Relative Motion Analysis of Two Particles Using Translating Axes (learn to solve any problem)
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- čas přidán 4. 03. 2020
- Learn how to solve relative motion analysis of two particles problems, step by step. By the end of the 4 examples, you should be able to solve any problem you face in this chapter.
Two planes, A and B, are flying at the same altitude...
At the instant shown, car A has a speed of 20 km/h...
The man can row the boat in still water with a speed of 5 m/s...
At the instant shown, cars A and B are traveling at velocities...
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.
🔹When to use sin or cos: czcams.com/users/shortsvynnKlJD_Jo
Just found your channel a couple days before my first exam and its incredible how you can clear up these topics in just aboout 10 minutes. Thank you for these videos
You're very welcome! I appreciate you taking the time to comment, very nice of you :)
@@QuestionSolutions the best channel :)
No one would ever fail maths if they had a teacher as clear and intelligent as yourself!
Thank you very much for your kind words. I really appreciate it. :)
Another underrated tutorial channel. You are a great help. Thank you !
Many thanks! Glad to hear this helped you, and I appreciate your comment :)
I spend almost a whole semester just to understand dynamics but this tutorial makes me understand it in just 11 minutes. Thanks a lot !
Great to hear! I hope you do great in your semester. Best wishes :)
amazing video,it should hit 1M , bro i am praying for you best life, u just cleared major concepts in just 11 minutes, ma sha allah, may allah guide you and help u!
Thank you so much. I am really glad this video helped you. Best of luck with your studies!
Jeff Hansen's videos and your videos for recap and problems to work on with help have done wonders for my studying. You guys are amazing!!!
Really glad to hear these videos are helpful! Best wishes with your studies :)
So beautiful, so elegant, just explaining like a wow ❤
Thank you so much! ❤️
you are great. you made it very clear to me in 10 minutes.
I am happy to hear that. Many thanks!
Literally the best videos online for my engineering courses - they help SO much
Thank you very much. I am really glad they are helpful to you. Keep up the great work and best wishes with your studies!
you have been seeing my comments on all your dynamics videos but you are so amazing! Im actually understanding these concepts now!!
I try my best to look through all the comments every day when I can. :)
I am really happy when students learn concepts better, that's why I started this to begin with. It's like an addition to your textbook/course material. I really appreciate you taking the time to write a comment, I enjoy reading them, so thank you very much!
your channel is the best to watch ever..........May God bless you
Thank you very much!
Absolutely excellent!... exceptional!... What can I say.... Phenomenal!.….. Thank you very much!
You are very welcome!
What a legend! great work my brother! You made me an Engineer! keep up the Best!
I am really glad to hear that! Keep up the awesome work.
this is one of the best channels on youtube, very high quality. keep it up bro
Thank you very much, I really appreciate your kind words!
I ... I'm sooooo glad I found your channel... you've gained a subscriber for life ❤✌
I am really glad to hear that, thank you so much! Best of luck with your studies. ✌
if only i found you sooner but still not too late to cover what I missed. thank you
You got this! Keep up the great work and do your best.
Man this is great, online learning problems solved ! Thanks soo much
You're very welcome!
You are the best teacher ever in my life thank you!
You're very welcome and best wishes with your studies!
These videos are great for me to practice my school work. I search for the main topic that we are studying and I always pick your videos because you go over the important formulas and then work through some great example problems. I think you do a great job and I hope you continue to do what you do. My only wish is that you had showed your method of solving that equation at the end of the third example. I am struggling with that one but I will keep working on it and see if I can get the right answer. Thank you!!
Hi, thank you so much for your wonderful comment and I am really glad these help you. :) There are a few reasons as to why I don't go through solving equations. First, I am really trying my best to keep these videos as concise and short as possible while giving out the most information. So going through the steps of solving those types of equations extends the time, and also, it doesn't really pertain to the chapter, solving these equations should be covered in linear algebra. It's sort of like having to explain the Pythagorean theorem every time I use it, it doesn't really fit into the video if that makes sense. 😅 Second, there are a lot of these online solvers that actually solve these equations with step by step solutions, (i.e. WolframAlpha, and symbolab). Third, in my experience, your exam for this course should not contain equations that take very long to solve, since the exams test your fundamental knowledge of the material covered in this course rather than what is covered in your other courses. You might face these equations in linear algebra or calculus classes, but ridiculous integrals or weird equation solving shouldn't be a part of your exam (this is usually the case but it could be different for you). I really do appreciate your comment and your feedback, if I find that an equation might be troublesome for students in the future, I will attach a solved version in the description below. For some videos, I have already done this. 👍
Thank you for such a great tutorial!!
