Suppose the gate was submerged such that the free surface was height H above the top of the gate with the section above the gate being a vertical wall (think upside down P). Would the vertical portion of the pressure force still be the weight of the water within the control volume? The horizontal portion would still be similar to the current solution where it would be gamma*width*total height*gravity.
A problem similar to what you are describing starts at about 17:00. In the case of an "upside down P", the vertical hydrostatic force is NOT just the weight of the water. You have to account for the upward hydrostatic force on the bottom horizontal surface of the control volume. I hope that helps. ua-cam.com/video/LdbEpRXUpOQ/v-deo.html Here is another solved problem that considers this situation: ua-cam.com/video/Awp-iP3ktDI/v-deo.html
In problem 2, the normal to the gate had vertical components, but the summation of forces resulted in no vertical forces because of the weight of the water. If that is a correct interpretation, please explain? (This seem to be the intuitive result, but contradicts the idea that the force is normal to the surface?) Thank you. Your videos are great for self-teaching.
Thanks for the kind words. I'm not 100% sure which problem you are referring to. If you give me a time where I discuss F_v, I will try to help. (I don't think F_v is zero in any of the problems, so I was unsure.)
@@r2k314 The free body diagram at 23:50 shows the forces on the water adjacent to the gate. F_v in this free body diagram is equal to the vertical component of the pressure distribution (which acts in the opposite direction in the gate). I hope that helps.
FAB=FH. So, this force is considered, indirectly. When you take moments on the gates, you have to consider the forces of the water on the gate. I hope that helps.
If I'm understanding your question: There is no vertical force on "the top of the free body". The vertical force acts upward and equals the weight of the water, which acts downward.
@@FluidMatters if we considered only the upper left quarter of the circle instead of the semi circle then there would have been vertical force acting on the upper part of the quarter circle then why not for the semi circle?
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thank you for your great explanation
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Thank you sir for your videos.
Suppose the gate was submerged such that the free surface was height H above the top of the gate with the section above the gate being a vertical wall (think upside down P). Would the vertical portion of the pressure force still be the weight of the water within the control volume? The horizontal portion would still be similar to the current solution where it would be gamma*width*total height*gravity.
A problem similar to what you are describing starts at about 17:00. In the case of an "upside down P", the vertical hydrostatic force is NOT just the weight of the water. You have to account for the upward hydrostatic force on the bottom horizontal surface of the control volume. I hope that helps. ua-cam.com/video/LdbEpRXUpOQ/v-deo.html Here is another solved problem that considers this situation: ua-cam.com/video/Awp-iP3ktDI/v-deo.html
In problem 2, the normal to the gate had vertical components, but the summation of forces resulted in no vertical forces because of the weight of the water. If that is a correct interpretation, please explain? (This seem to be the intuitive result, but contradicts the idea that the force is normal to the surface?) Thank you. Your videos are great for self-teaching.
Thanks for the kind words. I'm not 100% sure which problem you are referring to. If you give me a time where I discuss F_v, I will try to help. (I don't think F_v is zero in any of the problems, so I was unsure.)
@@FluidMatters Thank you for your help. At 23:50. Note the vertical forces do not have a component on AB, compare with 18:58.
@@r2k314 The free body diagram at 23:50 shows the forces on the water adjacent to the gate. F_v in this free body diagram is equal to the vertical component of the pressure distribution (which acts in the opposite direction in the gate). I hope that helps.
Hello Professor, may i know why F(AB) is not considered in the moment ?
FAB=FH. So, this force is considered, indirectly. When you take moments on the gates, you have to consider the forces of the water on the gate. I hope that helps.
why is there not a weight force of the gate when using the FBD of the gate in the first example?
Because the weight of the gate is neglected, as stated in the problem statement and verbally at 5:32. I hope that helps.
@@FluidMatters thank you very much I must’ve just missed it
Where is the vertical force acting on the top of the free body in no 2 prblm 7:55
If I'm understanding your question: There is no vertical force on "the top of the free body". The vertical force acts upward and equals the weight of the water, which acts downward.
@@FluidMatters if we considered only the upper left quarter of the circle instead of the semi circle then there would have been vertical force acting on the upper part of the quarter circle then why not for the semi circle?
At 6:57 inclined forces at the upper portion should have vertical components ..isn't it? Can u pls clarify?
The line of action of Fh and Fv will pass through the centre of circle. But you have located somewhere.
Individually Fh and Fv do not pass through the centre, only the total resultant vector.
how did u get the depth as 1 m in the first example, the image was in 2d
The problem statement asks for the forces are per unit depth, i.e. per meter of depth into the page. So, you use a width of 1m.
@@FluidMatters thank you so much for making these videos sir - you’re God-send. Love from Texas sir.