F = ma Normal and Tangential Coordinates | Equations of motion| (Learn to solve any question)
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- čas přidán 6. 08. 2024
- Learn to solve F=ma problems with normal and tangential coordinates.
Learn the basics of F=ma: • F=ma Rectangular Coord...
Learn normal and tangential components: • Curvilinear Motion: No...
A girl having a mass of 25 kg sits at the edge of the merry-go-round...(1:15)
The 0.8-Mg car travels over the hill having the shape of a parabola...(2:50)
The block B, having a mass of 0.2 kg, is attached to the vertex A...(5:45)
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Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.
My guy is the New Organic Chemistry Tutor and a little more direct and quicker! U R the Champ!!!
Thank you for your kind words :) Really appreciate them!
I appreciate the way you walk through the math for each problem. You do a great job balancing efficiency and detail!
Many thanks! I try to make them as concise as possible 👍
By far the best videos for Dynamics on CZcams
Thank you very much!
You channel has such an insanely good quality. thank you
You're very welcome!
The visuals help alot
I am glad to hear that :)
Thank you so much for these videos, the problems are super similar to the ones that we are going over in class and the ones in my textbook. I really appreciate it and I will def be using your other videos these are great.
You're very welcome! Keep up the great work and I hope all the videos are helpful to you :)
best channel on youtube
😅Thank you!
In the last question, I did not understand why the tringle change between N and T. For example, you have used N (4,3,5), but with T (3,4,5), so could you explain it?
The ratio triangle is an alternative to showing an angle. So it's the same as if we say T is at an angle angle of 53.13 degrees from the "n" axis, and force N is at an angle of 36.87 degrees from the "n" axis. Just as the angle changed, the ratio triangle changes because they are at different angles. If you want to get a refresh on ratio triangles, I cover an example in this video: czcams.com/video/NrL5d-2CabQ/video.html
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Thanks!
Hi, this is vikram from India,
I like your videos. Please increase video duration by including more problems with animations.
Hi Vikram. I usually pick 3-4 questions that cover the basics of every other question you will face in each chapter. By adding more of the examples, it will only go through the same steps. If the fundamentals can be learned from the questions I pick, every other question in the chapter can be done :) Also, I think a lot of people like videos around ~10 mins, short and concise. But I will keep your input in mind, and if a chapter requires more questions to be completed, I will include more. Many thanks for your input, I appreciate it.
@@QuestionSolutions Videos shorter than 15 mins for concept questions is quite good actually. If a topic requires more examples doing it in 2 parts would be better instead of video being 2x long. People generally look for these videos before exams(like me) and concentrated videos are way more appealing than one hour lectures.
hello! i have a question regarding the second problem. For the resultant frctional force what i did was saying that the work done by all non-conservative forces must be equal to the change in the total mechanic energy of the system: Wncf= delta-Em . Because the frictional force is the only non conservative force applied we can say that it is equal to delta-Em. for the delta-Em we can call the highest point in the graph (x=0 , y=20) as point "B" . lets call: kinetic_energy(point)---Ek , potential_energy(point)--Ep. thus we get : Delta-Em= Ek(a)+Ep(a)-Ek(b)-Ep(b). because the height in a=0, it follows that Ep(a)=0 , and because the velocity is constant we get Ek(a)-Ek(b)=0 . Therefore : delta-Em= -Ep(b) . because of what we knew from before we say that : the work of the frictional force must be equal to -Ep(b) . we know that the work of the frictional force ( lets call it Wf ) is equal to abs(f).d.cos(alpha). we also know that the value of Ep(b) is M.H(b).g = 800x9.81x20=156960. Now, we need the value of alpha to find cos(alpha) and the value of d. alpha is always 180 degrees, because the friction force always points in the opposite direction of the movement and thus cos(alpha)=-1. now we only need D. and this is the hard part: we know that the lenght of an arc of a graph is given by the integral of (1 +(dy/dx)^2).dx - this in the interval from Xb to Xa, that is, from 0-80. We have the Y/X relationship so we only need to do the maths. I did them and i got the value of 82.72m for the arc-length ( wich is equal to d). the minus from the cos(alpha) cancels the minus from -Ep(b) and so the abs(f) is positive (as expected of course) and equal to : (156960)/82.72 = 1897,5N .... I don´t know what i did wrong, can you help me!? thanks, great video as always!
I am really sorry but I have no clue 😅 It's really hard to find where something went wrong because it could be something as simple as a number error, or something deeper like a methodology error. Again, really sorry I couldn't be more helpful.
Thank you Sir.
May Lord always bless you.
