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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!

Yea, if you're going mechanical, check my playlists page, I have a playlist for a bunch of different courses, freshman through senior level. Good luck.

Glad you enjoyed it!

In that scenario, I think the top view will have 2 diagonal lines instead of horiz/vert pairs. And you'd still also have the same horizontal line on the front view. Instead of the final 3d image looking like there's a flat "shelf", the front most 2 points of the shelf would connect directly up to the top point of the triangle, so you'd have a triangular slanted face instead of a horizontal and vertical face. A diagonal line on 1 view, will be a visible face on the other 2 views. Your new diagonal line will correspond to a triangle on the front view, so you'll see that same triangle on the top view also.

@@BrianBernardEngineering I actually managed to get it figured out! Thank you and great video!

You're welcome. Have a great semester.

It depends on whether you count "ground" also called the "fixed link" as part of n or not. If you count ground as a link, it doesn't move, so you need to subtract 1 from n. If you don't count ground as a link, and n only includes parts that move, then 3n is correct, no need to subtract 1.

@@BrianBernardEngineering Ok thank you so much now I got it.

Thanks so much, have a great semester!

Different radius. The version I use is when the slanted radius is known, so you use slanted distance for height. The version you use with vertical height would require knowing the radius in the horizontal direction, which you usually wouldn't know.

Thank you so much. I always thought this video was pretty good for the topic, but it has - by far - the worst youtube metrics of any video on my channel, and I cannot figure out why. Maybe it got accidentally shown to a bunch of business majors or something and they all hated it?

no, you're amazing!

Thanks so much. Have a great rest of your semester.

Lift will be determined by whatever is being controlled. If its connected to a slider or switch or any other mechanism, how far does that other piece need to move in total. Minimum radius, the cam will perform better the larger it is, but larger is more expensive and takes up more room. So, use the largest minimum radius that you can afford or have space to accomodate. The degrees for rise, dwell, fall ... this is based on time. Do you want it to rise and fall at the same speed, or do you want one of them to be faster? Do you want it to stop and stay at the top position for any amount of time (yes dwell), or do you want it to return immediately (no dwell). This will depend on the purpose of the machine. If you just want it to rise and fall one right after the other, you'd just make rise 180deg, fall 180deg, with no dwell. Or if you want it to fall faster than it rises, Rise 300deg, fall 60deg to make it lower 5x faster than it rises. I hope this helped.

alpha, do you mean gamma? I scanned the video and didn't see an alpha, but I might have missed it if you could give a time stamp. I solve for gamma using 2 law of cosines, which could conceivably be combined into 1 giant equation, but I think its easier to solve it in 2 steps than combine the steps. This method should work even when theta is >180. The red and green triangles will be overlapping a little since the green will be below the x-axis, but the method should still work, solve for length of the Lbd, then solve for gamma.

@@BrianBernardEngineering Sorry for my mistake. Yes, I meant gamma. If the method works even for theta > 180 that's perfect. Thanks you're a saviour!!

Thanks so much. I kind of drew a blank about what new graphics videos would be useful to make, so I moved on to other courses. Might come back to this class in the spring. In the meantime, checkout my graphics playlist if you haven't seen it yet, about 16 videos maybe so far.

@@BrianBernardEngineering auxillary views and especially section views would be amazing

@@daviyauntalbot Those are on my list and 2 of the next 3 topics planned for my Graphics playlist, along with Dimensioning. Unfortunately though, I don't think I'll get to making these until the spring semester. I've got kind of a backlog of other things ahead of them. So - not in time to help you with your course now. But hopefully to help other students later.

I have static force analysis of a crank slider coming out in a week. After that though, it might be awhile before next one sorry. If you haven't seen my mechanisms playlist yet, there's 12 videos there that might have more of what you are looking for. Have a great semester.

50 cos(30). Normally cosine is used for x component, but that's because we normally measure angles to the x axis, and cosine is 'adjacent/hypotenuse', so adjacent is the x direction. But for the 50lb force in this problem, the 30deg is measured from the vertical y axis, so cosine is adjacent to y, so cosine is for vertical component.

when i read this comment at the time, i thought you were referring to the 'resultant' force part of my answer which i did correctly, but you were probably asking about the moment, which i did wrong. this was a typo that you correctly noticed. i just didn't notice what you meant. thank you for pointing it out, even if i was slow you understand you. i appreciate it a lot.

Thanks so much. Have a good semester!

diameter does not affect "speed" as in meters/second, but it does affect "volumetric flow rate" as in liters/hour. Use Bernoulli eqn to find speed by comparing the top of the tank to the bottom exit from the tube. Then multiply speed by area of the pipe, pi r^2. You'll then have m/s*m^2 which gives you m^3/s, volumetric flow rate. Then unit conversion to liters per hour. Warning - this flowrate only applies at this moment in time. As the tank empties out, the flowrate will slow down, since the height gets lower as it empties, so the exit velocity will decrease, so flowrate decreases. I have another video called "Find time taken to empty a tank" which addresses this exact warning and how you can overcome it, if you are really interested in finding time, not just flowrate.

@@BrianBernardEngineering thank youuuu ❤️ its fine...because i will use power head pump with power 1000liter/hhour to return the water from bottom tank move to the over tank...to make the water will flows continuosly...so the water in over tank will not deacrease or going to empty ❤️❤️❤️ Now i find the missing key that diameters doesnt effect the speed of the flows as in meter/second ...big thanks for you ❤️❤️❤️

Thank you. have a great semester!

