I swear, when I had my first statics class in college, I was regretting choosing engineering, but now that I know I have access to such helpful resources, I am not worried at all. Thank you so much for your help. Please make more videos on mechanic statics, we engineering students can really use it!!
Don't fret, statics is not hard as long as you understand the fundamentals. Once you are comfortable with expressing forces in cartesian form, taking moments, etc, everything else just falls into place. Keep up the good work and best wishes with your engineering degree.
Thank you so much for this explanation sir! Im wishing for this channel to grow larger. You really deserve some praise! You just saved many engineering students 😂
Thank you.................. i watched all your videos Keep it up......... Very useful and informative video..... it is commendable..... Best wishes FROM INDIA 🇮🇳 ..........
Hello . You make me clear in a very good way I am the students of Industrial and manufacturing engineering. But I have a question that for F 2 why we have to find the x component. And in F 1 why we dont find the Y component
Thank you so much, you are really the best lecturer on this chapter. You have the talent to teach, please keep it up. May I know the software you are using for visuals/graphics and how you can use it that way, very interesting.
In regards to the F2 of the problem in 3:11 , shouldn't the angle with respect to the y axis be 45 degrees as well? Because if we make use of the 45 degree angle that's with respect to the x-axis to look for the x-component, we could also make use of the same 45 degree angle that's with respect to the x-axis to look for the y component with the use of sin. However, it gives us a different result compared to making use of the given 60 degree angle with cos. if F2x = (100N)cos(45), then F2y = (100N)sin(45). However, F2y = (100N)sin(45) doesn't give the same result as F2y = (100N)cos(60)
What you are saying cannot be done, because these angles are given with respect to individual axes. In addition, they must be given with respect to the positive x,y,z axis, or you have to do the math and calculate the angle with respect to the positive axis. They are not interconnected, which is why cosine must be used when solving coordinate direction angles. So you can't use sine, you have to use cosine to figure out the answer. What you are saying can be done with transverse and azimuth angles (like the one shown at 5:37). I hope that helps.
sir, I have a doubt ....in 4:02 ...in f3 vector why we are subtracting 108- 45 degree and writing it as 250cos(135) j with respect to positive y axis....why cant we write directly as (minus) -250cos(45) j with respect to negative y axis ...could you please clarify it?
It makes no difference, the only thing is that you need to remember to use a negative sign whenever it's not with respect to a positive axis. It's easier to remember to subtract the angle from 180 since it gives you ability to understand what you're doing. Either way, you will get the same answer. The proper method is to subtract the angle from 180 but it's up to you :)
1:19 can we use the cordinate direction angle with sine instead of cosine ? Like can we use the sine law of triangle to find the angles instead of cosine in the video
@@RivanMwaanga-qf4qn For coordinate direction angles, the angle must be given with respect to the positive axis. So notice that in the problem you're asking about, we're given the angle for the y-axis with respect to the negative y-axis. We need the angle from the positive side of the axis. So to get that, we need to subtract 45 from 180. I hope that helps!
on 3:15, since the force runs in the negative z axis, shouldn't we subtract 120 from 180 to give us an angle of 60 between the positive z-axis and the force? We did this for the j component, so why not for the k? Thank you!
Maybe I am not understanding your question properly, but we didn't subtract anything for the j component. Since the angles were already given to us with respect to the positive x,y,z axes, nothing needs to be subtracted. We only ever subtract from 180 degrees if the angle was given with respect to a negative axes. For coordinate direction angles to work, we have to use the angle with respect to positive axes. Now if we look at the 120 degrees, we see that it is indeed given to us measured from the positive axis. You can tell the positive sides by looking at the labels for each axes, so for z, it's labeled on top, that's the positive side, for y, it's labeled on the right, so that's the positive side, and so forth. Lastly, when we do the simplification, we see that the the z component is indeed negative, and you can see that at 3:29, it's below the x-y plane. If we had used cosine 60 for the z-component, we would end up with a positive force value, which would be incorrect.
@@QuestionSolutions Ah! I apologize! You are correct, the angle given for k is with respect to the positive z axis. I had a total brain fart and imagined the angle given was with respect to the negative z axis. That was my question, whether we should subtract the given angle 120 from 180 as I thought it was respect to tue negative z axis. Thank you for responding! I love your videos by the way, they have helped me so much!
