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Vector multiplication in static equilibrium

  1. Sep 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A 1 200-N uniform boom at 65 degrees to the vertical is
    supported by a cable at an angle of 25 degrees to the horizontal
    . The boom is pivoted at the bottom, and an object of weight = 2 000 N
    hangs from its top. Find the tension in the support
    cable

    2. Relevant equations


    3. The attempt at a solution
    I have the answer but I just don't understand how it works

    image.jpg

    What does blue square time red square mean? The red square means horizontal tension and the blue square means the vertical distance between the cable attaching point and the ground (as shown in the diagram below), why do you multiply them with each other?
    20150912_235957.jpg
     
  2. jcsd
  3. Sep 12, 2015 #2

    SteamKing

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    Do you understand how to calculate the moment of a force about the pivot point of the boom?
     
  4. Sep 13, 2015 #3
    Net torque = 0
    Tension from the cable = the hanging weight
    like this? I don't know.

    I don't understand how you can calculate the torque just by using horizontal tension and the vertical distance instead of a force that is perpendicular to the surface.
     
  5. Sep 13, 2015 #4

    SteamKing

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    What surface? You're trying to calculate the torque created on the beam by the tension in the cable, which has two parts:

    1. The moment arm is the vertical distance (3 * L / 4) * sin(65°) between the attachment point of the cable and the pivot, and the force is the horizontal component of the tension in the cable, T*cos(25°).

    2. The moment arm is the horizontal separation of the attachment point of the cable to the boom and the pivot point, (3 * L / 4) * cos(65°), and the force is the vertical component of the cable tension, T * sin (25°).

    The calculation of the total torque about the boom pivot is split in this manner due to the definition of the moment of a force about an axis of rotation.

    In the diagram below, the moment of the force F about O is M = F × d:

    IMG00002.GIF
     
  6. Sep 13, 2015 #5
    ok I have finally get it. Thanks!
     
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