That last question was brutal. as always thank you. Your videos will be viewed by generations of engineering students to come, because of how clean they are and how clear the explanations are.
Hey man, just wanted to say thank you for your videos. I'm not sure I would have passed my statics class without the examples and the clear way you lay out information. I'm glad to see you are still making videos. If you make videos about stress, strain, and other strength of materials topics I'll be here to watch! Thanks again.
Thank you for taking the time to write a comment like this, I really appreciate it. I will definitely create videos about stress, strain, and I have been requested multiple different topics to cover, so hopefully, in due time, I can do them all. Truly happy for you, and I wish you the best on your future endeavors! Thanks again.
@@QuestionSolutions please sir make videos on mechanics of materials and structural design and even hydrology/fluid mechanics. I will make a gofund me page for this let me know.
I cant express my love on how many times this channel helped me, clean and straight to the point!!!!! Btw the 2force member actually really helped turn a lot of the complex questions into easier ones, i hope my prof actually teach this.
I am really glad to hear it! The 2 force member should definitely be in your textbook, and hopefully your professor covers it, but if they don't, you can always ask during office hours.
Always manage to do a better job at teaching me the material than my professor. I recommended ur account to other students cuz of how much of a blessing ur Statics playlist is!!
This video summed up what my professor couldn’t. I am a bit foggy on solving the three equations at the end. I thought B(y) would be the starting point but then there is theta. Idk how to isolate the 3 unknowns there. Or better yet which equation to start with. But still thank you for the video, it helped a great deal. I subscribed and will continue to watch
So you can use any of the variables to start off the solving process. In general though, exams usually give you easy to solve equations. The process is to isolate for one variable, plug that into the next, then isolate again for the same variable in the next equarion and plug that into the last equation. I really encourage you to revise simultaneous equation solving with elimination and substitution. It's really important for most courses :)
Hey, in the last question, at the very last part, with 3 equations with 3 unknowns, how do we actually solve it? I've tried many times before, but wasn't able to manually do it without a calculator😅 You're a great teacher!
Yes, that one is more tedious, because you will have to use trigonometric identities. Remember that tan θ = sinθ/cosθ. So if you divide everything by cosθ, it will probably help you solve this question easier. If not, another way is to graph it. :)
Sir，thank u for your great video, I would like to ask why did you count the tension of the rope at 6：39？isn't it also an external force for the member BC just like the previous example at pin C at 4:32?
I just want to point out that the first example is a pretty rare and unique problem where we are specifically told a certain force is applied to a pin so we had to look at internal and external forces that are applied to the pin and that are applied to the whole body. In the 2nd example, it's the same as someone applying a random force just a certain small distance away from the body. So we didn't calculate forces applied to a specific pin, rather, each member itself. So the forces at 6:39 are forces applied to the body.
Please kindly re-watch 1:36. The reason we knew that was because it's a two-force member. It's good to recognize them, so you can have 1 single force instead of having one vertical and one horizontal. Makes the math simpler, but only if it's a two force member.
I have a question, in the third question why we cant take the bar AB to calculate first instead to CD and I got different answer. Ur vid is great keep up the good work !
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.
Very informative video. Thank you. I have a question, at 4:45 ,is it that whenever we have such a case( a force applied at a pin joint) and we are considering equilibrium of the two pin jointed members separately, the external forces acting at the joint isn’t included
This is a very rare and specific situation. We only treat it the way we did because of the wording in the question, "The 600 N force is applied to the pin." I just wanted to showcase what needs to be done if such a question arises. Otherwise, you can solve it like any other problem shown on the video assuming the forces are applied to the frame.
It's not external when looking at a specific beam, so for example, when looking at ABC. However, looking at the whole system as a whole, the 600 N force will create a moment about point E. That's why when we look at member ABC at 4:09, we don't include it. I explain the process behind it around that time as well so you can understand how to solve problems with this type of wording.
Amazing video, man! But I didn't quite understand why you ignored the 600KN force at joint C. Is it because if we're dismantling the frame, any force applied at the joints will become an internal force and won't be considered?
We are told that the force is applied to the pin. This question is actually a tricky one because of the wording and it's quite rare. I just wanted to show it so that if you ever get a curveball thrown at you, you can figure it out.
Informative !! thank you so much for the video. I have a small doubt. I did not understand why you have not considered the 600N external force while drawing FBD. Can you please clarify
@@shyamdev6928 Are you referring to the 600N force applied to the pin? This question is sort of tricky because it specifically states that the 600 N force is applied to the pin only.
So the y-component would face straight up. Imagine this object is free to spin about point C. If we push up at the location where the y-component was applied, would it turn clockwise about point C, or counter-clockwise? It would turn counter-clockwise. Since we picked clockwise to be positive, and this would cause a counter-clockwise moment, it is negative.