You're very welcome!
My professor makes his lecture videos unavailable to watch after a certain day and time. Having helpful videos to refer to when I'm confused is great and it is nice that they aren't that long
Are you unable to download your professors videos to watch later? Anyways, I am happy these help you and I wish you the best of luck with your studies.
@@QuestionSolutions yeah the site we use to watch the videos doesn't allow us to download them
@@gamingtrials4484 That's too bad. I would think it'd be helpful for students to re-watch the lectures.
I'm so happy to find your channel and watch your videos before that I couldn't understand Dynamics thanks u r the best :D I hope if you make videos about CENTRAL -FORCE MOTION and SPACE
MECHANICS
Glad to hear it helped! I hope I can get to those topics too :)
@@QuestionSolutions another thing I couldn't get how you solved the equation in the last question 0.448Vb=5costeta and 0.894Vb=5sinteta :)
@@mezonoor4022 Here are the 2 equations graphed: www.desmos.com/calculator/whvvqenssu If you click where they intersect, you will see the 2 values 👍
Hi this is vikram from India,
I have seen all videos. Animations are good. Later I came to conclusion that definitely your channel will be come as 3BLUE1BROWN. IF you develop your channel. Please try to solve each and every problem in Hibbeler.
Hi Vikram, I appreciate your kind words. I will do my very best :) Best of luck with your studies!
YOU ARE INCREDIBLE ❤ thank you 🙏
You're very welcome! ❤
You literally just saved my life
😅 glad to hear!
best yt channel for UNI topics
You are the best
Thanks for watching!
thank you I have value idea
I'm preparing myself for the final exam
You're very welcome and best wishes with your exam!
Thank you for your explanation! do you have any other resources where i can find more advanced exercises? or maybe u solved some on your channel
Your textbooks will have plenty of other practice problems.
Thx a lot🌸💜.
You're very welcome! 💜
Sir, again thank you for your wonderful videos. Regarding question at 3:58, I read that if you were to determine car B relative to A, you must treat A as a rotating frame and use a more complete acceleration equation. Is that correct? Why? Do you plan to make videos about relative motion and rotating frames? Thanks again.
No, you can do it the same way we did any of these problems without making it complicated. You just switch the equations around like I showed at 0:53. I believe there is another video about relative motion, but that's with rigid bodies, and that does involve rotating frames and so forth. Please check the playlist :)
thank you so much
can you teach us how we can solve two equation with two unknow . one of the unknown is the angle theta please
There are a lot of videos on how to solve simultaneous equations, so please kindly look for them on youtube. If it's too tough to solve, an easy way is to simply graph it :)
Hey!, in the boat example, could we assume that the time that the boat will take to travel the 25 meters in x-axis is the same as the time it takes to travel the 50m in y-axis? because then we would have relation of the components of it´s velocity! thank you, great video.
Are you asking about travelling independently in the x and y directions? If so, no, because the boat can travel the horizontal 25 m much quicker, especially since the velocity vector of the river is also pointing to the right. So for the boat to go to right in the x direction would be much much quicker than going straight up 50 m in the y direction. 👍
@@QuestionSolutions but that would imply that the boat's path would not be the hypothenuse of the triangle I think
@@joaobrites5931 No, because the times are different, that is why the boat goes at an angle. If there was no velocity of the river, you'd just go straight up, since nothing is pushing you to the right side. But the river has a velocity, which means while you try to go up, the boat is getting pushed to the right side. I think you are overthinking this. Imagine you are swimming in a river. Isn't it much easier to swim with the flow then to go straight up? The flow helps you swim faster. The same as big ships using the ocean currents to gain speed, or using the wind on sails to go faster. It's added velocity. 👍
Thanks man
You're welcome!
If I just have found this channel last semester , I wouldn't have suffered b'cus of statics 🥺😪
🥺 Maybe this semester, they can help? Best of luck with your studies!
I have a question, on the river example, how do u know 5m/s is the Vb/r? Can you make a comparison of this to other examples? (like the plane one). I'm having trouble with the reference point, It's not as obvious as the other examples. Thank you so much!
