Resultant of Three Concurrent Coplanar Forces
Вставка
- Опубліковано 9 сер 2016
- Demonstration of the calculations of the resultant force and direction for a concurrent co-planar system of forces.
This video demonstrates the tabular method for 2d systems.
I'm watching in 2024❤
Same here
Same here
Same here
Me too got a test tomorrow 😅
Frr
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!!
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!
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?
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 is much more easier than what our prof. taught us !! thank you so much!
GOOD lecture of high standard , thank you very much hope to see more in the subsequent days.
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
You are one of the best tutors I have ever met on UA-cam
This was wayyyyyyyyyyy more easier than all of those laws that was needed to solve for this, thank you so much
You do a much better job of explaining this than my tutor. Thanks.
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.
THANK YOU SO MUCH YOU JUST SAVED MY LIFE IN THE HSC!!
Thanks very much, you have made me understand what i didn`t understand in class. Am humbled.
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
Imagins 6yrs ago and it's very useful for me today thanks man 🙏
I realy like your video .
U save my life those who going to darkness. Thank u sir g
thankyou, you're a life saver from my physics class, you taught me very well than my prof, we love you :)
🥰
Great tutorial! This will help me out on my exam!
I love your accent and thank you you saved me I literally cried in class today because I couldn’t understand this and we have a test over it TOMORROW. YOU ARE A LIFE SAVOR
Awesome, I'm glad I could help. Good luck for your test!
thank u very much for helping me,love u my teacher,may god bless u
Thank you so much..after searching a lot got your perfect video
Thank you very much, i was totally confused when my lecturer taught me this. Thanks again for your help😊
Very understandable and organised. Thanks!
Really really great video!!! Thank you so much, it is very much appreciated.
Great that you have found it useful. You're welcome.
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
Felt very good. I understood. Thanks you Sir 😊
Thanks for the lesson way more easier than the way my physics teacher taught me
This was very helpful.Thank you.
Thanks for making me understand this topic
I UNDERSTAND WHAT I DONT UNDERSTAND I MY CLASS,THANKS.
Shout out to those who watching at exam time
shout out those watching at the time of quarantine
Meeeeh
Yessss. Let's get them grades!
@@moodymoe100 yes sirrrrrrrr!!!!!!
Late
thank u so much, i have mideterms coming up and this rly cleared stuff up
Awesome, good luck for your midterms.
thank you so much, this really helped me so much!
Thank you so much sir. Helped me big time.
Thank you for your explanation
This has helped a lot, thanks!
I greatly appreciate. No wonder a day teaches for one to meet his destination.
Wow you made this seem a lot more simple than when my teacher explained jt😂
Thanks for your valuable help
How did u find the 150° angle?
Why do you choose negative angle when it is on the positive side of the horizontal line?
thanx mate i was able to clear my internal exams thanx to you
Great to hear!
Thanks a lot it was so helpful 👏🏻😊
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
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.
aren't you supposed to subtract them because they are not going on the same direction??
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.
thanks for your teaching, help a lot
No worries, thank you for your appreciation.
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
Thanks to this man for making our maths classes easier
what if you only have angles that dont have a relation to the x or y axis?
Thanks for the video, i learn a lot from it.
I have a question, what will be the direction of a force whose vertical component is 0N and horizontal is 30N
loved the explanation.... I just subed!
Thank you for subbing and thanks for watching!
THANKS SIR.
HOW TO SOLVE THE ONE WITH 4 FORCES ACTING ON THE POINT
perfect video
Very great tutariol, how can we post some questions to you for help
this guy just save me, thanks !
Great vid !!!
How did you calculate the square root of cos and sin cause I got a different answers
I have now understood thank you
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.
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.
Great vid helped alot
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
Thank you very much ☺
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?
Thank you so much T.T
Awesome Man!
An absolute legend this man!!!
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.
QUESTION:
what will you do when after adding the x-comp and you are getting a negative number and if same happens to the y-comp
Hi Albert, You would add the vectors head to tail; with the x-component pointing left (horizontal with the page) and the y-component pointing down (vertical with the page). You would have a triangle that looks like this... www.researchgate.net/figure/RightAngle-Triangle-Hopper-Scale-19-Therefore-from-the-right-angled-triangle-Fig_fig2_281274006 (ignore the y on this image)
You could find the angle theta using the tan(theta)=O/A relationship
Then to find the angle with respect to the x axis you would add 180 degrees to theta above.
Note: If the vector is in the "third quadrant", the angle with respect to the x-axis should be between 180 degrees and 270 degrees.
For the magnitude: you would do the same as in the video. Note that the square of a negative is a positive. For example (-4)^2 is equal to 16.
This is perfect thanks so much. God bless you and make you great in Jesus name amen
It's helping a lot
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....
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.
Perfect!!!
How did the angle become a negative at the final answer
Thank you so much sir👌
In the beginning should it not be 5*cos(150) and 5*sin(150) instead of 30 degrees?
Good lecture
Thank you so very very very much
Thanks bro you are awesome
Nice video sir..
very nice.thanks
Thank you so much
May god bless you ❤
You can write it -18.5 or 341.5
Thanks man helped alot
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.😊🙌🏽
Sir What about the direction of components of other two forces..? X component is negative in 3 quadrant please guide me
Perfect!❤️❤️
How do you know which one to use ? as in sin cos, or tan ?
x component is always cos and y component always sin
I'll also add that this is true, only if the angle is measured from the x axis.
thank you so much......................
How do u know that the angle is negative?
Oh MAN YOU'RE A FUCKING STAR...
Thank you so MUCH...
Wow
Very explained