Hey do you have fluid mechanics tutorials? I used to watch your videos for engineering dynamics and statics and passed them. I would like to see your videos for fluid and soil mechanics
Hello, at 9:45 why isn’t the answer for Ay = 9.91 kN in the upwards direction? I’ve done it on my calculator a bunch of times but do not get the same answer. I think the direction on Dy is in the wrong direction, since Cy = 1.52kN in the upwards direction and 7*sin(60) = 6.06kN in the downwards direction, surely Dy should be 4.54kN in the upwards direction to balance out and have net 0 newtons in the y direction … please advise thanks
I am not sure if you have the timestamps correct or if you're asking about the right question? Ay is already shown to be upwards, and I am not sure where you got the 9.91 kN from? Ay is 3.09 kN. Please let me know where you're referring to so I can help you out better. Again, even for Dy I don't know what you're referring to. It's shown to be upwards, we got a positive value, which confirms that it is indeed a force in the upwards direction.
@@bnyaminsabah It's always really nice to read a comment like yours. Thank you for taking the time to write one. I wish you the best in your studies. Keep up the great work!
On the first question what would the reaction forces be for E? Would Ex be 950N and Ey be 600N. If so, when I drew the FBD of member EDC the forces did not add up. One way I got Cx to be 350 while the video shows 650. I'm confused.
To find the reactions at E, you would have to write a moment equation about point A and then go through the same process. Also, the question is asking for the components of the force pin C on ABC, not EDC.
Please watch this video first: th-cam.com/video/jQDEOwrR4UU/w-d-xo.html The simple answer is that you don't need to know, since it's just an assumption. If you get a negative answer, then the direction is opposite to your assumption (the magnitude is the same).
What a helpful video! Sir, I have a question. In 04:27 you describe Cx and Cy each by member of ABC and EDC. Is it correct to understand that they are not the action and reaction, but just internal forces by each member?
Think of it as the force of each member, like member ABC, and EDC applied to the pin, just broken into components. This is one of those videos where the wording in the question is what sets it apart, because we are told that the 600 N force is applied to the pin. Usually, they are forces applied to members, not pins.
I still have some doubts about the method of drawing the force on the pin. When I specify a direction, such as the force extending along the positive x-axis, sometimes it is a needle arrow (the arrow at the needle extends outward), and sometimes it is an arrow needle (the arrow extends towards the needle), how do I determine whether to extend outward or towards the needle. For example, at one minute and 30 seconds in the video, the force in the x-direction is an arrow that extends outward (with the pin reversed). If I draw an arrow pointing to a pin, can I?)
So what you're referring to are vectors. Vectors are lines with an arrowhead. We draw them to show a force in a certain/specific direction. Usually, we make an assumption as to the direction. If you get a positive answer, then your assumption was correct. If you get a negative answer, your assumption was incorrect, and it's opposite to your assumed direction. It might be helpful to watch this video: th-cam.com/video/Ixv1QYUAMWk/w-d-xo.html and this video: th-cam.com/video/jQDEOwrR4UU/w-d-xo.html
Is this Engineering? I'm wondering because as a 6th grader, I am teaching myself Electromagnetism and Engineering on Edx, and I'm wondering if this is engineering statics. I am also wondering because I've done problems that look similar to this, like finding Tension in ropes. I'm currently learning Trusses right now, and I'm piecing together information. Thanks. And yes, I'm a 6th grader, and yes I'm taking college courses on Edx, and yes, I'm self-teaching myself. I would be grateful in your favor if someone could respond :D
Yes, this playlist includes topics covered in a first year engineering statics course. The dynamics playlist includes first year dynamic topics, and the thermodynamics playlist covers first year topics relating to thermodynamics. All topics are engineering topics, but they also get shared by a few other faculties as well, like physics and chemistry. Awesome job for doing all of this at your age. Keep up the great work!
tq fr ure replied!! at 3:04 u said point E only have Ey n Ex ...while in the calculation at 3:15 you use moment about E to find Fa... are we not supposed to use moment about point E ?because point E only has 2 support reactions
@@hazwelnut. Okay, thanks for the timestamps. So I think there is a misunderstanding. There is NO moment reaction at a pin, since at a pin, the member can freely rotate clockwise or counter-clockwise. In other words, a counter moment is not created. What we did was just write a summation of moments about point E. This just means if we are standing at point E, what sort of moments will be felt due to the moments created by other forces. You can write a summation of moments about any point you want, it could even be at the middle of beam, or wherever you please. Please see: th-cam.com/video/QNNnPZ68STI/w-d-xo.html
I don't know, it's not possible for me to answer your question since I can't see any of the steps you took, where you made a mistake, or whether you wrote down the proper equations 😅
First, I sum the moment at point B, then I assume Ay upward(clockwise) moment arm = 8 Second, the 16 kn pointing downward(counter clockwise) moment arm = 4
Now I am in statically indeterminate frames 😔 it's something way different than we used to do in structural analysis 1 and statics dynamics and strength of materials
Everything you mentioned is already covered in the playlist. Look at the moments video if you would like to see problems solved involving moments, look for the equilibrium video if you want to see problems solved about equilibrium 👍 the playlist for statics has it all :)
@@QuestionSolutions Oh okay because in the video you said that you can only use two combinations of members: - ABC and DBF - EDC and DBF So, I wasn't sure if EDC and ABC would be valid. Is it?