@@Pguzman42 I am very happy to hear that it's all cleared up. Keep up the awesome work and I wish you the best with your studies! And thank you for your kind words :)
So you can just look to see what angles are given. If you are given 3 angles, then you have 3 components, if you're given just 1, then it's in a single plane, and that plane is dependent on the angles given. So look at 2:21, you can see we are given just 1 angle, with respect to the positive z-axis, so we know it must lie in the z-y plane. Even though we are given 1 angle, we know that with respect to the y-axis, the angle must be 90-40 = 50 degrees.
No because the forces are given with coordinate direction angles. The F' method is only applied to forces given with transverse and azimuth angles. Please make sure to realize the difference, or you won't get the correct answers.
Understood the concept in just 10 minutes while I struggled understanding it from my prof for weeks already. I hope teachers should also learn how to teach.
For real, some lecturers don't know how to teach, sometimes I ask my self what really they are doing. TH-cam is the best platform for well explained lectures
In the third example. Shouldn’t Fx2 equal Fx1 because I observed a rectangle lying on a the xy-plane and the two short sides of a rectangle are the same length and the two long sides are the same length.
I don't know how to answer your question because I am not understanding what you are saying with the observed rectangle. Are you saying components of F1 is shared for the components of force F2? If you can maybe draw a diagram of what you mean? Sorry, I am unable to follow your question.
@@QuestionSolutions okay. So go to the purple triangle in the xy-plane that has the 30 degree angle, and you will see a line on xy-plane that looks to be parallel to the y axis. Follow that line until you stop at the x axis. The line looks to be perpendicular to the x-axis.Then start from that point and follow that line until you get to the origin we’re the hook is. Next, make you way along the y-axis and stop at the line that is perpendicular to the y-axis. Finally, follow that line until it stop that line that parallel to the y-axis. When you go around that path, you will see that it looks like a rectangle was drawn.
@@darrylcarter3691 Okay, I see what you mean. But if you look at the original diagram, you can see that the line parallel to the y-axis lying on the x-y plane ends shorter than the x-component of force F1. So while, it's true, it forms a rectangle, it doesn't line up with force F1, so the x-component of force F1 is not equal to the x-component of F2. That was the original question right?
Hi, so you have to look at the diagram really carefully. Specifically, look at the axes shown. Locate where, x, y, and z axes are. After you do that, look at force F1. I draw the components of force F1 at 2:49. See how it has an z-component and a y-component. They are along those axes. It's in the z-y plane. Now look at force F2. It's NOT in the x-y plane, it's also in the z-plane. Look at time 3:30. See how the force is below the x-y plane. That means it's not lying in the x-y plane, but it has all three components. You can also tell because the angle between the force and the z-axis is 120 degrees. That means it goes past the 90 degree mark, so it must be below the x-y plane. Let me know if you need further help. Best wishes with your studies.
Not sure how you got 570, and I can't say without seeing how you typed this into your calculator. The value of 610 is correct though. So double check your work. Also, there are tons of online resources to check your work, so utilize them to figure out where you made a mistake. See: www.cymath.com/answer?q=sqrt((575.52)%5E2%2B(131.52)%5E2%2B(-153.55)%5E2) I hope that helps!
I assume you are referring to question 1? Please kindly use time stamps when asking questions, it's easier that way :) Force F1 lies on a single plane, so it lies on the zy plane, and is not given with coordinate direction angles. So we can't use it, we just need to break the force into components using sine and cosine. Please watch this video: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html Thanks!
@@yusufmoola6471 Yes, it is the hypotenuse. The diagram might be hard to tell, but draw it out with a protractor, you will see that it is indeed the hypotenuse.
@@sanatani_manushy77 No, because the other angle with respect to the y-axis would be 135 degrees. Remember, the 45 degree angle is with respect to the y-axis, so 180-45 = 135 degrees. It's not 30 degrees. You are trying to add up an angle that's with respect to the y-axis with the z-axis :)
That's incorrect. You need to look at it with respect to the angle given. You can use 60 degrees (not sure why you would because that's extra work), but if you do, you need to use cosine for the j component and sine for the k component. Remember, always look with respect to the angle. Also, "why you have taken 40 degree bcz the angle is not mentioned wrt y" I am not sure what you mean here? You don't need the angle with respect to an axis to solve it. You can use any angle given as long as you know whether to use sine or cosine to get it.
