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Statics: tension in cable holding gate closed

  1. Aug 3, 2015 #1
    Problem580Figure2.jpg Problem580Figure2.jpg Problem580Figure2.jpg Problem580Figure2.jpg 1. The problem statement, all variables and given/known data
    The tension in the cable is 800LBs. Find the depth of water that produces this tension. The gate is hinged at B; the cable is attached at A. (Figure 1)



    2. Relevant equations


    3. The attempt at a solution
    OK, so this is just 2 opposing lever arms: the tension in the cable is pulling the gate closed (clockwise) and the pressure of the water is trying to open the gate (counter-clockwise).
    So we start on the easy side.

    Given the geometry, we know that the gate forms a 3/4/5 triangle and therefore the angle between the gate and the vertical is 36.8 degrees (Figure 5).
    An 800-LB pull on the cable translates into a 480-LB pull perpendicular to the gate (Figure 2).
    The cable's lever arm is therefore (480)(2.5)=1200 Ft-LBs.

    This is opposed by the force of the water.

    OK, so I cheated and looked at the answer in the book, and I still cannot see how they got this answer.

    The book says that the water is 5.18 feet deep; we call this h.
    h/3 is 1.727 feet; this is the centroid of the body of water.
    So 1.727 feet from the bottom takes us .727 feet into our 3/4/5 triangle (Figure 4), which means that at this point the lever arm for the water is 1.6 (Figure 3).
    The force of the water is (1/2)(gamma x h)(h) which is (1/2) (62.4 x 5.18)(5.18).
    This equals 837 Lbs/Ft squared.
    The torque exerted by the water is therefore (837)(1.6) = 1339 Ft-Lbs

    And we were expecting 1200.

    Where did I make a wrong turn?
    Thank you in advance for any help:smile: Problem580Figure1.jpg Problem580Figure3.jpg Problem580Figure4.jpg Problem580Figure5.jpg Problem580Figure2.jpg
     
  2. jcsd
  3. Aug 3, 2015 #2
    Could you draw the 5.18ft on your diagram? It is not clear where this dimension is taken from.

    Also, how wide is the gate?
     
    Last edited: Aug 3, 2015
  4. Aug 4, 2015 #3
    The depth of the water, which they are consistently calling h in the book, is supposed to be 5.18 feet to create the 800-LB tension in the cable.

    For some reason, I am unable to upload another diagram, so I parked the diagram here:
    http://www.firstbyteashville.com/images/problem580fig6.jpg
    In this diagram, which applies to a similar problem, they show a 6-foot-deep pool of water.

    In other similar problems in the book, when they do not specify otherwise, we assume the gate is 1 foot wide. So I have made that assumption here.


    Thank you again in advance.
     
  5. Aug 4, 2015 #4
    First, I think the force you calculated is too high. You calculated the total horizontal load acting on a vertical surface that is 5.18ft high. You want the portion of the force acting on the gate only. See if you can come up with a better number for this.

    For the center of pressure yp there is a formula that involves the moment of inertia I0:
    yp = yc + I0/[A(yc)]

    Have you seen this formula before?
     
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