@@ImmortalAJ-r2t I don't use snapshots but I think that can work if you use high-res images. Most latex generators can export as a PDF, so I export as pdf. Then you get vector equations. You can use a photo editing program to get it to proper size as well.
@@QuestionSolutionsProbably yes , I encountered a semicirclular frame that has a semicircular distributed load w =sintheta. In statics hibbeler chapter 7. How can I resolve a semicirclular load distribution ?
I am not entirely sure of your question, since nothing was discarded. We needed to find the reaction forces at A, and the easiest way to do that was to take the moment about point E. So what we want to do is take the moment about a point that eliminates as many unknowns as possible. When we take about it about point E, we can eliminate the reaction forces about that point since their lines of action goes through that point. Please see: th-cam.com/video/QNNnPZ68STI/w-d-xo.html if you need a refresh on moments.
So take a look at the 16kN force with respect to the purple dot in the middle. If the beam can freely rotate about that purple dot, when we apply the 16kN force, would it turn clockwise or counter clockwise? It would be counter clockwise, so it's negative because we assumed clockwise moments to be positive.
sir please upload rigidbody kinetics and rigid body kinematics i watched already but now I can't find the video. please reply with the exact link sir please.
@@driesvanranst3440 I actually go through this in details since students tend to get confused here. It's very important that the question states "600 N force is applied to the pin" so please watch from 4:12. I even draw out the pin diagram to explain this. Also no, it's not 0N, the diagrams are correct 👍 If you still don't understand after re-watching that part, I will do my best to re-explain it the best I can :)
@@driesvanranst3440 No, we solved for Cy, we wrote a moment equation about point B and it turns out, Cy is 0 since no other forces are effecting the beam. What process did you use to get Cy to equal 600N?
@@QuestionSolutions I'm very confused. I used 2 different ways of solving the problem. First time I got Cy = 0N, second time Cy = 600N (up). I solved by using members ABC and EDC. What I supposed: When drawing the EDC and ABC members separately, at least one of the diagrams should include the given 600N (down) at point C. When I put the 600N on the EDC diagram, the result is that Cy = 0N. When put on the ABC diagram, the result is Cy = 600N (up).... Would be easier if I could show you my Free Body Diagrams.
I completely disagree on the first problem. If this were the case then the pin itself would not be in equillibrium. It's more accurate to distribute the force evenly to both members as an external 300N load and then Cy would be 300N
So this question tends to be a confusing one for a lot of students, but the key is in the wording. "The force is applied to the pin", very few questions you'll encounter will say that. You just have to read problems really carefully and apply your knowledge to solve it. I encourage you to speak with your professor during their office hours and get an insight into this problem if it's confusing so you can clear up any misconceptions. If this type of problem shows up on your exam, you'll have a good foundation to tackle it. Best wishes :)
@@QuestionSolutions I deffinetly will I'm probably wrong and I'm not opposed to being wrong but just looking at the diagram of the pin its clear if Cy is 0 that the pin would be pulled down. Part of the reason I'm confident is I also made a simpler similar problem myself of two two force members With pin supports on either end Creating an isocscles triangle with the bottom angles being 45 degrees and A 100N load being applied at the top vertically down on the pin. Breaking up the Diagram into the two members. The only way for Cy,Cx to be the same for both two force members was if I split the 100N load between Both Two force members so 50N down for both. In which case Cy would be 0 and Cx 50 N to the left for the left Member and the right for the right member. If you Treat the 100N load like it doesnt exist when you break up the two members you get a completely wrong answer. To me this makes sense cause even if you apply a load on the pin. Assuming it's symmetrical Aka Both members are equal distance from where the load is applied the load will be distributed evenly to both of them down.
@@chrismarklowitz1001 One thing you can do is write to Hibbler (author of the book) and directly ask him to get more clarification. But yeah, I definitely would ask your professor or TA and ask this question with a force applied just to the pin. It's a very good question, but often tricky for a lot of students. Good to teach it before it goes on the exam 👍 Also, are you understanding the diagram properly? The pin is between EDC and ABC. It can't fall out because Cy is 0. It's between the two beams. The force is countered at the supports.
I think you might be misunderstanding. There is no moment, because the beam is in equilibrium, there is no movement. It's pinned so it cannot rotate about point B.
@@QuestionSolutions Doesn’t pinned support mean that it’s locked in the x and y axis, but free to rotate? I thought that fixed support was the only support type that prevented rotation. Is there a particular reason why this beam can’t be treated as an overhang beam after the load is calculated at D? Or am I missing something fundamental about frames and machines?
