U SAVED MY LIFE. I WAS REALLY BLUR THE WHOLE TIME WHEN MY LECTURER EXPLAINS THIS DURING MY MECHANICS CLASS BUT BRO U SAVED ME. NOW IM DOING MY HOMEWORK IN THE SPEED OF LIGHTNING AFTER WATCHING THIS. GOD BLESS U.
You are doing a deed which is the highest of ranks..you are serving knowledge!.. May god bless you for doing this in this century,where people are just spoiling others. Thanks a lot ,this helped me more than you think: )
I am doing engineering level 3 and I joined in late when they already finished this topic. I am grateful that I finally have learnt from you. Thanks for the video again. Keep on posting more great stuff.
What a brilliant video, I failed this topic first time around now I’m revising for it again and my lecture it’s the best. So this video has helped me a lot !
Thanks sir for the refreshment because it’s been like six years now I studied this. If I am not confused I believe the angle for the resultant force should be 360°-18.5° since all the angles of the other forces were taken from the posive x-axis for the calculation
Isn't the angle supposed to be read from the positive x-axis in an anticlockwise direction?? Which will make the reference angle for 18.5 degrees to 341.7?? I need clearance please
Yes, you are correct. 341.5 would be a suitable answer (-18.5 is also a suitable answer). The angle is to be read from the positive x-axis, in an anticlockwise direction based on the convention I have used. Some people may give you the requirement that the angle must be postive, and between 0 and 360 degrees.
This is a great video but I have a question?? I’ve been given a problem with a negative angle -30* would I assume this would be negative x and y axis so 330*?
Thank u so much this video had helped me but , if i may ask why did u add all the forced and their angles together I thought there was a formula for this how can I use the formula to solve it
Not sure if I am answering your question but I will have a go. The reason for adding the forces together is because this could lead onto finding out the overall force acting on a system. This would/could then help us to find out loads and stresses within the system. You might notice that I am quite hesitant to write a formula. I have been stung in the past by being taught a formula only (it was a math exam when I was 17), but then I was given a question in a test that required proper understanding of what I was doing. As a result i wasn't able to even try the question. For this reason, I try to teach based on understanding the basic idea. However, here is a formula that may be useful for you Sum of forces in x direction = F1 * cos (theta1) + F2 * cos (theta2) + F3 * cos (theta3)+..... Sum of forces in y direction = F1 * sin (theta1) + F2 * sin (theta2) + F3 * sin (theta3)+..... Resultant = [ (sum of forces in x-direction)^2 + (sum of forces in y-direction)^2 ]^0.5 Angle = atan ( (sum of forces in y-direction)/(sum of forces in x-direction) ) **Note that for the angle you need to be aware that for tan there are multiple solutions between 0 and 360 deg and you will need to find the correct value.
Thanks for the video. But while working I didn't see you taking into consideration the quadrant of the various functions (sin, cos) 🙄Like in the second quadrant cos is negative
Please check your 5N force components. Why did you take 5cos150 as your x component? It should be y component as per the position of the angle 150deg. As the sine component always lies opposite to the angle whereas the other side is given to the cos component. Please check the resolution of the 5N force. I might be wrong too. Please correct me in that case. THANKS
You haven't considered the direction of sin and cosin, Why? I mean in ur video there is 15N acting downwardnso while adding the x component , it would be -15Cos(300) right?
Hi Mary, sorry for the delay. The direction is taken care of if we use all angles as measured from the x-axis because the sine curve is a negative when the angle is bigger than 180 and less that 360.
It would be more clear if I put it as the angle anti-clockwise as that is normal. At that time of making the video, I had made that decision because I had calculated the angle and was wanting to make it clear that angle was as measured clockwise from the axis. Both answers are right (measuring clockwise or anticlockwise); but different people have different conventions that they follow. In your case, you have been taught the convention that anti-clockwise measurement of angles is required. Using this anticlockwise is positive convention is good as it will set you up to use matrices in the future (should you learn this).