5:00 What if the question asks you to find the unit vector? Do we just divide the coefficient of the components with the magnitude? e.g. 195.7/407.03 for x component?
I am a student of mechanical engineering .this vedio is very helpful for Engineering Statics.❤️❤️❤️
I am really glad to hear that! I hope all the statics videos will be helpful to you. Best of luck with your studies. ❤
@@QuestionSolutions thank you bro
Very good engalish ...you must have learned it from .....vedios
How have your mechanical engineering studies been going?
I swear, when I had my first statics class in college, I was regretting choosing engineering, but now that I know I have access to such helpful resources, I am not worried at all. Thank you so much for your help. Please make more videos on mechanic statics, we engineering students can really use it!!
Don't fret, statics is not hard as long as you understand the fundamentals. Once you are comfortable with expressing forces in cartesian form, taking moments, etc, everything else just falls into place. Keep up the good work and best wishes with your engineering degree.
How is it going so far🤣
How has engineering been going?
Thank you so much for this explanation sir! Im wishing for this channel to grow larger. You really deserve some praise! You just saved many engineering students 😂
That's awesome to hear :) Thank you very much for your nice comment.
I am a mechanical engineering student. and this video makes everything so simple, my professor can't teach this like how you did it. thanks a lot.
Glad it helped! Keep up the awesome work and I wish you the best on your journey to become an engineer.
superb explanation and great animations. I wished I found this channel sooner. God bless you.
Thank you very much!
your voice is just perfect to listen, and your discussions are very clear. thank you very much!!!!
Thank you very much! 😀
Extremely helpful! Thank you, God bless you!!!
You're very welcome!
Was struggling with this for 2 weeks. 10 minute video just saved me.
Really glad to hear that :) Keep up the great work!
This is so explanatory.. Thanks for those short examples, and the fast solutions.🔥
You're very welcome! Best wishes with your studies.
Extremely HELPFUL....I understood the entire concept, hence feeling confident with these kinds of calculations...
Really glad to hear that :) Best wishes with your studies!
Thank you so much. You have no idea how helpful this was to me and am sure a lot of engineering students.
Glad it was helpful and you're very welcome :)
Bro Simplified my LIFE in a few minutes, how hilarious.
Appreciate it pal
Glad to hear it!
This is good teaching and well illustrated. Thank you so much.
You're very welcome. Thank you for taking the time to comment :)
Am student of electrical eng I found ur videos so helpful for Engineering statics❤
Glad to hear that :) Keep up the great work and best wishes with your studies. ❤
Beautifully explained. Great work.
Thank you very much!
THANK YOU FOR THIS. I'm studying engineering (pre year) and your channel is so helpful to me in my studies thank you again :)
I am really glad to hear that! Keep up the good work and I wish you the absolute best with your studies. :)
Your explanation is very nice. I follow you from Egypt ❤❤
Awesome! Thank you! ❤
i cant explain to u how much u saved me thank you very much
I am very happy to hear that! Do your best and good luck with your studies.
So much better than my teacher
Glad to hear the video was helpful :)
Thank you..................
i watched all your videos
Keep it up.........
Very useful and informative video.....
it is commendable.....
Best wishes FROM INDIA 🇮🇳
..........
Thank you very much, best wishes to you as well :)
@@QuestionSolutions Please explain, how to quickly solve vector algebra in mechanics.........
@@rishikumarshukla986 What do you mean by vector algebra? Solving for components?
Thank you so much!! You just saved my life
You're very welcome! You got this, do you best :)
I'm in the first grade in engineering and this video helpp me a lottttt please keep going 🥺❤️❤️
I am glad to hear there. There is a playlist of statics that should pretty much cover everything you need for the first year :)
@@QuestionSolutions i will definitely look at this 💪❤️
@@manarawad3339 Great! Leave a comment if you need clarifications on any of the questions covered in the videos. Best wishes with your studies ❤
This is captivating and very helpful
Thank you very much, I am happy to hear it helps :)
this is so freaking detailed summarizes 8 weeks of classes lol
I hope it was helpful to you. Keep up the great work and best wishes with your studies.
Hello . You make me clear in a very good way I am the students of Industrial and manufacturing engineering. But I have a question that for F 2 why we have to find the x component. And in F 1 why we dont find the Y component
Please give me a timestamp to your question. I don't know where you're referring to. Thank you :)
Thank you so much, you are really the best lecturer on this chapter.