@@kuklangaren I think for these types of problems, the best way is to try and imagine this beam moving. It's pinned at B, but notice we have a roller at A and C. So let's say we push down at C, can it rotate? No, why? Because the roller at C will prevent it from turning clockwise. Now let's say we push at A. Can it rotate? No, why? Because the roller at A will prevent it from turning counter clockwise. So what that means is, it doesn't matter where we cut this beam, it will be in equilibrium. This principle is so important when we look at trusses. Especially when we make cuts through a member. So while we do have a pin at D, this whole collective beam will be in equilibrium so moment is equal to 0. Another tip you should keep in mind is that if you're doing a statics course, your object is always in equilibrium, it will not be moving. All equations, which means F_x = 0, F_ y= 0, F_z = 0, M = 0 will be zero. When you do a dynamics course, then your object will be moving, and you'll have to think about it differently.
@@QuestionSolutions Thanks for the quick replies! What you’re saying makes sense, but I guess what I’m trying to ask is why this beam is calculated differently from a similar looking three support continuous beam? In such a beam, we would still have one pinned support and two roller supports, yet the middle support does have a moment. In order to solve such a beam, you first have to find that moment. Why is that?
Are you talking about 11:04? If so, it's because the wheel is pulling it down, so the spring force is also downwards. The spring creates tension downwards at the top, and tension upwards at the bottom. 👍
So it's a right angle triangle with a hypotenuse of 2 m. Remember that sin is opposite over hypotenuse, and cosine is adjacent over hypotenuse. Draw out the triangle on a piece of paper, label the angle, and the hypotenuse side as 2 m. You can then write the other sides as functions of sin and cosine :)
That last question was brutal. as always thank you. Your videos will be viewed by generations of engineering students to come, because of how clean they are and how clear the explanations are.
You're very welcome :) I really appreciate your comment, very kind of you.
👍 agreed
Hey man, just wanted to say thank you for your videos. I'm not sure I would have passed my statics class without the examples and the clear way you lay out information. I'm glad to see you are still making videos. If you make videos about stress, strain, and other strength of materials topics I'll be here to watch! Thanks again.
Thank you for taking the time to write a comment like this, I really appreciate it. I will definitely create videos about stress, strain, and I have been requested multiple different topics to cover, so hopefully, in due time, I can do them all. Truly happy for you, and I wish you the best on your future endeavors! Thanks again.
@@QuestionSolutions please sir make videos on mechanics of materials and structural design and even hydrology/fluid mechanics. I will make a gofund me page for this let me know.
I cant express my love on how many times this channel helped me, clean and straight to the point!!!!! Btw the 2force member actually really helped turn a lot of the complex questions into easier ones, i hope my prof actually teach this.
I am really glad to hear it! The 2 force member should definitely be in your textbook, and hopefully your professor covers it, but if they don't, you can always ask during office hours.
Always manage to do a better job at teaching me the material than my professor. I recommended ur account to other students cuz of how much of a blessing ur Statics playlist is!!
Thank you so much, really appreciate the shares! Best wishes with your studies.
THANK YOU! I have my statics mid term tomorrow, and I was really stuck.
You got this! Best wishes on your mid term.
This video summed up what my professor couldn’t. I am a bit foggy on solving the three equations at the end. I thought B(y) would be the starting point but then there is theta. Idk how to isolate the 3 unknowns there. Or better yet which equation to start with. But still thank you for the video, it helped a great deal. I subscribed and will continue to watch
So you can use any of the variables to start off the solving process. In general though, exams usually give you easy to solve equations. The process is to isolate for one variable, plug that into the next, then isolate again for the same variable in the next equarion and plug that into the last equation. I really encourage you to revise simultaneous equation solving with elimination and substitution. It's really important for most courses :)
GOAT of static channels...
Thank you!
Still got a low grade man ain't no professor using that easy questions... No one in my class could solve our exams man.
@@alaybey9771 Sorry to hear your exam didn't go too well. Hopefully, you can pull your grades up in the next semester!
Hey, in the last question, at the very last part, with 3 equations with 3 unknowns, how do we actually solve it? I've tried many times before, but wasn't able to manually do it without a calculator😅 You're a great teacher!
Yes, that one is more tedious, because you will have to use trigonometric identities. Remember that tan θ = sinθ/cosθ. So if you divide everything by cosθ, it will probably help you solve this question easier. If not, another way is to graph it. :)
Taught this better in 10 minutes than my prof did in 2 hours
😅Thank you very much. Best of luck with your studies.
Excellent job! Thank you!!
You're very welcome!
This is such an amazing video! THANKS!
You're very welcome!
Sir，thank u for your great video, I would like to ask why did you count the tension of the rope at 6：39？isn't it also an external force for the member BC just like the previous example at pin C at 4:32?
I just want to point out that the first example is a pretty rare and unique problem where we are specifically told a certain force is applied to a pin so we had to look at internal and external forces that are applied to the pin and that are applied to the whole body. In the 2nd example, it's the same as someone applying a random force just a certain small distance away from the body. So we didn't calculate forces applied to a specific pin, rather, each member itself. So the forces at 6:39 are forces applied to the body.