Hello Tierra, I dealt with them by using an angle from the positive x-axis. The sine of an angle is positive from 0 degrees to 180 degrees, and negative from 180 degrees to 360 degrees. The cosine of an angle is positive from 0 to 90 degress, and 270 to 360 degrees; and negative from 90 degrees to 270 degrees. Check cos150 and you will see that this is a negative value. (x component for 5N force) Check sin150 and you will see that this a positive value. (y component for 5N force) Check cos300 and you will see that this is a positive value (x component for 15N force) Check sin300 and you will see that this is a negative value (y component for 15N force) Hope this helps
Hello Adam, it was to make it clear how to find each term that is being added when you find, for example, the sum of forces in the y direction. You can skip this step and speed up your calculation time by directly adding those terms together in a single equation.
Yep, you will get a different angle. You are using the tan equation to find an angle. Usually, I just draw the triangle that results in order to find an angle relative to either axis; then measure the angle from the x-axis. ------- tan equation below ------- Tan (angle) = opposite/adjacent In some countries they use Tan (angle) = perpendicular / base.. I dont like this naming convention because it confuses some people.
Good point: its because I use both interchangeably; and the benefit of using 5sin30 like I have in the video is that sometimes it can be quite alot faster (or more convenient) than using the angle relative to the x-axis. Part way through creating this video, around the 3 minute mark, I saw that it would be easier to demonstrate an answer that finds the x and y components of the vectors when they are written in polar notation (using magnitude and the angles measured from the x-axis).
Cornelis Kok so if you were to use the original 5sin30 etc, would you just be able to do the exact same table, except with all values relative to something else, and what would that something else be, I’m a little confused. Thanks for your help
@@jacobfewings4068 yup, you could do the same table. One difference is that you would need to deal with determining if the component is negative or positive. For this question, the x component of the 5N force is negative. Sum of forces in x direction = 10cos45-5cos30+15cos60 Also, the y component of the 15N force is negative Sum of forces in y direction =10sin45+5sin30-15sin60 If you check the totals for the equations above you should get the same answer as the video.
If you are representing the forces as phasors (like I am in the video above) then you would. Some people, including me, sometimes solve these questions by treating each of the forces as the hypotenuse of a triangle and use the values obtained at @2:15.
We can use this method for vectors that are 3d space (x, y, z) coordinates. It would mean that the co-ordinate is not coplanar. In this case, the force in the z direction component, for each of the vectors, is zero for all of the vectors.
Thanks for the tutorial, I really appreciate .. but why is the reference angle negative that is -18.5 I thought we make it positive... To get an accurate angle
No worries. The angle I have written as I am measuring the angle from the x-axis and assumed that anti-clockwise from this position is positive. This convention is fairly common. Since the direction is 18.5 degrees clockwise from the x-axis; I have written this as negative. Hope this helps.
he get does degree from the reference angle which from Qudrant 1 cos going counter clockwise to the respected vector and then he subtract the given angle by. 90, 180, 270, and 360. (e.g F2 got 30 degree, so from the x axis from quadrant 1 counter clockwise to x axis of quadrant 2 therefore 180-30=150 degrees.
Hi Md ZI, there are different methods to solve this type of question. The 'tabular method' refers to the table being used. There are four other methods I am aware of: graphical method; analytical method; geometric method; unit vector. A special case of the unit vector method is to use matrices. For the second question, can you help me to understand what you mean by asking it in a slightly different way?
It was in degrees. You have the option to change between radians, degrees, and grad. Do you know what model of calculator you own? It usually is written on the top right corner...
