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You're welcome

you're definitely not the only one :-) I wish you all the best for your engineering studies.

Happy to have been able to help. I wish you well with the rest of your studies.

Thanks for the feedback. Glad you found it helpful.

Glad I was able to help. Thanks for the feedback. I wish you all the best with your studies.

Glad it was helpful!

8 years later, I’m still here :) Thanks for commenting. Glad the video helped. Good luck with your studies.

You may take moments about any point you like. However, there is usually one (or maybe two) points that are most convenient for ease of solution algebra. In this case I take moments about point B because there are two unknown reactions at point B. They will have zero moment about point B which then enables me to determine the unknown reaction at C straight away (i.e. one equation with only one unknown).

@@TerryBrownMechEng If we take moment about point _A_ ,sum of the couple moments and force moments respect to _A_ also gonna equal to zero ,right?Same for point _B_ also?

@@yigitcan824 yep 👍

hi @jonathansanchez2854, sorry for the delayed reply. How are you going with your understanding? Your question indicates that you may have some misunderstandings. If you still need me to reply, please let me know. cheers.

sorry, it's a bit complicated to explain here without the aid of pictures. Key thing to remember is that the hinge cannot transmit moment from one side to the other and therefore the internal bending moment will be zero at the hinge

The strut CD is a 2-force member. The force at D is equal, opposite to and co-linear with the force at C.

Glad you liked it and I hope it was helpful. OK. Do you mean putting the units next to every term in every equation? I find that gets a bit cumbersome and also makes it hard to fit the equations on the limited ‘whiteboard’ page size. I always include the units for every value calculated. Thanks for the feedback though. Cheers.

Well, we could have done that, but then we would have to find the perpendicular distances for Rcy and Rcx. We already have the perpendicular distance between Rc and point B, since Rc is normal/perpendicular to BC, so no need to break Rc into components.

@@TerryBrownMechEng Thank you so much

Hi, yes, I’m still around, but missed your comment. Sorry. Yes, Rbx will be negative and that is the answer that I get at about the 13min mark. Your last sentence indicates you may have some misunderstanding about moments though.

@@TerryBrownMechEng I guess I I may be conflating horizontal equilibrium (LHS forces=RHS forces), where a moment is force x distance about a point, so I assume the reaction force will be the same but in opposition? I'm studying mechanics 1, we're looking at forces through a dissected beam currently. I did have an "A-ha" moment last week realising that the bend above and below the horizontal axis of the beam (tension and compression) indicates which direction the moment goes in (as often the reaction forces feel instinctively opposite to what I feel, I guess a reaction is always going to be opposite to the applied force) But thank you for replying, you must truly have passion for this subject.

Sometimes our instinct/intuition is wrong. Good and correct free body diagrams and use of the equilibrium equations will tell us the ‘truth’. Keep thinking more deeply about what you’re doing, and asking questions, and you’ll get it.

@@TerryBrownMechEng I'm studying distance education, so just reading textbooks really. So your videos help thank you.

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+k

It's - k

Hi Abam, sorry, I don't have any other videos of this problem. Also, I don't really understand what you mean by "moment of A from 4 metres". A is a just a location on the beam. When we talk about moments, we talk about the moment of a force (or sum of moments) about a particular point, e.g. A, B or any other convenient or relevant point.

these textbook style problems are simplified to provide relatively simple problems that students can apply the methods of engineering mechanics to solve relatively quickly and easily. When you see forces depicted as an arrow at a point, like in this problem, the arrows actually represent the force caused by some other body/component contacting or connected to the body/component we are analysing. Instead of showing the connected/contacting body/component and making you solve a multi-body problem, the textbook just shows the force that is applied. We do a similar thing in the solution when we replace the angled surface at B with a force at 30 degrees. The 200 N applied force at 20 degrees could have come from a smooth contact with another surface at 20 degrees to horizontal.

it's given in the diagram at the start. Vertical distance between the pin connections A and C.

the 0.2 dimension is the distance from the centre-line of the beam to the top of the beam (i.e. the beam is 400mm deep)

Thanks for the feedback. I will do so. I have a few new ones that are unlisted at the moment that I will make public soon. Do you mind telling me whether you are one of my UTS students or not?