In question 2 how do we know F AB points diagonally up and not down? Great video!
Please kindly re-watch 1:36. The reason we knew that was because it's a two-force member. It's good to recognize them, so you can have 1 single force instead of having one vertical and one horizontal. Makes the math simpler, but only if it's a two force member.
I have a question, in the third question why we cant take the bar AB to calculate first instead to CD and I got different answer.
Ur vid is great keep up the good work !
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.
God bless you!!
Thank you!
The best teacher♥️. Thank you very much.
You're very welcome! ❤
Thank you so much man
You're very welcome!
Very informative video. Thank you. I have a question, at 4:45 ,is it that whenever we have such a case( a force applied at a pin joint) and we are considering equilibrium of the two pin jointed members separately, the external forces acting at the joint isn’t included
This is a very rare and specific situation. We only treat it the way we did because of the wording in the question, "The 600 N force is applied to the pin." I just wanted to showcase what needs to be done if such a question arises. Otherwise, you can solve it like any other problem shown on the video assuming the forces are applied to the frame.
Helpful video thanks for sharing this information, helps for testing! 🤩
Glad it was helpful! Best wishes with your tests.
For the first problem you said the 600N force is not an external force, so why then did you used it to solve for the reactions at A?
It's not external when looking at a specific beam, so for example, when looking at ABC. However, looking at the whole system as a whole, the 600 N force will create a moment about point E. That's why when we look at member ABC at 4:09, we don't include it. I explain the process behind it around that time as well so you can understand how to solve problems with this type of wording.
Thank you for this 🙌🏻
You're very welcome :)
Amazing video, man!
But I didn't quite understand why you ignored the 600KN force at joint C.
Is it because if we're dismantling the frame, any force applied at the joints will become an internal force and won't be considered?
We are told that the force is applied to the pin. This question is actually a tricky one because of the wording and it's quite rare. I just wanted to show it so that if you ever get a curveball thrown at you, you can figure it out.
@@QuestionSolutions
Thank you very much.
I didn't see it that way.
THANKS FOR THE VIDEOS BRO really helpful💓💓💓💓💓
Glad you like them! 💓💓💓💓💓
Informative !! thank you so much for the video. I have a small doubt. I did not understand why you have not considered the 600N external force while drawing FBD. Can you please clarify
Thank you. Please give me a timestamp so I know where you're referring to. Thanks!
@@QuestionSolutions 4min 32 seconds
@@shyamdev6928 Are you referring to the 600N force applied to the pin? This question is sort of tricky because it specifically states that the 600 N force is applied to the pin only.
Hello Professor, question about 7:02 where calculating moment at C. Why is the y-component of F(AB) negative? Thank you your videos sure help a lot!
So the y-component would face straight up. Imagine this object is free to spin about point C. If we push up at the location where the y-component was applied, would it turn clockwise about point C, or counter-clockwise? It would turn counter-clockwise. Since we picked clockwise to be positive, and this would cause a counter-clockwise moment, it is negative.
Great, thank you for the response. I have my test today and you have helped a lot!@@QuestionSolutions
Hey do you have fluid mechanics tutorials? I used to watch your videos for engineering dynamics and statics and passed them. I would like to see your videos for fluid and soil mechanics
I don't have fluid mechanics videos. It's on my list to do, but it probably won't be done for quite some time.
Hello, at 9:45 why isn’t the answer for Ay = 9.91 kN in the upwards direction? I’ve done it on my calculator a bunch of times but do not get the same answer. I think the direction on Dy is in the wrong direction, since Cy = 1.52kN in the upwards direction and 7*sin(60) = 6.06kN in the downwards direction, surely Dy should be 4.54kN in the upwards direction to balance out and have net 0 newtons in the y direction … please advise thanks
Also, if you take a look at 9:02, the Dy is in the correct direction, so it must be a simple error.
I am not sure if you have the timestamps correct or if you're asking about the right question? Ay is already shown to be upwards, and I am not sure where you got the 9.91 kN from? Ay is 3.09 kN. Please let me know where you're referring to so I can help you out better. Again, even for Dy I don't know what you're referring to. It's shown to be upwards, we got a positive value, which confirms that it is indeed a force in the upwards direction.
This video is goated 🐐
Thanks!
Great video 🌼
Thank you!
@@QuestionSolutions no thank you your videos are so nice and usefull 😊💛
@@bnyaminsabah It's always really nice to read a comment like yours. Thank you for taking the time to write one. I wish you the best in your studies. Keep up the great work!
Appreciate your efforts ❤️❤️
Thank you so much 😀 I saw you didn't upload anything recently, have you been busy?
@@EveryEngMechanical Prefer not to say 😅
@@QuestionSolutions 😂okok
@@EveryEngMechanical ❤
On the first question what would the reaction forces be for E? Would Ex be 950N and Ey be 600N. If so, when I drew the FBD of member EDC the forces did not add up. One way I got Cx to be 350 while the video shows 650. I'm confused.