I will reply with an answer I gave before because it's the only way I can explain it. I cannot make a comparison because this type of wording only applies to questions where something is in water, or a sail boat with respect to wind sort of thing. I think this question is sort of a play on words. I found that many questions of this type, where a boat is in water, and we're given the velocity of the boat with respect to still water is taken as the velocity of the boat with respect to the river. A way to think about it is, when the water has no flow, so 0 m/s velocity, the boat has a velocity of 5 m/s, so it's already with respect to the water because we are saying when one has this velocity, the item has another velocity. So in this case, when the river has a velocity of 0m/s the boat has a velocity of 5m/s. The key being "in still water." I actually don't know if that's a good explanation, but it seems like one way to think about it.
@@QuestionSolutions thank you so much!
I am just now seeing this.
Your playlist basically carried me through my midterms, and I'm forever grateful haha!
@@lolhavefun9201 You're very welcome!
hello sir! how do you input the equations on a graphing calculator? for an instance the 0.448Vb = 2+5cos(theta). I'm quite confused on how to come up with the solutions on the boat example.
For normal graphing calculators, make VB = x, and Theta = y. So your equation would look like this: 0.448x=2+5cos(y). Here is an example: www.desmos.com/calculator/un08zfcl8u
Look to see where they intersect and make sure it's in degrees, not radians.
Let me know if you need further clarification.
@@QuestionSolutions Thank you for replying my comment sir! I got the answer and how to input the graph. Thank you so much sir! You are such a savior!
@@josegregosilao414 Awesome! I am very glad you got it :)
thanks
You're welcome!
for finding the Vb and the theta... is there any way to solve it without using desmos ?
You can, but it'd be way too time consuming. You will most likely never get a question where you have to solve something like this on an exam, so don't worry too much.
hello, how can i get the Vb= 5.57 m/s? where the value of the theta in 2+5costheta
Easiest would be graph the 2 equations and see where they intersect.
Great day sirr!, I've got a small short issue with the last question with regards to the normal acceleration at 10:06 , how was the angle trasnlated? The dotted line, from my current understanding, is just a parralel line to the normal accelration so essentially a replica and this normal acceleration makes a y=x line essentially which makes a 45 degreee angle split between the y and x axis and so when this normal accelration line was "translated" down it formed a perpendicular 90 degree angle with the 30 degrees orignally in the question? or well to be honest I am actually confused so what ever i mightvee said right now i am not even sure about haha, hope you could help clear that out. By the way i am not sure if u rememebr this absurd hahah username but Yeah ive contacted you before and so on , hope youre having a great day
I will try to explain. The dotter line is parallel to the normal acceleration vector. Remember that normal acceleration always points towards the center of the curve. The components of normal acceleration however will be along the x and y axis (this you can see from the light orange vectors). So what we need is to figure out the angle. So the easiest way to do it is to draw a parallel line, get the angle, and we can do that using alternate interior angle law to do so. I think you might have forgotten these: www.mathsisfun.com/geometry/parallel-lines.html
If you didn't, this is exactly what was applied to figure out the angle. Also, if you want a get an in-depth idea of this, draw it out on paper using a straight edge and you can easily see it :)
I replied to your email 👍
@@QuestionSolutions Do you mean you replied to me by sending me an email? Because I for some reason can not find the reply haha sorry
@@Diggydogdraddy Yes, please check your junk/spam :)
How do u solve 2 equations 2 unknown on cal or just any solutions?
You can use substitution, elimination, a matrix, or even graphing to solve two equations. Here is an example of how I solved one: www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0
A simple but more in-depth explanation is shown here: www.mathsisfun.com/algebra/systems-linear-equations.html
You can also find more resources by searching "simultaneous equation solving" on CZcams or any search engine.
Most calculators can solve equations up to 3 unknown variables. Please look at your calculator model and their respective manual.
Pls do the kinetics of the particle work and energy
I will definitely get to it, and I've been meaning to, but these days have been really troublesome with work and what's going on recently. Stay safe, I will make it as soon as I can. Sorry for the wait.
@@QuestionSolutions thanks alot
Do you have a link where I could learn how to solve simulataneous equations that have sin and cos in them?
No, unfortunately, I don't have a link, but the general idea is to use trigonometric identities. The easiest way if you're on a time crunch is to graph them or if all fails, use an online solver for a quick answer. 😅
@@QuestionSolutions thanks so much
@@Robtmmartine77 You're very welcome
Hello. What is the best way to solve the two equations for the boat question non-graphically? Bit stuck on this. Many thanks!
It's very tedious to solve that problem without graphing it. You will need to do 3 substitutions involving tan(θ/2), sinθ=(2y/y^2+1), and cos(θ)=(-y^2+1)/(y^2+1). Not worth it, and in all likely hood, you will never get a question where you have to solve a system of equations like this on an exam in a statics course. So don't worry too much about it.
