Statics: tension in cable holding gate closed

AI Thread Summary
The discussion centers on calculating the depth of water required to produce a tension of 800 lbs in a cable holding a gate closed. The tension creates a torque of 1200 Ft-Lbs, while the water's force generates a torque of 1339 Ft-Lbs, leading to confusion about the discrepancy. The book states the water depth should be 5.18 feet, and the participants discuss the geometry and assumptions regarding the gate's width and the calculation of forces. Clarifications are sought on the dimensions and the application of formulas related to center of pressure. The conversation emphasizes the importance of accurately determining the forces acting on the gate to resolve the problem.
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1. Homework Statement
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)

Homework Equations

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:
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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?
 
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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.
 
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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