To find the reactions at E, you would have to write a moment equation about point A and then go through the same process. Also, the question is asking for the components of the force pin C on ABC, not EDC.
you are the best
Thanks!
Thank you this is amazing
You're very welcome!
May you plzzzzz explain chapter 9 CENTER OF GRAVITY, CENTER OF MASS,
AND CENTROID OF A BODY your explanations is amaazinggggg
Yes, I will add it to my to-do list, not sure when I can get to it though. Sorry! :(
@@QuestionSolutions thank youuuuuuu
Good explanation ,
Glad to hear!
How did you solve the 3unknowns in the last problem? Thank you
I think I graphed them.
I was searching for Frames in Image Compression, and found this video in the results, tbh these Frames are more interesting than those Frames 😂
That's really funny 😅 Really happy you found this even a little bit interesting.
Hi , i have a questions for this lesson! It is how can we know exactly the directions of reactions at pin supports ???
Please watch this video first: th-cam.com/video/jQDEOwrR4UU/w-d-xo.html
The simple answer is that you don't need to know, since it's just an assumption. If you get a negative answer, then the direction is opposite to your assumption (the magnitude is the same).
Thank you!
You're very welcome!
question 2 was clever
😀😀
What a helpful video!
Sir, I have a question. In 04:27 you describe Cx and Cy each by member of ABC and EDC. Is it correct to understand that they are not the action and reaction, but just internal forces by each member?
Think of it as the force of each member, like member ABC, and EDC applied to the pin, just broken into components. This is one of those videos where the wording in the question is what sets it apart, because we are told that the 600 N force is applied to the pin. Usually, they are forces applied to members, not pins.
@@QuestionSolutions Thanks a lot:)
@@chimchar2200 :)
@@QuestionSolutions this was the best replay for all my confusions. thanks, a lot!
I still have some doubts about the method of drawing the force on the pin. When I specify a direction, such as the force extending along the positive x-axis, sometimes it is a needle arrow (the arrow at the needle extends outward), and sometimes it is an arrow needle (the arrow extends towards the needle), how do I determine whether to extend outward or towards the needle. For example, at one minute and 30 seconds in the video, the force in the x-direction is an arrow that extends outward (with the pin reversed). If I draw an arrow pointing to a pin, can I?)
So what you're referring to are vectors. Vectors are lines with an arrowhead. We draw them to show a force in a certain/specific direction. Usually, we make an assumption as to the direction. If you get a positive answer, then your assumption was correct. If you get a negative answer, your assumption was incorrect, and it's opposite to your assumed direction. It might be helpful to watch this video: th-cam.com/video/Ixv1QYUAMWk/w-d-xo.html and this video: th-cam.com/video/jQDEOwrR4UU/w-d-xo.html
Is this Engineering? I'm wondering because as a 6th grader, I am teaching myself Electromagnetism and Engineering on Edx, and I'm wondering if this is engineering statics. I am also wondering because I've done problems that look similar to this, like finding Tension in ropes. I'm currently learning Trusses right now, and I'm piecing together information. Thanks. And yes, I'm a 6th grader, and yes I'm taking college courses on Edx, and yes, I'm self-teaching myself. I would be grateful in your favor if someone could respond :D
Yes, this playlist includes topics covered in a first year engineering statics course. The dynamics playlist includes first year dynamic topics, and the thermodynamics playlist covers first year topics relating to thermodynamics. All topics are engineering topics, but they also get shared by a few other faculties as well, like physics and chemistry.
Awesome job for doing all of this at your age. Keep up the great work!
Thx !!
You're welcome :)
Why should we neglect the reactions at pulley at 6:24?
Because we are looking at the external forces. The effects of the pully on the pin doesn't help us solve the problem 👍
Second question the distance is 0.5 how did you get 0.8 from where did u add sir
So the radius of the pulley is 0.3m. So you get 0.5 m + 0.3 m = 0.8 m. :)
thanks for your feedback sir hope you keep updating more videos
appreciate it sir @@QuestionSolutions
In 1st example (at 3:10) how force at point A produces negative/clockwise torque?
If you push at A, the frame would spin counter-clockwise. Since we assumed clockwise to be positive, that would give us a negative value.
Hi, fr the first question... why there's a moment at point E when it only hve to support reactions namely Ex n Ey?
Please give a timestamp so I know where you're referring to. Thanks!
tq fr ure replied!! at 3:04 u said point E only have Ey n Ex ...while in the calculation at 3:15 you use moment about E to find Fa... are we not supposed to use moment about point E ?because point E only has 2 support reactions
@@hazwelnut. Okay, thanks for the timestamps. So I think there is a misunderstanding. There is NO moment reaction at a pin, since at a pin, the member can freely rotate clockwise or counter-clockwise. In other words, a counter moment is not created. What we did was just write a summation of moments about point E. This just means if we are standing at point E, what sort of moments will be felt due to the moments created by other forces. You can write a summation of moments about any point you want, it could even be at the middle of beam, or wherever you please. Please see: th-cam.com/video/QNNnPZ68STI/w-d-xo.html
@@QuestionSolutions Thanks !! That clears up my confusion. Have a good day sir. Keep it going!
