of
    the three tensions and the weight of the signal at C.

    RonL
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  5. #5
    Super Member malaygoel's Avatar
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    Quote Originally Posted by ThePerfectHacker
    Good that I understand.
    But how do I solve them?
    There are 5 variables over here and 2 equations.
    Let us analyse first equation(you have two unknowns)
    Draw a 2D coordinate system in the plane with B as origin and x-axis parallel to the basei.e. road I think. you know the angles of the three vectors with the axes. For b to be in equilibrium, components of the forces along both the axes is zero(you will get two linear equations for two unknowns).
    After solving the first equation, you will two unknowns in the second.

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    Okay I got it now, thank you.
    I wrote them out in components form, thus I get 4 equations and 4 variables.
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    Thumbs up Hooray for applied maths

    Looks like we have a good thing going here with the applied math section.
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  8. #8
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    Quote Originally Posted by ThePerfectHacker
    Two traffic signals are temporary suspended from a cable as shown. Knowing that the signal B weighs 300 Newtons, determine the weight of signal C.

    (I know how to compute vectors and how to find them. All I need to know is which force vectors are used. And what is the equation. Just write out the vectors used.)
    There are four unknown forces--the tensions on the 3 sections of the cable, and that one due to the weight of signal C.
    To find that force due to the weight of signal C, find a way to eliminate or to not include the 3 tensile forces, leaving you to deal only with the given 300N signal B and the unknown signal C.
    This said way is to take moments at the intersection of the tensile forces at AB and CD. The 3rd tension is not even a problem here.
    The system is static, and in equilibrium. Imagine cutting the cable at two points: anywhere along AB and along CD, so you have a free body diagram of the system.
    Extend AB and CD to intersect at point, say, E. The figure now looks like a triangle BEC. Draw the forces acting at points B and C.
    At point B: the tensile force along AB, and the downward vertical force 300 newtons.
    At point C: the tensile force along CD, and the downward vertical force, say, X newtons.

    Since the FBD is in equilibrium, then the moments of all the forces about any point are equal to zero. (Moment = force times perpendicular distance.) Take the moments about point E.
    ---Moments due to the two tensile forces are zero--because the perpendicular distance of each from point E is zero.

    ---Moment due to the 300N at point B:
    The direction of the 300N is vertical, so its perpendicular distance from point E is a horizontal distance. Call this horizontal distance between points B and E as "a" meters.
    Hence, moment due to 300N is (300*a) newton-meters, counterclockwise.

    ---Moment due to the X newtons at point C:
    The direction of the X is vertical also, so its perpendicular distance from point E is a horizontal distance too. Since the given horizontal distance between points B and C is 3.4 m, then the horizontal distance of X from point E is (3.4 -a) meters.
    Hence, moment due to the X is X*(3.4 -a) newton-meters, clockwise.

    Then, the summation of moments being zero, all clockwise moments equal all counterclokwise moments,
    X(3.4 -a) = 300*a
    So,
    X = (300a)/(3.4 -a) newtons ------------(1)

    You need to find "a" then. One way of doing that is by using "equations of the lines" of AB and CD, by their point-slope forms.
    (y -y1) = m(x -x1) ----(i)
    m = (y2 -y1)/(x2 -x1) = (y1 -y2)/(x1 -x2) ---(ii)
    Blah, blah, blah.

    You should get a = 0.8353 meter.

    Therefore, depending on what precision you want, you should get X = about 92 to 98 newtons. ------------answer.

    (All the given dimensions are in one decimal place only, so you are to use numbers in your computations in one decimal place only?)
    Last edited by ticbol; September 8th 2006 at 08:38 PM. Reason: line CD, not BD.
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  9. #9
    Super Member malaygoel's Avatar
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    Quote Originally Posted by ThePerfectHacker
    Good that I understand.
    But how do I solve them?
    There are 5 variables over here and 2 equations.
    You can apply Lami's Theorem

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    Malay
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