Sure, the short answer is we can use either. The long answer: For the x component of the 15N force, we have: 15 cos 60 = 7.5N in the positive x-direction; and 15 sin 60 = 12.99N in the negative y-direction (pointing down). This method treats the 15N vector as the hypotenuse of a triangle, then we need to deal with finding if the direction is positive or negative. We would then have the x-component as 7.5N and the y-component at - 12.99N. The negative is because it is pointing in the negative y-direction. In the table that is shown in the video @5:16, instead of treating the vector like a triangle, I have treated it as a rotating vector. The sine function provides the y-component of a rotating vector, and the cosine function provides the x-component. This video shows that the height of rotating vector can be used to create the sine wave: th-cam.com/video/QFi16s4RXXY/w-d-xo.html also this one (be careful, this one has classical music) th-cam.com/video/miUchhW257Y/w-d-xo.html Basically, there are two ways to do this and in this video I used the first method to describe the components, and I used the second method for the components in the table. I hope you will find that: 15cos300 is equal to 15cos60, which is the x-component; and that -15sin60 is equal to 15sin300, which is the y-component. Please let me know if this helps :-)
Not for this one. If you look at the diagram drawn @7:34, you will notice that the angle is going downwards from the x axis. In this case you would subtract 18.46 from 360 to get an angle of 341.54 measured anti-clockwise from the x-axis Or you can subtract 18.46 from 0 to have an angle of -18.46 measured anti-clockwise from the x-axis (note that the negative means the angle is measured clockwise from the x-axis)
The figure does help to point out the sign convention and I have used a symbol at 11:00 to indicate the sign convention. I do agree with John that including a diagram to indicate the sign convention could be useful to avoid any doubt as I have seen several questions about the angle and the negative sign with the 18.5 degree angle in the comments section of this video.
Yes, the 3.42 is minus. The minus is inside a square, so it becomes positive. So (10.24)^2 is positive, and (-3.42)^2 is positive. Sqrt [ (10.24)^2 + (-3.42)^2 ] is what I could have written too...
It was due to an assumption that a positive angle would be anti-clockwise from the x-axis and a negative angle would be clockwise from the x-axis. Also, the symbol I used (@11:00) before the 18.5 means that I have assumed that a positive angle is anti-clockwise from the x-axis
Be careful with that idea. I'll say, yes BUT only if your angle is being measured from the positive x-axis. I do not prefer using that because I find that I am slow at answering questions if I keep changing angles as being measured from the x-axis. There are two ways that I am showing in this video. Method 1: where the angle is always measured from the x-axis Method 2: taking the vector and treating it like the hypotenuse of a triangle, then using that to find the x- and the y- components. I prefer method 2 because it is often faster.
You can treat it as negative, and you will still get the same answer. sqrt( (10.42 )^2 + (-3.42)^2). Note that a negative number squared is a positive number.
U SAVED MY LIFE.
I WAS REALLY BLUR THE WHOLE TIME WHEN MY LECTURER EXPLAINS THIS DURING MY MECHANICS CLASS
BUT
BRO
U SAVED ME. NOW IM DOING MY HOMEWORK IN THE SPEED OF LIGHTNING AFTER WATCHING THIS.
GOD BLESS U.
Yuri Ikutsuki Glad I could help. Thanks for your message.
Cornelis Kok same
@@CornelisKok Fake account lmao
Wowohhhz
This video has helped more than any textbook or lecture slide could ever do. Studying online isn't bad after all thank you!!
thanks a lot for the tutorial a lot has been learned....You have just educated a kid in Gaborone,Botswana,Africa through the power of internet
Glad this has been helpful for you!
How did it go with physics?
I'm watching in 2024❤
Same here
Same here
Same here
Me too got a test tomorrow 😅
Frr
You are doing a deed which is the highest of ranks..you are serving knowledge!..
May god bless you for doing this in this
century,where people are just spoiling others.
Thanks a lot ,this helped me more than you think: )
If you look around closely, you will find that everyone is serving knowledge. One just have to open himself to get the effect!
Cheers mate
*G
this is much more easier than what our prof. taught us !! thank you so much!
The way my lecture calculated this was so confusing me, but now I get it, I will surely use this method..thank you so much
This was wayyyyyyyyyyy more easier than all of those laws that was needed to solve for this, thank you so much
You are one of the best tutors I have ever met on TH-cam
I am doing engineering level 3 and I joined in late when they already finished this topic. I am grateful that I finally have learnt from you. Thanks for the video again. Keep on posting more great stuff.