In the third question why I can't get your answer in Ay when I use AB for solving, my answer is 8kn
I don't know, it's not possible for me to answer your question since I can't see any of the steps you took, where you made a mistake, or whether you wrote down the proper equations 😅
First, I sum the moment at point B, then I assume Ay upward(clockwise) moment arm = 8
Second, the 16 kn pointing downward(counter clockwise) moment arm = 4
Now I am in statically indeterminate frames 😔 it's something way different than we used to do in structural analysis 1 and statics dynamics and strength of materials
I hope you're still doing well on your courses. Keep up the good work and do your best :)
I have question about the first question, why we not consider BD as the two force members?
It's pinned at B, and it's not bent or anything like that. So it can't be a 2 force member. 👍
@@QuestionSolutions thank you for replying! It's more clear now!
@@fitrizuhairi1898 Glad to hear :)
can u do more l specific videos on just solving topics on moment , vectors and equilibrium from hibbeler textbook 13 edition
Everything you mentioned is already covered in the playlist. Look at the moments video if you would like to see problems solved involving moments, look for the equilibrium video if you want to see problems solved about equilibrium 👍 the playlist for statics has it all :)
In Q1 why can't we use members EDC and and ABC?
You can use whichever ones you want to get the answer. I just showcase one method of getting to an answer. :)
@@QuestionSolutions Oh okay because in the video you said that you can only use two combinations of members:
- ABC and DBF
- EDC and DBF
So, I wasn't sure if EDC and ABC would be valid. Is it?
In which software you write mathematical equations
I use word with the latex functions. If some of the equations don't look nice, I use online latex generators.
@@QuestionSolutions sir do you take snapshot of equation or convert it into vector file
my AI file looks blurry in AE
@@ImmortalAJ-r2t I don't use snapshots but I think that can work if you use high-res images. Most latex generators can export as a PDF, so I export as pdf. Then you get vector equations. You can use a photo editing program to get it to proper size as well.
@@QuestionSolutions thanks sir i will try
Hello, could you do a curved frames that have a radius R?
Is it a curve that can be considered a 2 force member, like the one at 1:48?
@@QuestionSolutionsProbably yes , I encountered a semicirclular frame that has a semicircular distributed load w =sintheta. In statics hibbeler chapter 7. How can I resolve a semicirclular load distribution ?
in the first example why is it possible to take moments then completely disregard those forces? How did you decide which joint to take moments about?
I am not entirely sure of your question, since nothing was discarded. We needed to find the reaction forces at A, and the easiest way to do that was to take the moment about point E. So what we want to do is take the moment about a point that eliminates as many unknowns as possible. When we take about it about point E, we can eliminate the reaction forces about that point since their lines of action goes through that point. Please see: th-cam.com/video/QNNnPZ68STI/w-d-xo.html if you need a refresh on moments.
At 9:43 why is 16(4) negative but all the other ones are positive?
So take a look at the 16kN force with respect to the purple dot in the middle. If the beam can freely rotate about that purple dot, when we apply the 16kN force, would it turn clockwise or counter clockwise? It would be counter clockwise, so it's negative because we assumed clockwise moments to be positive.
from what reference are these questions
Whatever books I use, I write it in the description 👍
sir please upload rigidbody kinetics and rigid body kinematics
i watched already but now I can't find the video. please reply with the exact link sir please.
Check the dynamics playlist here: th-cam.com/play/PLXePpKFSUW2ZXw_D5h0TTyac-KGlFRxnS.html
In the ABC member of first example, I don't get why we should draw 1250N in the Free Body Diagram, but not the 600N.
Actually I think Cy should be 600N pointing upwards instead of 0N. Could it be you made a mistake here?
@@driesvanranst3440 I actually go through this in details since students tend to get confused here. It's very important that the question states "600 N force is applied to the pin" so please watch from 4:12. I even draw out the pin diagram to explain this. Also no, it's not 0N, the diagrams are correct 👍 If you still don't understand after re-watching that part, I will do my best to re-explain it the best I can :)
@@QuestionSolutions But at 5:42, you wrote Cy = 0N. Shouldn't that be 600N (upwards)?
@@driesvanranst3440 No, we solved for Cy, we wrote a moment equation about point B and it turns out, Cy is 0 since no other forces are effecting the beam. What process did you use to get Cy to equal 600N?
@@QuestionSolutions I'm very confused. I used 2 different ways of solving the problem. First time I got Cy = 0N, second time Cy = 600N (up). I solved by using members ABC and EDC. What I supposed: When drawing the EDC and ABC members separately, at least one of the diagrams should include the given 600N (down) at point C. When I put the 600N on the EDC diagram, the result is that Cy = 0N. When put on the ABC diagram, the result is Cy = 600N (up).... Would be easier if I could show you my Free Body Diagrams.