@@QuestionSolutions Thank you very much for the quick response! But won't this method of solving lead to a situation where you are always going to have these unknowns (sin and cos)? I have seen these types of questions in past exam papers where you are not allowed to use graphical caculators. Is there another method of solving you would recommend?
@@finlaysemple6938 So sin and cosine aren't the unknowns, you would just get repetitions of pi when you solve (theta and "v" are the unknowns).
The equations for this problem look deceptively easy because there aren't any trigonometric identities to substitute to get an answer quickly. In the questions you see on exams, they are probably much easier to solve. The goal in statics isn't to teach you how to solve math problems like these 😅 You might see a question like this in calculus or algebra. Either way, the y-substitution method will work. I believe wolfram alpha will show you a method that works as well (again, I think it's the y substitution method they use). It's possible your professor knows a method too, ask around during office time. 👍 Best wishes :)
I got mad respect for you !! ಥ_ಥ
Many thanks :)
For the boat example, I really struggle to understand why the velocity given for the boat is taken as the velocity of the boat with respect to the river, when it says that that's the velocity in still water. Could someone please help me out with understanding that?
I think this question is sort of a play on words as well. I found that many questions of this type, where a boat is in water, and we're given the velocity of the boat with respect to still water is taken as the velocity of the boat with respect to the river. A way to think about it is, when the water has no flow, so 0 m/s velocity, the boat has a velocity of 5 m/s, so it's already with respect to the water because we are saying when one has this velocity, the item has another velocity. So in this case, when the river has a velocity of 0m/s the boat has a velocity of 5m/s. The key being "in still water." I actually don't know if that's a good explanation, but it seems like one way to think about it. Good luck with your studies! :)
@@QuestionSolutions Thank you!
You're very welcome!
Thanks for asking this question, I was looking through the comments to see if someone asked or I was the only one who didn't understand
@@QuestionSolutions I am still unsatisfied, should the relative velocity of the boat to still water be the same as its relative velocity with flowing water?
For the problem with the boat, how do you solve for theta and vb?
The easiest way is to graph the two equations and see where they intersect.
@@QuestionSolutions aaah, that makes sense. I was trying to by solving for vb then substitute one in the other, but wasn't sure where to go after that. Thank you again! Having a rough semester with my dynamics professor, so I'm needing all the resources I can get.
Yes, it's quite difficult to do through substitution. On the bright side, it's very unlikely you'll need to solve a system of linear equations similar to that on an exam :) I wish you the best with your dynamics course, do as many questions as you can!@@paugust
Why is Vboat not 5 m/s? In the problem it states that the man can paddle the boat at 5 m/s in still water. Wouldn't Vb/r be something different than the velocity at which he can paddle in still water?
So if the question is asking for the speed of the boat, the answer wouldn't be in the question itself right? So the speed of the boat can't be 5 m/s. 😅
So here, vb/r is what the speed of the boat is with a flowing river and a man paddling it. It is the real speed of the boat at those specifications.
Thanks a lot❤️
You are very welcome! ❤️
3:31 why did you find this angle and not the opposite one?
So at the location we found the angle, there are 2 lines intersecting. That point is the start of our coordinate system, which is why that angle is shown. So the resultant is the hypotenuse of the triangle, which starts at the center of our coordinate system. That's the angle that's always represented.
Why does in the airplane problem, sin is incorporated with i. On the other hand, in the boat problem, the sin is paired with j?
Please see this video I made: czcams.com/users/shortsvynnKlJD_Jo
It's pretty much the same thing, I just used a force to break into components. You always have to look from the perspective of the angle.
What's j and if for?
i and j represent x and y components.
Where can we get this in like slide form or pdf?
There aren't any slides or pdf. It's just a video I created to help students out. You can always take screenshots if you need 👍
@QuestionSolutions
Thank you. The way you explain is really helpful. I'm currently doing engineering dynamics your videos help me understand a whole topic in just a few minutes.
@@user-si5qr7zd7x Thank you! I'm glad these videos are helpful to you. Keep up the awesome work and best wishes with your studies. :)
For the boat question. How was you able to solve it.
I think I graphed it 😅 You can also use wolfram alpha.
@@QuestionSolutions because I used the 0.448vb = 2 + 5sin(theta), divided both sides by 0.448 and I got v(b) = 4.46 + 11.2cos(theta), and the 0.894vb = 5sin(theta) and when I divide both sides by 0.894 and got v(b) = 5.59sin(theta). Then I plug this in and got 5.59sin(theta) = 4.46 + 11.2cos(theta) and switched it to 5.59sin(theta) - 11.2cos(theta) = 4.46. The problem how am I going to solve for theta if the equation doesn’t equal 0. I tried the wolfarm alpha. They show me the graph but I was wonder how to solve it without graphing it.