Thank you very much!
Hello, I just dont understand what you have to do when the angle is negative. Can you always consider that as being positive and why ? So does that mean that I can consider every single angle not only that particular angle ( in this case theta ) as being positive ?
So if you get a negative value for the angle, it just means that the angle is measured from below the axis (opposite to the sense you thought to be positive). You can apply the opposite angles to show that the other angle is also the same, just positive. I think I covered a negative angle problem in the 2nd question. Sometimes, it's really helpful to draw it out, then you can get a better sense of what a negative angle is. To answer your question about is it always positive: when you break forces into components, you can always figure out the positive angle even if you get a negative angle, and use the positive angle to break it into components.
At 0:24 is that the approach every time? Thought normal would just be into the curve nd tangential is direction of velocity?
n 3D problems, you will need to establish a coordinate system. You can definitely point the tangential axis towards the direction of acceleration, it's totally up to you.
Could you please explain why you used tan theta to find angle. Sin theta made more sense to me since it involves both slope of the curve and 80 meters.
Thank you for the video!
Please kindly provide a timestamp to the location you're referring to. Many thanks!
@@QuestionSolutions It is at 3:49.
@@arathyrengith5152 It's an equation that should have been covered in calculus but the essence is that dy/dx is simply the slope and tan is opposite over adjacent. In a nutshell, if you need to find the angle using dy/dx, you use tan.
@@QuestionSolutions Thank you for explaining!
@@arathyrengith5152 You're very welcome!
Can you rotate the b and n axis for last example and solve? Meaning, T and N forces would not need multiplied by angle, on weight in both force equations?
You can change the coordinate system (rotate, translate, etc), as long as you break forces that don't align with the axes into components.
@@QuestionSolutions I did this and ended up with N = (mg)cos 36.86 for the binormal direction and T - (mg)sin36.86 = ma for normal however I did not get the right answer and I'm stumped. I'm unsure what I did incorrect and is there anything you see immediately that I'm missing?
In the last question What if I laid the the coordinate system along the slanted side of the cone?
You can do that, though breaking components might be a bid harder. Try to pick coordinate systems that make life easier :)
In 7:52 please can you explain which direction the normal force is acting on the block. Is it on the surface of the cone. I'm confused
So if you look at the same time stamp you sent me, at the bottom left, you can see the free body diagram. So the normal force (green arrow) is acting perpendicular to the surface of the box, along the surface of the cone.
Since the block is not moving along the plane tangent to the cone, why cant we set the parallel component of gravity to the tension? It yields a different result.
Thanks,
Jacob.
Please give me a timestamp so I know where to look. Many thanks!
i think in this case it has (a)n so it cant be use like that
Hello, at 7:06 why didn't you place the axes along the same direction as the cone and tension?
You can if you like, but breaking the components might be a bit harder that way. It's up to you, many ways to solve the problem :)
Can we use the rotation about a fixed axis in 2.45 ?
Unfortunately, no, because we don't have enough givens to do it that way. Would have been easier though :)
hey, my question probably would be ridiculous but I will ask anyway, could you please let me know why in the first question N is negative?
There is a small typo, it should be so that down is picked positive for the first equation. However, the answer is still correct. The value of the normal force is positive, and it faces up to counter the weight of the girl.
So regarding the first problem with the merry go round, does this mean that the normal acceleration can also be applied to centrifugal force? I thought that the acceleration equation could only be used for centripetal accelerations?
You might be overthinking this. Acceleration always has 2 components, doesn't matter if you're on a straight path or a curved path. The only difference is, if you're travelling along a straight path, normal acceleration is 0. When you're travelling along a curve, whether it's on a curved road, a merry go round, etc, your normal acceleration will not be zero. That normal acceleration is based on the velocity and the distance to the center of the curve. Did you cover centrifugal forces in this chapter? Most of the time, it's not covered so it doesn't confused students because to use it, you need to think of this problem with a rotating frame of reference, which we definitely don't need to.
@@QuestionSolutions ohh ok thanks a lot! I haven't taken a formal engineering dynamics course yet, but I will be in a couple weeks. I thought that by watching this series, it would definitely help my brain to warm back up from a long summer. Thanks again for responding!
@@Stutteringjohnfan2015 Ahh okay. Awesome that you started beforehand, hopefully, these videos will be helpful when you go through your course :) Best wishes!
@@QuestionSolutions thank you!!
@@QuestionSolutions just wanting to clarify something, since the girl can only rely on friction to stay put on the merry go round, then that means the friction must be greater than or equal to the normal acceleration of her position? Secondly, would the acceleration still be pointing towards the center of rotation even though it opposes the friction force keeping her on the platform?
