If you're referring to the top most force, applied at D, that is NOT considered when writing a moment about D. Please carefully look at the color coded boxes and lines, each corresponding to which moment is being calculated. The 5 kN used for the calculation is the one applied at point E.

@@QuestionSolutions ohh I see it now,my mistake.Thankyou

It's completely up to you, so at 2:56, we chose clockwise as positive. Notice that the moment arrow (the curved blue arrow is facing clockwise), so it's a positive MC. But look at 5:20, again, we chose clockwise as positive, but our moment arrow, the blue one, is counter-clockwise. So it's negative.

@@QuestionSolutions Ohh i see, thanks for the answer

@@user-hv6ef9ie1g You're very welcome!

For the first question, there are no x-forces acting upon the beam, so there isn't anything for us to consider. For the 2nd question, it depends on the direction you chose the moment to be positive. So for the 2nd example, we chose clockwise as positive. So looking at 6:16, look at the 320 kN force. It's pointing down, and imagine the beam is free to rotate about point A, which way would the beam rotate? If you push down at the location where the 320kN force is applied, it will turn clockwise about point A, so it's a positive moment. The same is true for 20 kN force. But look at the BY force. If you imagine using your finger to push up at that location, which way would the beam rotate? It would rotate counter clockwise, so it's a negative moment.

You're welcome!

No, if you want the y-component, you will need to use sin60 and get the component. Otherwise, what you are getting is the resultant force FDC, which is why sin60 was already used in the equation. Again, just to clarify, what you find as FDC is FDC, NOT any component.

@@QuestionSolutions what you find as FDC is FDC, NOT any component. That explains 4 questions i have :') thanks a lot mate, i have test tomorrow lookin back at some work my professor doesn't explain sh*t I only find ready answers (results) . again thanks <3

You're very welcome!

You can, the choice is yours on which side you want to use.

You're very welcome. I hope you did great on all your courses, keep up the great work!

You're very welcome!

negative FDE?

@@onuhvictoria7996 Please give me a timestamp so I know where you're referring to and the reason why you think it's negative.

I don't know what you said, sorry!

No, I don't.

Thank you very much! ❤

Well, I hope you get many good sleeps in your future! And thank you, glad this video was helpful :)

Congrats on graduating and well done for remembering!

Clockwise was taken as positive.

Thank you very much, for the kind comment and the recommendations on topics to cover. :)

No, and I don't think I can explain it in a comment. This goes back to statics, when we write equations of equilibrium. In a very simple sense, T is the variable assigned to the EF and GH links. Through the process of writing a f=ma equation for the vertical forces only, you can directly solve for T. Nothing else needs to be done, it gives a value for the force in the members EF and GH. If you want the y-component, you will need to use cosine 30 to get it and if you want the x-components, you would need to use sine30 to get it. I doubt this comment is helpful because your question requires a lot of previous fundamental knowledge to answers. If you have time, please watch some early statics videos about equilibrium of forces. It should give you a good foundation for your question. Or, it might be to good to speak with your professor or TA and ask them to cover this question during their office hours.

You're very welcome :)

I'll add it to my to-do list. :)

So sometimes, you have to extrapolate data. But, I also list 4 resources that give you the pressure based on the temperature. Please check the description :)

❤️

So the position vector goes from B to A. In other words, to the left and up. We chose up and to the right to be positive. Since the x-component goes to the left, it's negative.

@@QuestionSolutions ohhh, thank you so much sir :>>

You're very welcome!

For which angle did you get 85?

@@QuestionSolutions example 1 the angle alfa between Fr and 450N

@@lordgt2679 I am not sure how you got 85 for that angle. That isn't a complementary angle either. You probably made some numerical error somewhere down the line. Here is the triangle solved: www.calculator.net/triangle-calculator.html?vc=&vx=700&vy=450&va=&vz=497.01&vb=&angleunits=d&x=Calculate

@@QuestionSolutions hello i used sine law 700 sin45/497 = sin alfa.

I think mine is the complementary angle

I definitely will do more thermodynamics videos in the future, but for now, entropy is the last topic covered. :)

You can solve these problems in multiple ways. Use the method that gives you the correct answer in a short time.

Thanks!

Usually, you can approach these questions in multiple ways. When you do enough questions, you sort of get a feel for the way you should go at it, and if you're wrong, or you get stuck, you can always backtrack and do a different method.

So it means you assumed the opposite direction to what it really happening. You can just need to write another line with a positive value and an arrow indicating the proper direction. So whenever you get the same magnitude but a negative answer, that means the magnitude is correct, just the assumed direction was wrong.

It's under general plane motion, in the chapter that discusses impulse and momentum. See: bit.ly/3Kfep8f

You can solve these problems in many different ways. I showcase one method, but as long as you get to the same answer, the path you take doesn't necessarily matter much.

You're very welcome! 💜

Wow, thank you! I am really glad you found it helpful.

Please watch this video first, and then if you still have the same concern, send me a comment, thanks! kzbin.info/www/bejne/goHHZamNbctsd8U

Please see this video, it's less than 60 seconds, and you will understand it much better: kzbin.infovynnKlJD_Jo?feature=share

Please see this video, it's less than 60 seconds: kzbin.infovynnKlJD_Jo?feature=share

You're very welcome!

You can, there might be a numerical error in your answer. As long as proper moment equations and equations of equilibrium were written, you will end up with the same answers regardless of the bar chosen.

The cut was made above the reactions so they were not needed. No calculations were made to figure out the reactions at A, or G. Also, where did you get that BA is equal to 50kN? At 7:07, we find that BA = 45kN.

@@QuestionSolutions Thanks for your quick response! I wanted to verify that the force in member AB equaled the vertical reaction force at A. I understand that it is not directly relevant to the section method. I just checked my calculations for some of forces in the Y axis, and I totally forgot to include the 5kN on point E. Now I have the correct solution for AB. Thanks again for the brilliant video!

@@milesbrack9188 Okay, glad to hear it was cleared up. Keep up the good work and best wishes with your studies :)

👍

You're very welcome!

You are doing integrals incorrectly. The integral of x, is x^2/2. So here, the integral of w is w^2/2. See www.symbolab.com/solver/integral-calculator/%5Cint%20xdx?or=input I encourage you to review integrals because they are a key part of dynamics and will be used a lot :)

@@QuestionSolutions Hi sorry I forgot to write the division by two, of course the integral of w is (w^2/2) but my question was after the integration is done the input of of 0 to w which are the intervals of the integration. Because [w^2/2] with the insertion of w would be (w^3/2).

@@sjfn9846 So you are solving definite integrals incorrectly. So let's say we solve an integral, and we have x^2/2 and the lower and upper bounds are 3 and 2. In that case, we plug in (3^2/2-2^2/2). Now, instead of 3 and 2, we have x and 2. Again, all we do is just plug them in. So we get (x^2/2-2^2/2). It doesn't become x^3/2-2^2/2. Here is a simple example: www.symbolab.com/solver/definite-integral-calculator/%5Cint_%7B2%7D%5E%7Bx%7D%20xdx?or=input I hope that helps :)

@@QuestionSolutions I got it now thank you so much for such a nice channel and the fast replies, you’re amazing at describing this to such an easy level.

@@sjfn9846 You're very welcome. I just want to make sure, you understand how to do the limits on a definite integral now? If not, please let me know. Otherwise, best wishes with your studies!

Also in this question why did we ignore the friction

@@mariaalhijazeen4974 We didn't ignore the friction, it was used to find Fmax, at 10:55.

I use after effects to animate.

Thank you. I use illustrator for the diagrams and after effects for animations.