Bro still in high school😢 and we do this in Fm
What a brilliant video, I failed this topic first time around now I’m revising for it again and my lecture it’s the best. So this video has helped me a lot !
Thank you, I hope you did well in your assessment.
Imagins 6yrs ago and it's very useful for me today thanks man 🙏
Thanks for the lesson way more easier than the way my physics teacher taught me
Thanks to this man for making our maths classes easier
This is the best method, hands down. Cheers
You do a much better job of explaining this than my tutor. Thanks.
thanx mate i was able to clear my internal exams thanx to you
Great to hear!
Shout out to those who watching at exam time
Thanks very much, you have made me understand what i didn`t understand in class. Am humbled.
thankyou, you're a life saver from my physics class, you taught me very well than my prof, we love you :)
🥰
Thank you so much..after searching a lot got your perfect video
THANK YOU SO MUCH YOU JUST SAVED MY LIFE IN THE HSC!!
Thanks sir for the refreshment because it’s been like six years now I studied this.
If I am not confused I believe the angle for the resultant force should be 360°-18.5° since all the angles of the other forces were taken from the posive x-axis for the calculation
Thank you very much, i was totally confused when my lecturer taught me this. Thanks again for your help😊
I greatly appreciate. No wonder a day teaches for one to meet his destination.
GOOD lecture of high standard , thank you very much hope to see more in the subsequent days.
I realy like your video .
U save my life those who going to darkness. Thank u sir g
shout out those watching at the time of quarantine
Meeeeh
Yessss. Let's get them grades!
@@moodymoe100 yes sirrrrrrrr!!!!!!
Late
Later😢@@ratava2285
thank u so much, i have mideterms coming up and this rly cleared stuff up
Awesome, good luck for your midterms.
Great tutorial! This will help me out on my exam!
I UNDERSTAND WHAT I DONT UNDERSTAND I MY CLASS,THANKS.
thank u very much for helping me,love u my teacher,may god bless u
You helped me.I didn't attend those lectures.now I understand
Great to hear, glad I could help.
@@corneliskok7794 where is your review finding forces vid? Cos sin....
Shout out to those watching before physics externals
Why do you choose negative angle when it is on the positive side of the horizontal line?
Really really great video!!! Thank you so much, it is very much appreciated.
Great that you have found it useful. You're welcome.
Wow you made this seem a lot more simple than when my teacher explained jt😂
you might of just saved my T levels cheers boss
Isn't the angle supposed to be read from the positive x-axis in an anticlockwise direction??
Which will make the reference angle for 18.5 degrees to 341.7??
I need clearance please
Yes, you are correct. 341.5 would be a suitable answer (-18.5 is also a suitable answer). The angle is to be read from the positive x-axis, in an anticlockwise direction based on the convention I have used.
Some people may give you the requirement that the angle must be postive, and between 0 and 360 degrees.
clear,neat and understandable for all pupil thanks a lot.
It's more than a help,, thanks for the great job
How did u find the 150° angle?
This is perfect thanks so much. God bless you and make you great in Jesus name amen
Can you give a like if I say he is better than most of our lecturers in colleges
Thanks for making me understand this topic
thanks for your teaching, help a lot
No worries, thank you for your appreciation.
This is a great video but I have a question??
I’ve been given a problem with a negative angle -30* would I assume this would be negative x and y axis so 330*?
Sorry for the delay. Yes, that would be right.
Felt very good. I understood. Thanks you Sir 😊
This has helped a lot, thanks!
An absolute legend this man!!!
this guy just save me, thanks !
Very great tutariol, how can we post some questions to you for help
This was very helpful.Thank you.
Thank u so much this video had helped me but , if i may ask why did u add all the forced and their angles together I thought there was a formula for this how can I use the formula to solve it
Not sure if I am answering your question but I will have a go.
The reason for adding the forces together is because this could lead onto finding out the overall force acting on a system. This would/could then help us to find out loads and stresses within the system.