You have the talent to teach, please keep it up.
May I know the software you are using for visuals/graphics and how you can use it that way, very interesting.
Thank you very much for the nice comment! I use after effects to animate and illustrator to draw the graphics. :)
Very helpful, thank you sir
You're very welcome!
uff i search out youtubefor this video since week TYSM sir
Happy to help :)
ok. you convinced me. i'll be subscribing.
Thanks! :)
Very nice and succinct explanation! Keep up the good work. Liked and subscribed.
Thank you for sharing.
Thank you very much! Best of luck with your studies.
this channel is saving me thank you so much!!!!
You're very welcome! :)
wow this is excellent!!! beautifully explained for every kind of person to understand. uli mwaume iwe😇
Glad it was helpful! Best wishes with your studies! :)
In regards to the F2 of the problem in 3:11 , shouldn't the angle with respect to the y axis be 45 degrees as well? Because if we make use of the 45 degree angle that's with respect to the x-axis to look for the x-component, we could also make use of the same 45 degree angle that's with respect to the x-axis to look for the y component with the use of sin. However, it gives us a different result compared to making use of the given 60 degree angle with cos.
if F2x = (100N)cos(45), then F2y = (100N)sin(45). However, F2y = (100N)sin(45) doesn't give the same result as F2y = (100N)cos(60)
help a brudda out.....I'm confused
What you are saying cannot be done, because these angles are given with respect to individual axes. In addition, they must be given with respect to the positive x,y,z axis, or you have to do the math and calculate the angle with respect to the positive axis. They are not interconnected, which is why cosine must be used when solving coordinate direction angles. So you can't use sine, you have to use cosine to figure out the answer. What you are saying can be done with transverse and azimuth angles (like the one shown at 5:37). I hope that helps.
sir, I have a doubt ....in 4:02 ...in f3 vector why we are subtracting 108- 45 degree and writing it as 250cos(135) j with respect to positive y axis....why cant we write directly as (minus) -250cos(45) j with respect to negative y axis ...could you please clarify it?
It makes no difference, the only thing is that you need to remember to use a negative sign whenever it's not with respect to a positive axis. It's easier to remember to subtract the angle from 180 since it gives you ability to understand what you're doing. Either way, you will get the same answer. The proper method is to subtract the angle from 180 but it's up to you :)
@@QuestionSolutions thank you sir😇
Thank you so much... It has really helped thumbs 👍 up
I am really glad to hear that. You're very welcome!
Excellent video
Thank you very much!
In 10 minutes I understood this topic
I am very happy to hear that!
Cartesian vector forces? More like "Cool videos that school us!" 👍
😅😅👌👌
tqsm this video help me. now i love static
Really glad to hear that :) Keep up the awesome work and best wishes with your studies.
wow this was amazing and helpful
thank you so much for your help
You're very welcome! I am really glad it was helpful. Keep up the great work.
1:19 can we use the cordinate direction angle with sine instead of cosine ?
Like can we use the sine law of triangle to find the angles instead of cosine in the video
Coordinate direction angles are always used with cosine.
very helpful thank you so much for this video
Glad to hear it was helpful! 👍
you are the man
Thank you very much!
very helpful video for me thanks sir
Glad to hear it was helpful :)
This was really helpful..
I am really glad to hear that :)
Sir I wanted to ask for the y component why are we using 135° i don't understand there
Please give a timestamp so I know where you're referring to. Thanks!
@@QuestionSolutions4:02 sir
@@RivanMwaanga-qf4qn For coordinate direction angles, the angle must be given with respect to the positive axis. So notice that in the problem you're asking about, we're given the angle for the y-axis with respect to the negative y-axis. We need the angle from the positive side of the axis. So to get that, we need to subtract 45 from 180. I hope that helps!
@@QuestionSolutions thank you much appreciated
Can you make videos on coplanar and 3-d force systems in equilibrium, please? Your videos are awesome.
I think I have a video on 3D equilibrium problems, please see: th-cam.com/video/EzquIxEoHRE/w-d-xo.html
thank you so much
You're welcome!
on 3:15, since the force runs in the negative z axis, shouldn't we subtract 120 from 180 to give us an angle of 60 between the positive z-axis and the force? We did this for the j component, so why not for the k? Thank you!
