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HOMEWORK 16.B - FA 23
17 thoughts on “HOMEWORK 16.B - FA 23”
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Think logically about what the value of the normal force at A should be when the gate is about to open. Combining this knowledge with the force (including the location) due to the water on the gate will help you solve for H.
Remembering the "2/3" rule for the location single-force equivalent load of a given linear load helps greatly with the logic for this problem, in addition to what Nathaniel said above.
Think about how the reaction force from A will have to reach an equilibrium with the force from the pressure to initially swing the gate open. This will help you identify the ratio for the height needed.
In order to find the relationshipo between the force from water and the force from A, we would relate them through the sum of moments, but how do we know what direction the moment of the water force should be without knowing explicitly what the height is?
The water pressure to the gate is perpendicular to the gate's surface and into the gate, this does not depend on the height. You can express the water pressure as one single force to the gate in the perpendicular direction, and use the right-hand rule to find the direction of the moment.
I'm having trouble understanding the Feq from the water pressure. Specifically how we are able to use a triangle to model the water pressure.
At first, It might seem like we have to use a rectangle to model the water pressure. However, This isn't true, as the water pressure at a particular point only depends on the point's depth. Think about putting your finger in an ocean. It doesn't feel the entire ocean's pressure and that point, but only the triangular force
It doesnt make sense to me how the water would flow out of the top of the gate if the majority of the hydrostatic pressure was below the pin location. It seems like the water would overcome the possible reaction force at Ax. The only way this would be possible is if there is water being pumped from the depth H.
As long as the location of the equivalent force is above point C and has sufficient magnitude (which it does if H is big enough), the gate can open. Think about the moment about point C.
Is anyone able to explain where the (1/2 p g H b)H comes for for the equivalent force
An interesting note about this problem is that it doesn't matter whether you define the moment from the water below C or above C, the sign works out either way due to (d-h/3) changing signs.
Conceptually, you can think about this as the center of pressure needing to be above C to open the gate, using math we can set up a sum of moments and solve with Normal at a being 0 given that there is no normal when the gate opens.
I am confused as to what is done with the normal at point A after solving for it. I understand using the moment about point C to find the normal force, but what is done with the normal at A after it is set to equal 0? It appears to me that everything would just cancel out to be 0, so I feel like I am missing something.
Make sure to include your equivalent force due to the water pressure in your moment equation. You are right that your normal force A would equal 0.
I feel like I've been able to solve the problem logically but I don't know how to get there mathematically. Why is F_{I} = (\rhogHb)H
I figured out my issue. For the problem make sure you use (d - H/3) instead of just H/3. Using H/3 makes the problem make very little sense.
I am having trouble starting the problem. How do I know where to position the force? I understand that it is dependent on Y-bar, but I do not know how to solve for it or where it would go.