Thank you. Have a great semester!

thank you for the kind words. have a great semester!

teaching is easy when you have great students ;)

Buoyancy is an "apparent force" sort of like Centrifugal force. It's useful help understand the functionality, but technically, it's not actually a force acting on the object. What we call buoyancy, is really the total combined net force after accounting for all pressure pushing perpendicular to the surface from every direction, force is pressure * area. Your understanding of the oil pushing down on the top surface of the block, and water pushing up on the bottom surface of the block is correct. When actually analyzing the real surface forces really pushing on the block, side forces all cancel, and the oil does not lift the block up, the net effect of the oil alone is to push down on the block. The net force of the water is to push up. When you calculate the net difference of these 2 forces, you should find that they equal the sum of the 2 buoyancy forces you calculate using the volume of displaced oil and volume of displaced water. If you calculated the force of the water pushing up on bottom of the block, it would not match the buoyancy force of the water, it would account for both water and oil, because oil is pushing down on the water, increasing the water's pressure.

All 6 of the shapes with blue numbers are able to move. Are you referring to #4, the green circle as fixed? This disk can still rotate, which makes it still a moving body. You can't really see it rotate because there's no markings on it, but I am considering it fixed in x/y, but with theta that can change. Thanks so much for the kind words, and good luck this semester.

@@BrianBernardEngineering I GOT IT! I had the wrong body count. In total there are SEVEN bodies. Six can move, Ground (1) is fixed. By the way, I am not your student. I am a really old person who is fascinated with this subject. Before this video of yours, I had maybe watched a zillion videos on this topic but never actually FULLY UNDERSTOOD this particular topic. Now I am going to go and look for a video BY YOU on the topic of Instantaneous Centers.

Good luck on your test! I've got 5 more heat transfer videos coming, on convection and radiation, that should be released over the next month so hopefully they'll be ready for when you get to those next sections of the course. They'll be added to my Heat Transfer playlist when they are published, so if you bookmark that, they'll be easy to find.

If you have a classmate that you think these videos would help Please send them a link. Getting discovered is hard. I appreciate personal recommendations so much.

It's probably not harmful to think of it that way, even if technically not fully correct. Similar to in electrical circuits, we think of current flowing from positive to negative, even though we now know its electrons that are moving, and they move the other way. There's a bunch of these sorts of things throughout engineering curriculums, that while not entirely correct, it's still kind of ok to think that way, because it may still be useful in understanding a bigger picture, even if the details are a bit off.

​@@BrianBernardEngineering Thanks for the reply. I agree. We should keep in mind as many possibilities as possible for some conclusion in order to be able to see what works to what extent. It should not be exclusive. Doubt about correctness leads to the right answer. I don't speak English very well, sorry if there are mistakes.

You're welcome. Good luck this semester!

You're very welcome. This is maybe the best example in any engineering class to demonstrate that if you understand the formula, you don't actually need to know the formula at all. You can just do it on your own.

Happy to help. Most of the course will be 2D problems. But there will be a little 3d sprinkled in here and there.

You're very welcome. I hope you do great!

That's very kind to say, thank you so much for watching.

I prefer to not really think of the joints as 'things that move', and only consider the links as moving objects. I usually view a joint only restricting the motion of 1 of the attached links, not 2. Consider a 4 bar linkage. The crank has its motion restricted in 2 directions by the joint connecting crank to ground. Then the coupler is restricted in 2 directions by the joint connecting coupler to the crank. So that joint, connecting coupler to crank, I don't think of it as restricting the cranks movement, because the crank is already restricted from its other end. This isn't the only way to think about these problems, but it's my way.

A little counterintuitive, but yes, it would drain faster with a pipe extension to make the exit lower. Same premise works for siphoning as well. If you want to empty out something like a fish tank using a siphon, don't have the exit of the siphon right below the tank, get a longer pipe so the exit can be all the way down on the ground, lower exit is faster exit velocity since you convert more potential energy to kinetic with greater change in height.

It's usually the difference in pressure that actually matters. Pressure inside something pushes outward, atmospheric pressure pushes inward. The difference is important since you are summing forces in the two different directions to get a net force due to gauge pressure. This is why you don't actually measure atmospheric pressure usually. Atmospheric pressures changes a bit everyday due to weather. Since we only care about the gauge pressure, it doesn't matter exactly what the reference points absolute pressure is.

Awesome. Unfortunate that it's rare, but glad I can be the exception. Have a great semester!

Thanks so much for the kind words. Also especially good for students to hear that this stuff can actually have a real world application, not just homework!

​​​But I think the calculation conversion you made from ft3/s to gal/min is wrong.😂 Correct me I been mistaken. But anyway, you did great explanation and easy to understand. Tq😂​@@BrianBernardEngineering

You're right - that is definitely the right way to do it. In practice I usually try to ensure all my units are consistent at the start so that I know the answer will work out to be the correct units, which allows me to get away with being lazy - but I fully concede that it is risky and lazy. I'm making a note to do better next time. Good luck this semester.

@@BrianBernardEngineering thanks! My semesters are long gone though! I am here for the joys of fluid dynamics!

Main difference with submerged exit is increased pressure. However far under the surface, that's extra pressure. Bernoulli Eqn from inlet to outlet, you would then have a lower change in pressure between inlet and outlet, same change in height between inlet and outlet, therefore velocity difference would have to be smaller. Flow rate would be slower.

thanks so much. that really means a lot!