Hi Sir. Thank you for your reply. Your videos are very helpful. Unfortunately, I wasn't one of your UTS students.

thanks for letting me know. And thanks again for the feedback. I sometimes wonder how worthwhile it is doing these videos as there are now many others doing similar and better e.g. czcams.com/users/TheEfficientEngineer and czcams.com/users/QuestionSolutions. Wishing you all the best for your engineering studies.

You're welcome!

do you mean the 200 N.m applied couple moment? I'm not sure why you think it is clockwise, the curved arrow is indicating anti-clockwise. If you mean the 200 N force applied at end B, then yes, it will cause a clockwise moment about point A and I have shown that moment as negative in my solution i.e. -200 x 7

yes. In these problems we are finding force-couple systems that are equivalent to the original external loads (not finding the reactions). If you determine the reaction forces/moments for each of these equivalent systems you should get the same answers for the reactions. It's actually a good exercise to do this to help with understanding. I do this in this video czcams.com/video/iJpRW6dZDzY/video.html

Exercise taken from Duff and Ross, “Freehand Exercises in Technical Graphics”, PWS-Kent, 1990

Eoin - not too late, I'm stil here :) . If you take moments about point B, the moment effect of the applied 200 N force about point B is zero (since the perpendicular distance from the line of action of the force to the point is zero). Try it and see that you get the same answers. This is a good way to check your equilibrium equations and calculations.

@@TerryBrownMechEng thank you very much!

Hi, thanks for the question. Taking moments about point A, the line of action of Rbx passes through point A, therefore the perpendicular distance is zero. And therefore Rbx has no moment about point A.

Hi Arinze, the negative answer that I get for RAx indicates that its sense is to the left (because in my free body diagram I assumed it to be to the right). If I had drawn RAx to the left, my value for RAx would have come out positive.

@@TerryBrownMechEng Ok, thank you very much sir. But let's assume that Ax is towards the right, would it then mean that Ax= AC?

@@arinzeanthony7447 I did assume it was to the right (in my free body diagram), and then in the solution the equation of equilibrium shows that it is equal in magnitude to the horizontal (or x) component of RC but in the opposite sense (I.e. negative and therefore to the left). Sorry for the delayed reply. I didn’t see the notification of your comment til now.

Glad it was helpful. Thanks for the comment and feedback. All the best with your studies in engineering.

Thanks for the comment and feedback. When you say “chaos”, are you referring to the fact that lots of reaction forces turn out to be zero? This does seem counter-intuitive at first, but if you look at the original problem, you can see that the bearing at A is not actually needed for equilibrium for the loads and cable support specified. There is nothing to cause rotation around the y-axis or z-axis which leads to the result that the reaction forces at A are zero. If we included the weight of the bent rod, then we would have weight of the section AB causing a moment about the y-axis and you would get a non-zero answer for the vertical reaction at A. Which forces do you think I have labeled incorrectly? They all look ok to me but sometimes it’s easy to miss your own mistakes. Also, which moments do you think I did incorrectly? Again, they all look ok to me. What values did you get for all the reaction forces?

Hi Brian, sorry, I'm not sure exactly what you mean by A and B supported. It's difficult to interpret without drawing/picture. Nevertheless, if there's a pin support at A that prevents vertical motion, then there will generally be a reaction force Ray. If there's a support at B, as well as A and D, the problem is likely to be statically indeterminate. If DC is pinned at both ends (i.e. it's a 2-force member), then Rc is equal and opposite Rd and directed along the line joining C and D, therefore Rdy = Rcsin45 =Rdsin45

Thanks! I have been wondering whether to continue making these videos as there are so many other people making similar videos. All the best with your engineering studies.