You might notice that I am quite hesitant to write a formula. I have been stung in the past by being taught a formula only (it was a math exam when I was 17), but then I was given a question in a test that required proper understanding of what I was doing. As a result i wasn't able to even try the question. For this reason, I try to teach based on understanding the basic idea.
However, here is a formula that may be useful for you
Sum of forces in x direction = F1 * cos (theta1) + F2 * cos (theta2) + F3 * cos (theta3)+.....
Sum of forces in y direction = F1 * sin (theta1) + F2 * sin (theta2) + F3 * sin (theta3)+.....
Resultant = [ (sum of forces in x-direction)^2 + (sum of forces in y-direction)^2 ]^0.5
Angle = atan ( (sum of forces in y-direction)/(sum of forces in x-direction) )
**Note that for the angle you need to be aware that for tan there are multiple solutions between 0 and 360 deg and you will need to find the correct value.
Very understandable and organised. Thanks!
Watching in 2024 thank you so much 🙏🇿🇲🇿🇲
Thank you so much sir. Helped me big time.
lifesaver, my maths lecturer only explained everything using 2 vectors but then gave us an assignment to find the resultant of 3 vectors
so thank you!
How did the angle of the final answer became negative from positive 🙏
Thanks for the video.
But while working I didn't see you taking into consideration the quadrant of the various functions (sin, cos)
🙄Like in the second quadrant cos is negative
thank you so much, this really helped me so much!
Thank you for the great lesson . How will we represent the angle if it has to be represented positively instead of negatively ?
you could add 360 degrees to the value to get an equivalent number.
-18.5+360 = 341.5
THANKS SIR.
HOW TO SOLVE THE ONE WITH 4 FORCES ACTING ON THE POINT
So to calculate the angles for the 5N,10N and 15N,do you have to start from the positive side of the x-axis?
If you use an angle from the positive x it gives you directly the sign where the force is going whether it is posive or negative.
thankyou for sharing this method!!
So does this mean that in the x component we are suppose to use a cosine and in the y component we use sin
Me watching this video in a lecture while the lecturer is busy reading exactly what's in the Slides.🙂
Please check your 5N force components. Why did you take 5cos150 as your x component? It should be y component as per the position of the angle 150deg. As the sine component always lies opposite to the angle whereas the other side is given to the cos component. Please check the resolution of the 5N force. I might be wrong too. Please correct me in that case. THANKS
You haven't considered the direction of sin and cosin, Why? I mean in ur video there is 15N acting downwardnso while adding the x component , it would be -15Cos(300) right?
Hi Mary, sorry for the delay. The direction is taken care of if we use all angles as measured from the x-axis because the sine curve is a negative when the angle is bigger than 180 and less that 360.
can i say the x component for vector 15 N is 15*sin30 and y component as -15*cos30 ?
Yes
How did you calculate the square root of cos and sin cause I got a different answers
Pls a simple question i thought the angle for the resultant force was measured anticlockwise from the x axis so why did you measure it clockwise
It would be more clear if I put it as the angle anti-clockwise as that is normal.
At that time of making the video, I had made that decision because I had calculated the angle and was wanting to make it clear that angle was as measured clockwise from the axis. Both answers are right (measuring clockwise or anticlockwise); but different people have different conventions that they follow. In your case, you have been taught the convention that anti-clockwise measurement of angles is required. Using this anticlockwise is positive convention is good as it will set you up to use matrices in the future (should you learn this).
How do u know that the angle is negative?
In the beginning should it not be 5*cos(150) and 5*sin(150) instead of 30 degrees?
I understand this solution. Only thing is, how come we didn't account for the negative values in the 2nd and 4th quadrant?
Hello Tierra, I dealt with them by using an angle from the positive x-axis. The sine of an angle is positive from 0 degrees to 180 degrees, and negative from 180 degrees to 360 degrees. The cosine of an angle is positive from 0 to 90 degress, and 270 to 360 degrees; and negative from 90 degrees to 270 degrees.
Check cos150 and you will see that this is a negative value. (x component for 5N force)
Check sin150 and you will see that this a positive value. (y component for 5N force)
Check cos300 and you will see that this is a positive value (x component for 15N force)
Check sin300 and you will see that this is a negative value (y component for 15N force)
Hope this helps
what was the point in writing down the individual components, like15sin(theta) etc?
