You're very welcome!

1 watt = 1 J/s. From there, you should be able to derive kJ/min to kW. Let me know if you are unable and I will help.

I hope you did well on your midterm!

Why?

Please use timestamps, I'm not sure where you're referring to.

You're very welcome! :)

Thank you for the kind words and for the feedback!

See this video, especially the 2nd example: kzbin.info/www/bejne/hKOvZpdjZ6iUmLM

You're very welcome! :)

Not sure where you're referring to, but it seems like you got what you needed :)

@QuestionSolutions no I didn't get what I needed. I need to write a spreadsheet today to calculate a Lenard Jones parameter for a radial distribution function and I'm worried I won't get it. The Matlab code on the course website says 5 errors and I dont know how to fix it.

@@JamesVestal-dz5qm So if you don't know how to fix it, it's probably a good idea to see your professor or your TA and get the help you need. Best wishes!

@@QuestionSolutions my professor loves to cap the office hour to 1 hour. Get out of my office I have another student to work with says the professor. But I haven't figured it out yet say I and he says try again in two business days!

@@JamesVestal-dz5qm Wow, never heard that before, must be a really busy professor. Is there a TA you can get help from? Or a friend who already worked on the project?

Why, thank you! I always love a good excuse to learn something new. Maybe I'll see you there?

:) Glad to hear!

So it has nothing, absolutely nothing, to do with whether a force is positive or not. Positives and negatives only tell you whether an assumption was right or not. What matters is the direction of the force. Is the force coming towards the pin or going away from the pin. Rewatch this part from 1:48.

I circle it at 6:50 😅It's the mass of the ring multiplied by the acceleration. That is 0.4 m away from point C, (perpendicular distance), which means it creates a moment. So whatever creates a moment must be written down on our equation.

@ I thought it was just the forces that create a moment and rhs was just moment of interim times angular velocity.

Thank you!

You're very welcome. In 2D problems, you can use clockwise as positive or counterclockwise as positive. You will get the same answer regardless of which you pick. It's totally up to you. In 3D problems, you will need to use the right hand rule to figure out the positive vector directions. Try it out :)

So we can't write that equation because we don't know the voltage in the 4 mA current source. If a current source is shared between two mesh currents, it's simply easier to write an equation for a loop so that the current source is ignored. You can write it with the current source, but you will need to assign an arbitrary voltage to the current source and then solve it at the end. That's a waste of time since writing an equation for a loop will get you an answer way faster and you're less prone to mistakes. Remember, at the end of the day, you are just writing KVL equations when it comes to mesh analysis. So if we don't know the the voltage, we can't write it.

@QuestionSolutions thank you

You're very welcome. Could you please give me a timestamp so I know where you're referring to, then I can take a look. Thanks!

Both will work and give you the same answer.

You're very welcome and I am happy to hear these videos were helpful to you. I will add mechanical vibrations to my list of things to cover, but it probably won't get done before your exams since I just started working on the circuit analysis playlist. On the bright side, there are lots of great videos on youtube on that subject so don't fret! :)

One day!

Thank you very much. I appreciate your comment :)

Sorry, but I don't know where you're referring to. I don't see the -122.5. Where is it?

You're very welcome!

So to figure out the direction of the vector, you have to use the right hand rule. While the angular velocity is clockwise, this is a 3D vector. If we curl our right hand fingers the same direction as our angular velocity, we see that our thumb points into the screen. So that's the negative z axis, or -k axis.

Does the question specifically say the force is applied to the pin? Does the diagram look the same? Really hard to say without seeing the problem/ diagram. I encourage you to see your professor during their office hours to get everything cleared up.

That's right, but even more easier to remember is to keep in mind that any force starting where the moment is calculated, cannot create a moment at that point. The same as force BC.

Thank you very much. Mechanisms of machines is a very broad topic. Are you looking for videos on a specific subtopic or just a general set of videos on the topic?

Thank you so much! :)

kzbin.info/www/bejne/goHHZamNbctsd8U I just found the video that explains.

Please see this video, and if you still have any questions, let me know: kzbin.info/www/bejne/goHHZamNbctsd8U

Aw, that's unfortunate to hear. But on the bright side, if you ever need a refresh, the videos will be here.

Thank you very much! I am really happy to hear you can understand this video :) Keep up the awesome work and I wish you the best with your studies!

You're very welcome and thank you for the feedback! To answer your question, the correct answer is {-128.34i + 128.34j - 256.68k}. There is a typo in the video where the K term is shown as positive, when it's actually negative. So while the magnitudes of your values are correct, the signs are wrong. It's probably just a small mistake in assigning positives and negatives to your i,j,k components, so if you have some time, take a look at this video: kzbin.info/www/bejne/fGmseaWdaNWWbck

So we want to represent length as a variable. If we use a specific distance, then the value we get is just for that distance. Let's say we did it at 2 m, then the all the values we calculated would be just for 2m. We want to use a variable for length so that we can plug in any distance we want and figure out the values at that spot. So if we wanted to figure out a moment at 3.5 m, all we do is plug that into our "x" value. So that's why we want the length represented as a variable.

You're very welcome!

Thank you very much, I really appreciate your comment. Glad to hear this was easy to follow :)

If you click the "join" button (next to the subscribe button), it should come up. If you don't see it there, if you go to the channel page, there is also another join button. Also, thanks for even considering to join this channel! :)

Thanks!

Happy to help!

You're very welcome!

So let's look at 00:45, where are we calculating the moment about? Couple moments don't work like the normal moments. You're not specifically taking the moment about a point, though you can think of it as taking it about the center between the two forces. So in the triangle, you're just figuring out the moment, not about a specific point, but rather the moment created by the opposing forces. If it makes it easier, you can think of the triangle pinned at the middle and the forces creating a moment that spins it about that pin.

So it should be: k[ (4 x -64.17) - (0 x 21.39)] = k[-256.68-0] = -256.68k. Hope that helps :)

The final answer in your video is positive 256.86 k. I just want to clarify sir if the final and correct answer is -256.58 k?

@@itsheynahhh Yes, -256.58k is correct. There is a typo in the video, that's why I made a pinned comment about it.

You're very welcome!

Yes, for relative velocity of particles, see: kzbin.info/www/bejne/paG5fIF5msmeiNE

Thank you! I hope the video was helpful to you, and I wish you all the best in your studies.

I'm glad to hear that :)

Glad it was helpful!

Great to hear!

So my guess is, you assume cos is for x and sine is for y. This is fundamentally wrong, and you must, absolutely must, remove that idea from your head. You have to look from the perspective of the angle every single time. Here, the opposite side to the angle gives us the x-component, and sine is opposite over hypotenuse. So you need to use sine. To get the y-component, which is now adjacent to the angle, you need to use cosine. See this video, it's less than 60 seconds and you will see what I mean: kzbin.infovynnKlJD_Jo?feature=share

Noted thank you ​@@QuestionSolutions

@@ZexMutsawu-nw2ov Awesome. Best wishes with your studies. Let me know if you need any further clarifications.