Maybe I am not understanding your question properly, but we didn't subtract anything for the j component. Since the angles were already given to us with respect to the positive x,y,z axes, nothing needs to be subtracted. We only ever subtract from 180 degrees if the angle was given with respect to a negative axes. For coordinate direction angles to work, we have to use the angle with respect to positive axes. Now if we look at the 120 degrees, we see that it is indeed given to us measured from the positive axis. You can tell the positive sides by looking at the labels for each axes, so for z, it's labeled on top, that's the positive side, for y, it's labeled on the right, so that's the positive side, and so forth. Lastly, when we do the simplification, we see that the the z component is indeed negative, and you can see that at 3:29, it's below the x-y plane. If we had used cosine 60 for the z-component, we would end up with a positive force value, which would be incorrect.
@@QuestionSolutions Ah! I apologize! You are correct, the angle given for k is with respect to the positive z axis. I had a total brain fart and imagined the angle given was with respect to the negative z axis. That was my question, whether we should subtract the given angle 120 from 180 as I thought it was respect to tue negative z axis. Thank you for responding! I love your videos by the way, they have helped me so much!
@@Pguzman42 I am very happy to hear that it's all cleared up. Keep up the awesome work and I wish you the best with your studies! And thank you for your kind words :)
love this!
Thank you!
Thanks very helpful
You're very welcome!
Great work man, helps a lot with statics.....what's your name fella?
Glad to hear these help, I'm Steven. 👍
Nice can you tell me please what program do you use ?
After Effects for the animations.
How can I find that the force vector is just in xz plane or xy plane so that its third component is zero?
So you can just look to see what angles are given. If you are given 3 angles, then you have 3 components, if you're given just 1, then it's in a single plane, and that plane is dependent on the angles given. So look at 2:21, you can see we are given just 1 angle, with respect to the positive z-axis, so we know it must lie in the z-y plane. Even though we are given 1 angle, we know that with respect to the y-axis, the angle must be 90-40 = 50 degrees.
very very helpful thank youuuuuuuuuu
You're very welcomeeeeeee
the angle between F' and F1, shouldnt it be 10? because (90-20-60) degrees ? on 6:14
The 20 degrees is between the positive x-axis and F'. It is not related to the angle between F' and F1. 👍
thanks for the video
You are very welcome!
You are amazinggg
Thank you :)
can you solve the first problem by using F2' and F3' ?
No because the forces are given with coordinate direction angles. The F' method is only applied to forces given with transverse and azimuth angles. Please make sure to realize the difference, or you won't get the correct answers.
@@QuestionSolutions so the sign to use transverse or azimuth angles is when they show a blue triangle?
Pretty much, yeah. 👍
Understood the concept in just 10 minutes while I struggled understanding it from my prof for weeks already. I hope teachers should also learn how to teach.
I am really glad to hear this helped :) I wish you the best with your studies and keep up the good work!
For real, some lecturers don't know how to teach, sometimes I ask my self what really they are doing.
TH-cam is the best platform for well explained lectures
In the third example. Shouldn’t Fx2 equal Fx1 because I observed a rectangle lying on a the xy-plane and the two short sides of a rectangle are the same length and the two long sides are the same length.
I don't know how to answer your question because I am not understanding what you are saying with the observed rectangle. Are you saying components of F1 is shared for the components of force F2? If you can maybe draw a diagram of what you mean? Sorry, I am unable to follow your question.
@@QuestionSolutions okay. So go to the purple triangle in the xy-plane that has the 30 degree angle, and you will see a line on xy-plane that looks to be parallel to the y axis. Follow that line until you stop at the x axis. The line looks to be perpendicular to the x-axis.Then start from that point and follow that line until you get to the origin we’re the hook is. Next, make you way along the y-axis and stop at the line that is perpendicular to the y-axis. Finally, follow that line until it stop that line that parallel to the y-axis. When you go around that path, you will see that it looks like a rectangle was drawn.
@@darrylcarter3691 Okay, I see what you mean. But if you look at the original diagram, you can see that the line parallel to the y-axis lying on the x-y plane ends shorter than the x-component of force F1. So while, it's true, it forms a rectangle, it doesn't line up with force F1, so the x-component of force F1 is not equal to the x-component of F2. That was the original question right?
