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Equivalent resultant force

  1. Jan 9, 2016 #1
    1. The problem statement, all variables and given/known data
    i am asked to find the equivalent resultant force for this question . for the distributed load , i divided into 4 parts , is my method correct?

    2. Relevant equations

    3. The attempt at a solution
    total force about y = 6sin30 + 3(3) + (5-3)(0.5)(3) + 5(4) + (7-5)(4)(0.5)

    total force about x = 6cos 30

    Attached Files:

  2. jcsd
  3. Jan 9, 2016 #2


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    What do you mean by "about x" etc.? You seem to be confusing forces and moments. The 6 sin 30 is a force in the y direction, but the 3(3) looks more like a moment, though I can't match it to anything in the diagram (there is no 3kN force anywhere), and the 5(4) might be the moment of the 5kN force at D about E.
    Please identify what each term represents.
  4. Jan 9, 2016 #3
    3(3) + (5-3)(0.5)(3) + 5(4) + (7-5)(4)(0.5) , each term here represent the different region for the load distribution diagram , i already stated it in the diagram
    sorry , i mean forec along x and y axis
  5. Jan 9, 2016 #4


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    You did indeed, but I still cannot see how you are getting these terms.
    First, is this equation supposed to be for forces in the x direction, forces in the y direction, or moments about some point? If moments about a point, what point?
    Secondly, how are you getting 3(3)? I cannot relate that to anything in the diagram. There is no force of 3kN. Perhaps this stands for the total force from the 3kN/m over the 3m from C to D, but that doesn't make sense because the force density increases from 3kN/m at C to 33/7kN/m at D.
    Thirdly, where does the (5-3) come from? Are the 5 and the 3 distances or forces?
    The 5(4) looks like the moment of the 5kN force at D about E. Please confirm.
    I can't make sense of the 7-5 term either.
  6. Jan 9, 2016 #5


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    You really should learn the formula for calculating the area of a trapezoid:

    It saves quite a bit of time. If b1 or b2 is zero, this formula reduces to that of finding the area of a triangle, A = (1/2)hb.

    You should be aware that the load shown at point D is a 5 kN concentrated load. It is a separate load from the distributed load.

    The value of the distributed load at point D is not 5 kN/m. The distributed loading changes by 4 kN/m over a distance of 7 m, so the value of the distributed loading at point D is calculated by linear interpolation. This step is unnecessary if you know how to calculate the area of a trapezoid, as shown above.

    There is also a couple of magnitude 5 kN-m acting CCW shown at point C, which doesn't seem to appear in your calculations.

    The problem statement as given is confusing, as haruspex noted. Please provide the exact wording for this problem from your text.
  7. Jan 9, 2016 #6
    so the total forecs along y should be = 3(7) + 5 + 0.5(4)(7 ) ? , since you aid that 5N is concentrated load....
  8. Jan 9, 2016 #7


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    Yes, though I would have done it as 5+(3+4)(3+7)/2, the (3+7)/2 being the average spread load.
    SteamKing mentioned the 5kNm torque, but since that is a pure torque it does not feature in the net force expressions.
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