Hello Adam, it was to make it clear how to find each term that is being added when you find, for example, the sum of forces in the y direction. You can skip this step and speed up your calculation time by directly adding those terms together in a single equation.
loved the explanation.... I just subed!
Thank you for subbing and thanks for watching!
For the resultant, do you have to go 'X' then 'Y' because the other way round would give the same magnitude R but a different angle right? cheers
Yep, you will get a different angle. You are using the tan equation to find an angle. Usually, I just draw the triangle that results in order to find an angle relative to either axis; then measure the angle from the x-axis.
------- tan equation below -------
Tan (angle) = opposite/adjacent
In some countries they use
Tan (angle) = perpendicular / base.. I dont like this naming convention because it confuses some people.
What was the point in working out 5sin30 and 15sin60 etc, if you just used the angle relative to the x axis?
Good point: its because I use both interchangeably; and the benefit of using 5sin30 like I have in the video is that sometimes it can be quite alot faster (or more convenient) than using the angle relative to the x-axis.
Part way through creating this video, around the 3 minute mark, I saw that it would be easier to demonstrate an answer that finds the x and y components of the vectors when they are written in polar notation (using magnitude and the angles measured from the x-axis).
Cornelis Kok so if you were to use the original 5sin30 etc, would you just be able to do the exact same table, except with all values relative to something else, and what would that something else be, I’m a little confused. Thanks for your help
@@jacobfewings4068 yup, you could do the same table. One difference is that you would need to deal with determining if the component is negative or positive. For this question, the x component of the 5N force is negative.
Sum of forces in x direction =
10cos45-5cos30+15cos60
Also, the y component of the 15N force is negative
Sum of forces in y direction =10sin45+5sin30-15sin60
If you check the totals for the equations above you should get the same answer as the video.
What is the direction please of your angle, is it 18.5 south of east ??
Yes, that is correct
Do you always have to read anti clockwise
If you are representing the forces as phasors (like I am in the video above) then you would. Some people, including me, sometimes solve these questions by treating each of the forces as the hypotenuse of a triangle and use the values obtained at @2:15.
thanks i ask one question how many direction ? or we use only three direction (x,y,z) there is no additional direction?
We can use this method for vectors that are 3d space (x, y, z) coordinates. It would mean that the co-ordinate is not coplanar.
In this case, the force in the z direction component, for each of the vectors, is zero for all of the vectors.
Thanks for the tutorial, I really appreciate .. but why is the reference angle negative that is -18.5
I thought we make it positive... To get an accurate angle
No worries.
The angle I have written as I am measuring the angle from the x-axis and assumed that anti-clockwise from this position is positive. This convention is fairly common.
Since the direction is 18.5 degrees clockwise from the x-axis; I have written this as negative.
Hope this helps.
Wondering how you got 150° and 300°.
he get does degree from the reference angle which from Qudrant 1 cos going counter clockwise to the respected vector and then he subtract the given angle by. 90, 180, 270, and 360. (e.g F2 got 30 degree, so from the x axis from quadrant 1 counter clockwise to x axis of quadrant 2 therefore 180-30=150 degrees.
@@andresmichael359 thanks. I get now
Am I wrong if I just leave the answer as 10.8N and - 18.5 degrees without writing it the way u did it?
How did the angle become a negative at the final answer
2:23
Excuse me, what's a tabular method?
Measured a the X-axis?:
10N
Hi Md ZI, there are different methods to solve this type of question. The 'tabular method' refers to the table being used. There are four other methods I am aware of: graphical method; analytical method; geometric method; unit vector. A special case of the unit vector method is to use matrices.
For the second question, can you help me to understand what you mean by asking it in a slightly different way?
Oh MAN YOU'RE A FUCKING STAR...
Thank you so MUCH...
What mode was your calculator in when u punched those numbers. I never got the same answers as you, and mine is in Norma.