@@darrylcarter3691 I can almost say that in almost all the cases, your dynamics exams will not have questions involving serious equation solving, since that's outside the scope of the course. For this question, I think the easiest way to solve it is to graph it, or you'll have to use trig identities and try and find an answer that way. 👍
@@QuestionSolutions Hey! Guess What? I just Figure out how to get the answer without graph it! I looked at the 0.448vb and 0.894vb, and I noticed that the vb magnitude is almost Parallel to The 5 m/s vector. So I figured I can use the Pythagorean Theorem to find vb. I did (0.448vb)^2 + (0.894vb)^2 and set them equal to 5^2; 5^2 of course is 25. I got (0.20 + 0.79) x (vb)^2 = 25 or 0.99(vb)^2 = 25. Divide both sides by 0.99 and got (vb)^2 = 25.252525; I decided to make it 30 since it seems close. Square rooted both sides and got vb = 5.5 m/s. You said 5.57 m/s which is kind of close. Than I use the horizontal equation component and got 84.6 degrees.
@@darrylcarter3691 Close enough 👍😅 Well done!
Is relative velocity in space different from relative velocity let's say of two cars?
You're only comparing one velocity to another, so notice that we didn't take into account gravity or any other force. As long as we have 2 velocities, we can compare one to the other, the same as if we had the velocity of 2 satellites in space.
yes i searched and found out a bit more on that. Thank you so much you dont realize how many students youre helping thank you dear friend.
@@jesussaquin6266 You're welcome!👍
btw how did you solve for the two unknows Vb and the angle?
@@jesussaquin6266 I get this question a lot 😅 Graphed it.
ı dont understand Vbr in 8.00 why is theta not 63,.4?
Because we don't know the angle yet. The angle we found, which is 63.4 was the angle φ which is shown at 7:31.
I still love you
:) Many thanks!
❤
❤️
How do I get speed and Angle by solving both equations?
I am unsure of the question you are asking. Could you give a timestamp to where you're having a hard time understanding? That way, I can give you a specific answer. Thanks!
8:06
Vb & Angle
@@ahmdkamel9924 Thank you for the timestamp. Here, you are just solving 2 equations (system of equations with 2 unknowns). Instead of "v" and "θ", think of it as "x" and "y" then it's easier to understand I think. You can use substitution, elimination, etc to solve, or any method you're comfortable with. If you need a quick answer, use wolframalpha.com. :)
I've done it more than once, but the Angle is wrong.
@@ahmdkamel9924 Okay, please see: www.desmos.com/calculator/kmmrynaue8 I graphed it for you. You can see the intersecting points at (5.556, 84.337). If the velocity you get is right but the angle is wrong, it is probable that you used an incorrect trig identity, or maybe made an error in calculation? I am not too sure.
software name please ?
Are you asking what I use to animate? It's after effects. 👍
Exactly!!
@@AmeerAlikh Okay great, I already replied, please check.
Did you solve that system of equations by hand? i been trying every trick in the book so far theta angle is making my hair fall off lol
Long ago, when I was in university, I've solved so many of these, but for videos, I use symbolab, desmos or Wolfram. Most of the time, your exams won't contain questions that involve ridiculously hard simultaneous equations since that's not what the course is trying to teach.
@@QuestionSolutions I was overthinking it and trying to solve theta from SE instead of just back sub for Vb after I squared everything and added the equations to get rid of theta, thanks for the videos tho, definitely a huge help that supplement my lectures.
@@uniform375 Awesome! Keep up the great work and best wishes with your studies.
va/b=va-vb,it esier to remember
👍
Which software have you used to make animation?
Illustrator to draw the diagrams and after effects to animate.
Why is this called using translating axis?
Did you google it? 😅
@@QuestionSolutions Haha no, trivially just popped up in mind
@@elements-24 😀 Simply, it means we change the coordinate system from one place to another, so like from point A to point B.
@@QuestionSolutions ok so it means translation of coordinate system not the rotation of the coordinate system, right?
@@elements-24 Yes. 👍
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Wooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooow 👍
at 7:50, what is the difference between the 2 angles?
So theta is the angle the velocity of the boat makes with the horizontal flow of the river. The phi angle is the angle of the boat with respect to the shore as shown at 7:33.