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Thanks, will do!

glad you liked it. i'll give TA Indy some tuna today. he's earned it for all his hard work!

Yes, I defined my xy coordinates positive Right and Up, default style. You could rotate these for a different problem where it makes sense, my choice was convenient, but ultimately arbitrary.

nice!

For any real application - I would recommend using multiple methods and checking how far apart your answers are. Both analytical methods like Lumped Capacitance and One Term Approximation, and also software calculated numerical methods - Finite Differences. If your methods all converge to similar answers, that is reassuring that you are safe to make assumptions and use easier methods. But if your answers are far enough apart that you aren't comfortable with that level of uncertainty - then you should not use oversimplified methods, use the most complicated ones you are able, since they use the fewest assumptions and will account for most information and hopefully be more accurate.

Many thanks ​@@BrianBernardEngineering

you're welcome!

This may just be a sign error somewhere. Like, instead of saying A=B, you are saying A+B=0 ... but if A=B, it's not A+B=0, should be A-B=0. If A or B is a function of h, then your answer for h would be negative. Your weight forces are pointing down, buoyancy forces pointing up. One of your volumes in calculating buoyancy will be a function of h, other volume a function of 0.30-h. Maybe a sign error here, in calculating volume, in calculating buoyancy force?

"I'm a cat!" - TA Serenity

Good luck, I hope you do great!

"Meow" - TA Indiana

you're welcome

Hmmm, let's check. From Bernoulli Eqn, the exit velocity would not change based on exit diameter, it's only a function of the drop in height. Same velocity, but double the diameter, would mean 4x the volumetric flow rate (area*velocity, and area is proportional to diameter squared, and 2^2 = 4).

Thank you so much 😀

The 3 circles on front view: outer circle is a solid cylinder in the back, middle circle is solid cylinder in front, small circle is a hole presumably all the way through. Need to add hidden lines for the hole on both top and right. top view has a few visible lines where the middle circle has its top cutoff.

Happy to help. I think you are the first student I've ever had from Costa Rica, thanks so much for leaving a digital post card!

"meow" - TA Indiana

awesome! have a great rest of your semester

If you check my channel page, I have a Graphics playlist with I think 3 different "missing lines" another 3 "missing views" and about 3 more "how to draw Isometric views" videos. Don't just watch the videos. Pause the videos, make the drawing yourself, then watch the video to check, not just your answer, but your method and thinking. If you only watch, you'll think you are learning, but you won't learn. You can't learn drawing by watching drawing - you need to draw yourself

Yes, mass moment of inertia I is definitely different for a hollow vs solid shape. As a simpler example to recognize this, look at the area moment of inertia of a ring vs a disk. They are both circles, but different moment of inertia. (if mass is the same, the ring has higher inertia, since more mass is further away from the center, disk has more mass closer to the center)

If you want more examples, check my Graphics playlist, I have a few other videos on this same topic, and they use different methods, so maybe a different technique will resonate with you. Good luck this semester.

@BrianBernardEngineering Thanks so much! I watched all the playlist and I aced my midterm!!! ❤️❤️

I think just the 1 missing line on the right view is all I see. The top view is just a big empty rectangle, which is looking at the top of the sloped surface. The right view missing line divides it in half, top rectangle is looking at the slope from the side, bottom rectangle looks at the short vertical face that is the short vertical line on the right side of the front view. Talking about drawings feels really complicated when I type it, hope that made sense.

I was using absolute pressures instead of gage pressures, is why I used atm as 101kpa instead of 0. This problem could be completed in terms of gage pressure if you wanted, you would reach the same answer. But be careful when looking up vaporization pressures for your liquid, since those will probably be published as absolute pressure. I've seen the vacuum experiments (papers about them). There are some really cool siphon experiments that reject easy explanations. My favorite is the Flying Drop experiments. Very cool.

anytime

your welcome. good luck during rest of the semester

youre welcome!

Unfortunately no - Navier Stokes is something I move though kind of shallow/quickly in my class, so I haven't made a video for it yet. Good luck though.