In the last problem, I'm confused, why is the weight not included in the summation of force in normal axis?
The weight is straight down, lying on the binormal axis, it doesn't have any normal axis components. That's why it was included in the summation of binormal axis forces, but not the normal axis forces. The other 2 forces can be broken into components and they have both binormal and normal axis components.
in question 2 we used t and n as coordinates, while in 3rd question we used b and n as coordination. can you help me understand how that works? and how to decide which one to apply?
So in the 2nd question, we actually have a tangential coordinate system because the car is travelling along a curve. Whenever an object travels along a curve, we have normal acceleration and tangential acceleration. To split these 2 accelerations, we use t and n coordinates. You can even use a normal x,y,z coordinate system to solve these problems, but you'd have to do a lot of geometry to make it work easier and there will be many steps involved. So in most cases, if we have normal and tangential accelerations, we use t and n coordinates, which makes the math simpler. In the last question, having a tangential axis doesn't do anything for us, since we actually want to write our equations upwards and to the right. So for upwards, the z-axis, is represented by the binormal axis. The normal axis is used to make the math simpler with the normal acceleration. It all depends on how quick/fast you want to get to an answer. You will find that in most cases, you can get to the same answer in different ways, but they might take longer. Over time, the more questions you solve, the faster you will get at picking which method to use :)
@@QuestionSolutions thank you for clarification. but it's kinda confusing, according to formulations t has to be in the x direction. while in 2nd question we put it as y direction, unless you have used a 2D coordination, can you explain how and when it's necessary to use 2D or 3D coordination? or we should not memorize the directions, and decide them according to the questions?
@@ano4309 You are absolutely right when you said "we should not memorize the directions, and decide them according to the questions." In every case, you will have to change the coordinate system and place it in a way that makes the math simple. So if you think about a very simple case, where a force is given with an angle, not lying on the x-axis (so horizontal), but at an angle of 45 degrees, if you place the coordinate system also at an angle of 45 degrees, thereby making the force lie on the x-axis, that would make the math simple right? That way, we don't have to break it into any components. If however, we place the coordinate system horizontal, then the force must be broken into x-y components using the 45 degree angle. This also depends heavily on what information is given to us in the question, and also on what we are trying to find. If a box is lying along an inclined slope, it's going to be a good choice to make the coordinate system be at an incline as well. If you have an object moving along a curve, usually, it's easy to do the math with normal and tangential axes. If you have questions where you need to consider normal acceleration and acceleration upwards (that is not tangential), then b and n coordinates are easier. If you don't have to consider normal or tangential accelerations at all, then the usual x, y, z, coordinate system is the easiest. Again, all of this is going to be based on what the question is asking for, and what information is given to you.
@@QuestionSolutions thanks alot!
@@ano4309 You're very welcome!
At 5:06, could I have done 800(9.81)cos63.44 ?
Yes.
is the normal (Fn) in the direction of the friction ?
Normal force is the force that's perpendicular to the surface an object lies on.
Sorry for asking. How to know that the t is using Sin, while the n is using cos?
Don't be sorry. So breaking vectors/forces into components is really important to know. If you have a few minutes, please watch this video, it goes through how to do it, and you'll also know when it's sine and when it's cosine. czcams.com/video/NrL5d-2CabQ/video.html
The first and second question explains it step by step. If you don't have the time, it's based on trigonometric ratios, so sine if the side is opposite to the angle, and cosine if the side is adjacent to the angle.
@@QuestionSolutions Thank You so much 🙏🙏🙏
@@JT00009 You're very welcome!
why did't you take normal axis aiming at the centre in the last question?
You can do that if you'd like. It's up to you how you want your coordinate system to look. You just need to make sure to break the forces into their proper respective components.
شرح رائع
I think this translates to "great explanation" so thank you very much! :)
in last question, why we cant use mg(cos53)=N?
or it has a ---> mgcos53 -N=ma? maybe it come from (a)n
@@Pp-je6ec Please kindly provide a timestamp so I can take a look at the equation you're talking about. Thanks!
Hi, I have a question for question#2 what does Mg stand for? is it milligrams?
Mg is megagrams, so 1 Mg = 1000 kg
There’s an error in 5:13. The angle is 26.56 but you have used 26.57 on the equation. 😅
😅 Yes, that's a typo.
why did we choose to have the b-n-t coordinate system to be with the horizontal axis? why did we not have it align with the forces of the block at 6:54
You don't have to. You can do it the way you mentioned as well.