It was in degrees. You have the option to change between radians, degrees, and grad. Do you know what model of calculator you own? It usually is written on the top right corner...
Cornelis Kok Oh right , yeah yeah I got it. Thanks.😊🙌🏽
15sin(60) should be negative right? Since it's in the negative y-axis. Am I right? Thank you.
Yes, you're correct.
uhm...sorry, but why are we using 15cos300 and 15sin300 meanwhile the components are 15cos60 and 15sin60? can i please get a clarity there.
Sure, the short answer is we can use either.
The long answer:
For the x component of the 15N force, we have: 15 cos 60 = 7.5N in the positive x-direction; and 15 sin 60 = 12.99N in the negative y-direction (pointing down).
This method treats the 15N vector as the hypotenuse of a triangle, then we need to deal with finding if the direction is positive or negative. We would then have the x-component as 7.5N and the y-component at - 12.99N. The negative is because it is pointing in the negative y-direction.
In the table that is shown in the video @5:16, instead of treating the vector like a triangle, I have treated it as a rotating vector. The sine function provides the y-component of a rotating vector, and the cosine function provides the x-component. This video shows that the height of rotating vector can be used to create the sine wave: th-cam.com/video/QFi16s4RXXY/w-d-xo.html also this one (be careful, this one has classical music) th-cam.com/video/miUchhW257Y/w-d-xo.html
Basically, there are two ways to do this and in this video I used the first method to describe the components, and I used the second method for the components in the table. I hope you will find that: 15cos300 is equal to 15cos60, which is the x-component; and that -15sin60 is equal to 15sin300, which is the y-component.
Please let me know if this helps :-)
Wow
Very explained
Do we not add 18.46 to 180 to find the glob/local angle?
Not for this one. If you look at the diagram drawn @7:34, you will notice that the angle is going downwards from the x axis.
In this case you would subtract 18.46 from 360 to get an angle of 341.54 measured anti-clockwise from the x-axis
Or you can subtract 18.46 from 0 to have an angle of -18.46 measured anti-clockwise from the x-axis (note that the negative means the angle is measured clockwise from the x-axis)
Thanks for your valuable help
You should include the sign convention on the angle theta
No, The direction is mentioned in the figure he made.
The figure itself justifies the direction.. this is amazing
The figure does help to point out the sign convention and I have used a symbol at 11:00 to indicate the sign convention. I do agree with John that including a diagram to indicate the sign convention could be useful to avoid any doubt as I have seen several questions about the angle and the negative sign with the 18.5 degree angle in the comments section of this video.
When you did the 10.24 squared plus 3.42 squared the 3.42 is still minus isn’t it? So it still has the minus attached when you do the square root?
Yes, the 3.42 is minus. The minus is inside a square, so it becomes positive.
So (10.24)^2 is positive, and (-3.42)^2 is positive.
Sqrt [ (10.24)^2 + (-3.42)^2 ] is what I could have written too...
how did you get negative 18.5 at the last?? I mean why negative
It was due to an assumption that a positive angle would be anti-clockwise from the x-axis and a negative angle would be clockwise from the x-axis.
Also, the symbol I used (@11:00) before the 18.5 means that I have assumed that a positive angle is anti-clockwise from the x-axis
Sir, is it really understandable that when we say x-component we use cosine? and y-component is sine?
Be careful with that idea.
I'll say, yes BUT only if your angle is being measured from the positive x-axis.
I do not prefer using that because I find that I am slow at answering questions if I keep changing angles as being measured from the x-axis.
There are two ways that I am showing in this video.
Method 1: where the angle is always measured from the x-axis
Method 2: taking the vector and treating it like the hypotenuse of a triangle, then using that to find the x- and the y- components.
I prefer method 2 because it is often faster.
Thank you Sir 🎉🎉🎉😮😮😮
why dd the negative 3.42 turn positive while calculating the resultant using pythagoras
You can treat it as negative, and you will still get the same answer. sqrt( (10.42 )^2 + (-3.42)^2).
Note that a negative number squared is a positive number.