I completely disagree on the first problem. If this were the case then the pin itself would not be in equillibrium. It's more accurate to distribute the force evenly to both members as an external 300N load and then Cy would be 300N
So this question tends to be a confusing one for a lot of students, but the key is in the wording. "The force is applied to the pin", very few questions you'll encounter will say that. You just have to read problems really carefully and apply your knowledge to solve it. I encourage you to speak with your professor during their office hours and get an insight into this problem if it's confusing so you can clear up any misconceptions. If this type of problem shows up on your exam, you'll have a good foundation to tackle it. Best wishes :)
@@QuestionSolutions I deffinetly will I'm probably wrong and I'm not opposed to being wrong but just looking at the diagram of the pin its clear if Cy is 0 that the pin would be pulled down. Part of the reason I'm confident is I also made a simpler similar problem myself of two two force members With pin supports on either end Creating an isocscles triangle with the bottom angles being 45 degrees and A 100N load being applied at the top vertically down on the pin. Breaking up the Diagram into the two members. The only way for Cy,Cx to be the same for both two force members was if I split the 100N load between Both Two force members so 50N down for both. In which case Cy would be 0 and Cx 50 N to the left for the left Member and the right for the right member. If you Treat the 100N load like it doesnt exist when you break up the two members you get a completely wrong answer. To me this makes sense cause even if you apply a load on the pin. Assuming it's symmetrical Aka Both members are equal distance from where the load is applied the load will be distributed evenly to both of them down.
@@chrismarklowitz1001 One thing you can do is write to Hibbler (author of the book) and directly ask him to get more clarification. But yeah, I definitely would ask your professor or TA and ask this question with a force applied just to the pin. It's a very good question, but often tricky for a lot of students. Good to teach it before it goes on the exam 👍
Also, are you understanding the diagram properly? The pin is between EDC and ABC. It can't fall out because Cy is 0. It's between the two beams. The force is countered at the supports.
At 9:45, why is the moment 0 at B? Why is this not treated as a continuous beam?
I think you might be misunderstanding. There is no moment, because the beam is in equilibrium, there is no movement. It's pinned so it cannot rotate about point B.
@@QuestionSolutions Doesn’t pinned support mean that it’s locked in the x and y axis, but free to rotate? I thought that fixed support was the only support type that prevented rotation. Is there a particular reason why this beam can’t be treated as an overhang beam after the load is calculated at D? Or am I missing something fundamental about frames and machines?
@@kuklangaren I think for these types of problems, the best way is to try and imagine this beam moving. It's pinned at B, but notice we have a roller at A and C. So let's say we push down at C, can it rotate? No, why? Because the roller at C will prevent it from turning clockwise. Now let's say we push at A. Can it rotate? No, why? Because the roller at A will prevent it from turning counter clockwise. So what that means is, it doesn't matter where we cut this beam, it will be in equilibrium. This principle is so important when we look at trusses. Especially when we make cuts through a member. So while we do have a pin at D, this whole collective beam will be in equilibrium so moment is equal to 0. Another tip you should keep in mind is that if you're doing a statics course, your object is always in equilibrium, it will not be moving. All equations, which means F_x = 0, F_ y= 0, F_z = 0, M = 0 will be zero. When you do a dynamics course, then your object will be moving, and you'll have to think about it differently.
@@QuestionSolutions Thanks for the quick replies! What you’re saying makes sense, but I guess what I’m trying to ask is why this beam is calculated differently from a similar looking three support continuous beam? In such a beam, we would still have one pinned support and two roller supports, yet the middle support does have a moment. In order to solve such a beam, you first have to find that moment. Why is that?
@@kuklangaren Maybe I am misunderstanding you. Are you asking about a single beam that is supported at 3 spots vs 2 beams pin connected?
Do you have any examples of this topic?
Sorry, which topic?
@@QuestionSolutions the frames and machines, Do you have more videos about them?
@@jjjjjjjjj2476 This is the only one, I try to cover 3 to 4 examples per topic. 👍
@@QuestionSolutions I appreciate that, thank you sir
You're very welcome
Sir
Why the direction of Fs is downward?
Is it known from the question?
Are you talking about 11:04? If so, it's because the wheel is pulling it down, so the spring force is also downwards. The spring creates tension downwards at the top, and tension upwards at the bottom. 👍
whts the name this book bro?
Any books used as reference is in the description 👍
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I love you
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Bro im kind of confuse on 11:29 on how it was 2 cos and 2sin 😭
So it's a right angle triangle with a hypotenuse of 2 m. Remember that sin is opposite over hypotenuse, and cosine is adjacent over hypotenuse. Draw out the triangle on a piece of paper, label the angle, and the hypotenuse side as 2 m. You can then write the other sides as functions of sin and cosine :)
Didn't get a thing :)
I am sorry to hear that :(
thank you but you explain things really fast
I try to keep them as concise as possible. 😅
This guy sounds Filipino based on his accent, but I could be wrong
I'm not Filipino 😅
very bad
Sorry to hear that!