@@QuestionSolutions yes
@@darrylcarter3691 👍
Please at 2:49 I don't understand why the force doesn't have an x component but F2 has a z component even though its in the x y plane
Hi, so you have to look at the diagram really carefully. Specifically, look at the axes shown. Locate where, x, y, and z axes are. After you do that, look at force F1. I draw the components of force F1 at 2:49. See how it has an z-component and a y-component. They are along those axes. It's in the z-y plane. Now look at force F2. It's NOT in the x-y plane, it's also in the z-plane. Look at time 3:30. See how the force is below the x-y plane. That means it's not lying in the x-y plane, but it has all three components. You can also tell because the angle between the force and the z-axis is 120 degrees. That means it goes past the 90 degree mark, so it must be below the x-y plane. Let me know if you need further help. Best wishes with your studies.
Thank you soo much I understand now
I wanted to ask if you have a video on finding position vectors in 3D should I send you a sample question?
@@adwoaaj5925 yes, I think this is what you're looking for: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
@@QuestionSolutions thank you
Any nitc student during mid sem
Best wishes with your semester.
At 7:54 when I calculated your numbers its just 570 not 610 why it is
Not sure how you got 570, and I can't say without seeing how you typed this into your calculator. The value of 610 is correct though. So double check your work. Also, there are tons of online resources to check your work, so utilize them to figure out where you made a mistake. See: www.cymath.com/answer?q=sqrt((575.52)%5E2%2B(131.52)%5E2%2B(-153.55)%5E2)
I hope that helps!
Hi. I keep getting different values for my sin and cosine solving. For example at 6:32 I got 61.70
Your calculator is set to radians, change it to degrees. 👍
Thank you so much. You really are amazz
Why do you use f1 different process from the others.
I assume you are referring to question 1? Please kindly use time stamps when asking questions, it's easier that way :) Force F1 lies on a single plane, so it lies on the zy plane, and is not given with coordinate direction angles. So we can't use it, we just need to break the force into components using sine and cosine. Please watch this video: th-cam.com/video/NrL5d-2CabQ/w-d-xo.html Thanks!
Nice
👍
Im not even studying math or anything, this just also help a dumbass like me program stuff im way too underqualified for
I am glad you find these videos helpful. Keep up the great work! Don't think negatively, you got this :)
For the last problem, how is f' the hypotenuse?
F' is the hypotenuse of the first triangle, not the second.
@@QuestionSolutions you said f’ is hypotenuse for the triangle with the 30 degree angle. But it looks like the y side is the hypotenuse.
@@yusufmoola6471 Yes, it is the hypotenuse. The diagram might be hard to tell, but draw it out with a protractor, you will see that it is indeed the hypotenuse.
I am sorry but isn't that 45 degree must be 30? Because 60 +45=105 degree but that is a right angle triangle.
Please give me a timestamp so I know where to look, thanks!
@@QuestionSolutions 3:39
@@sanatani_manushy77 No, because the other angle with respect to the y-axis would be 135 degrees. Remember, the 45 degree angle is with respect to the y-axis, so 180-45 = 135 degrees. It's not 30 degrees. You are trying to add up an angle that's with respect to the y-axis with the z-axis :)
7:53 how you have drawn this red dotted lines
They are each of the components, so 575 in the positive x-direction, 131 in the positive y-direction and -153 in the negative z-direction.
@@QuestionSolutions really great u are replying to everyone
@@saiprasadsatya3677 :)
2:28 sir i think the angle with y would be 60 degree and why you have taken 40 degree bcz the angle is not mentioned wrt y
That's incorrect. You need to look at it with respect to the angle given. You can use 60 degrees (not sure why you would because that's extra work), but if you do, you need to use cosine for the j component and sine for the k component. Remember, always look with respect to the angle.
Also, "why you have taken 40 degree bcz the angle is not mentioned wrt y" I am not sure what you mean here? You don't need the angle with respect to an axis to solve it. You can use any angle given as long as you know whether to use sine or cosine to get it.
@@QuestionSolutions ok thank you
Let me know if you need further help or if you understand it. Thanks!@@Ashjj1935
i love you please have my firstborn child
Haha 😀 Best wishes with your studies!
5:00 What if the question asks you to find the unit vector? Do we just divide the coefficient of the components with the magnitude? e.g. 195.7/407.03 for x component?
See this video where I cover unit vectors: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
@@QuestionSolutions thanks :)
@@famueduyou You're very welcome :)
Thank you so much
You're very welcome!