I would almost always recommend working in terms of gage pressure, which makes Patm = 0, and can be ignored.

Yes, this is correct. It would have been better if I had explicitly written "uniform flow" in my assumptions section, since I was making that assumption here, which makes the integral trivial.

I like the books by Randy Shih. He has many of them since he combines both hand drawing and software drawing in the same books, so there are different books for Inventor or AutoCAD or Solidworks. They are almost more of a 'workbook' style than a 'textbook', which I think is good for this subject where practice is more important than expansive knowledge.

Thanks so much for watching. If you watch the whole playlist, you'll get to see a story of TA Indiana saving you from a gang of hobos and escaping using a 4 bar linkage mini train car. It's my favorite video, but I admit that many students find it a bit distracting from the actual mechanism. Quick Return 4 bar linkage is the one with the story.

free delivery point - this lets pump spin at maximum speed without any change in pressure - in practice this can often lead to cavitation issues that can damage the pumps internal moving parts, so it seems like following the curve all the way down to the x axis might not always be mechanically advisable. If a system curve intersects a single standalone pump, it should always intersect 2 of the same pump in series. It will intersect at a different point, but since the single pump curve is inside the series pump curve, it's impossible for system curve to hit single, and not hit series.

​@@BrianBernardEngineeringthank you professor,much appreciated

haha, I don't know that I've ever noticed those before. Now that you point them out, it's like my first time ever seeing them. Sorry, I cannot explain it. But - I've never need to use them that I recall. Good luck.

Awesome, thanks so much for the post card. Good luck on your major exams!

Sorry, but i dont have the answer key either. thats why i made inventor drawings in this video to double check. i have a playlist with several other missing lines and missing view videos, so that will give you a few more worked out examples. not as many as a full sheet, but best i can do. good luck.

never mind i got it thank you for the video

Really glad I could help. And also, good work solving your own question!

You're welcome!

You're welcome!

Yea, edge cases get a bit wonky. You can have all the strength in the world, but if you try to apply it over any non-zero distance, you get much weaker.

@@BrianBernardEngineering ah yes. When an unstoppable force meets an immovable object.

@@BrianBernardEngineering hey, is it possible to graph the mechanical advantage of a 4 bar linkage?

@@FerociousSniper absolutely. Use theta, the angle of the crank as the x axis from 0 to 360, and y axis is your mechanical advantage. In your equation for mechanical advantage, you have 2 angles ... neither of which are what I just called theta, but they are both functions of theta. Rewrite those angles as functions of theta, and bam, you have yourself a way more complicated equation for mechanical advantage, that can now be plotted for theta. If you had angular velocity, you could do an extra step if you wanted x axis to be time instead of theta.

@@BrianBernardEngineering thank you so much!

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hello!

Yes, the problem shouldn't fundamentally change when using Left side view instead of right. You just have your miter line on the other side of the top view, but process is the same.

Thanks. Have a great semester!

Yea at 1:55 TA Indiana took a real spill. That's why I had my hands on him the whole first couple minutes, to help him stay balanced. Took my hands off him for 3 seconds and bam. He's a trooper though. After we finished filming, gave him a popsicle for all his hard work. That's his favorite food, popsicles, lol.

Your welcome. Have a great rest of your semester!

oh no, you're totally right. i even did it correctly an inch up the page when i found resultant forces in x and y directions, but i used the wrong one when doing the moment. excellent catch, you are correct, that should be 50cos30 times 3 for the moment. thank you very much for pointing this out, and for you, that's a great sign that you actually understand the problem and method!

centroid of the triangle will be true if 2 conditions are met. 1. the top of the water is touching your gate. if the top of the gate is submerged some distance below the surface, then the shape will be a trapezoid, not a triangle. 2. the cross section of the gate must be rectangular, so the width of the gate in the direction into the page must be constant. if the gate is actually a round hatch or any other non rectangular shape, then the problem gets more complicated.

I'm happy to help. have a great semester!