For the curved surface AB, determine the magnitude, direction and line of action of the vertical and horizontal component of hydrostatic force acting on the surface. L=1m
In this case, you would draw a different free body diagram of the water above the gate -- the weight of this water acts downward on the gate and is equal to the vertical force on the gate. I hope that helps.
5:35 Dear Sir, why do you count Fv pressure force.? Usually, I saw teachers concern about the weight of the water up to free surface to find vertical force. I only find in your all of the video you count vertical pressure force, why sir?
I'm not sure I understand your question. I think you are asking, why not calculate Fv using the missing water approach? I discuss this alternate method and it's drawback (finding the line of action can be tricky) here: ua-cam.com/video/yelqodaUfOE/v-deo.html
Note that Fv acts downward on the water (in the FBD) but upward on the gate -- the force equal is magnitude but opposite in direction. The question acts for the force on the gate (which is upward). I hope that helps.
@Fluid Matters Hi, thank you for replying. Another question, You have mentioned on the last part that Fv can be solved alternatively by multiplying the specific weight of the water by the imaginary volume on the other side of the gate. But if you do that, you get a different answer for Fv compared to your first solution. Why is that?
I don't know why I can't get it man although got AA in mechanics of solids and engineering dynamics and statics but I am struggling so badly with this... Sometimes FV equals the weight of water, sometimes it does not.. Sometimes we have 3 vertical components sometimes we do not 🤦🏼♂️ struggling so bad man soo baad
The difficulty of fluid is that every problem has different ways of solution, even in the same chapter! This a structural engineer has never seen this before
When the liquid is entirely above the surface, then the vertical force will be equal to the weight of the water. But when the water is below the surface (as in this problem) that cannot be true, since the vertical force on the gate is upwards. In these cases, the free body diagram has other vertical forces that need to be included in the force balance.
This is a two-dimensional problem. So, the calculated forces are per unit width into the page (width=1m). If the curved gate was say, 5m wide, you would multiply the force per meter of width by 5.
I do not understand why you calculate the weight of water for F vertical, the weight of water is not on the gate, we only calculate the forces on the gate.
My solution involves a free body diagram of the water adjacent to the gate (not the gate itself). F_v, F_1 and the weight of the water are the vertical forces that must balance. So, I end up getting the force on the gate (F_v) indirectly in this way. If you find this puzzling, it may help to review your basic statics. I hope that helps.
All the videos (and pdf downloads) for this introductory Fluid Mechanics course are available at: www.drdavidnaylor.net/
For the curved surface AB, determine the magnitude, direction and line of action of the vertical and horizontal component of hydrostatic force acting on the surface. L=1m
Appreciate all your time and videos, you've been tremendously helpful my whole semester. I'm leaving another comment to help the algorithm :)
Ha Ha! Thanks!
Vertical component can be directly found by volume of virtual water content in quarter circle multiply by specific weight...
But new concept.. ❤️❤️❤️
Yes. That is an alternate way to solve the problem. I talk about this approach at 12min 40s.
Absolutely oustanding!!
Thanks!
if the water were above the curve and the air below, it would be the same but instead of decreasing the weight of the vertical force I would add it?
In this case, you would draw a different free body diagram of the water above the gate -- the weight of this water acts downward on the gate and is equal to the vertical force on the gate. I hope that helps.
sorry if I'm wrong, but shouldn't you have multiplied the w force downwards by g=9.81? before subtracting it from f1?
W is the weight (not mass) of the water. So, you do not multiply by g.
thank you so much
Thank you so much for this amazing lecture.
Glad to hear it was helpful!
Thank you so much
Why you use h as 6m for calculating F1 instead of taking the value 3m?
The centroid of surface DB (and indeed, the entire surface) is located at a depth of 6m.
5:35
Dear Sir, why do you count Fv pressure force.?
Usually, I saw teachers concern about the weight of the water up to free surface to find vertical force.
I only find in your all of the video you count vertical pressure force, why sir?
I'm not sure I understand your question. I think you are asking, why not calculate Fv using the missing water approach? I discuss this alternate method and it's drawback (finding the line of action can be tricky) here: ua-cam.com/video/yelqodaUfOE/v-deo.html
thank you
ua-cam.com/video/_nkaOZI8tjI/v-deo.html
.
Sir thank you! Its help me to understand more my review.
i want to ask..why the area calculated by 6m multiple by 1.......where 1 is given in the question?
The question asks for the force per meter of depth into the page. So, the area is 6mx1m
Hi, from your FBD, your Fv goes downwards. But on the last part, you made it going upwards. Why is that?
Note that Fv acts downward on the water (in the FBD) but upward on the gate -- the force equal is magnitude but opposite in direction. The question acts for the force on the gate (which is upward). I hope that helps.
@Fluid Matters Hi, thank you for replying. Another question, You have mentioned on the last part that Fv can be solved alternatively by multiplying the specific weight of the water by the imaginary volume on the other side of the gate. But if you do that, you get a different answer for Fv compared to your first solution. Why is that?
@@jhunardchristianasayas7134 You get the same answer: vol*specific weight= pi*R^2/4*(9800)=3.14*36/4*9800=277.1 kN
@@jhunardchristianasayas7134 You get the same value: Fv= vol*specific weight= pi*6^2/4*9800=277.1 kN
Why did you take the width as equal to 1 meter? I didn't get this point
The problem statement says to find the force "per meter of depth (into the page)".
How did you compute the A that is downwards that is equals to 7.726m²?
This is explained at 8:12. A=6(6)-pi(6)^2/4=36-28.27=7.726 m^2. I hope that answers your question.
I don't know why I can't get it man although got AA in mechanics of solids and engineering dynamics and statics but I am struggling so badly with this... Sometimes FV equals the weight of water, sometimes it does not.. Sometimes we have 3 vertical components sometimes we do not 🤦🏼♂️ struggling so bad man soo baad
The difficulty of fluid is that every problem has different ways of solution, even in the same chapter! This a structural engineer has never seen this before
When the liquid is entirely above the surface, then the vertical force will be equal to the weight of the water. But when the water is below the surface (as in this problem) that cannot be true, since the vertical force on the gate is upwards. In these cases, the free body diagram has other vertical forces that need to be included in the force balance.
How did you get 1m at 13:00
All forces are per meter of depth (into the page). So, I take the depth as 1.0m.
👏
ua-cam.com/video/_nkaOZI8tjI/v-deo.html
can i ask why there is 1m of width ?
This is a two-dimensional problem. So, the calculated forces are per unit width into the page (width=1m). If the curved gate was say, 5m wide, you would multiply the force per meter of width by 5.
I do not understand why you calculate the weight of water for F vertical, the weight of water is not on the gate, we only calculate the forces on the gate.
My solution involves a free body diagram of the water adjacent to the gate (not the gate itself). F_v, F_1 and the weight of the water are the vertical forces that must balance. So, I end up getting the force on the gate (F_v) indirectly in this way. If you find this puzzling, it may help to review your basic statics. I hope that helps.