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.
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: )
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 !
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.
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*?
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 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.
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
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.
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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-QFi16s4RXXY.html also this one (be careful, this one has classical music) ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-miUchhW257Y.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 :-)
Can you please help me to solve this problem,Four forces A,B,C and D act at a common point zero as show in diagram below.The magnitude of forces are as follows.A is 5N,force B is 8N,force C is 6 N and force D has an unknown magnitude.
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).
I should have put -3.42 into the pythagoras theorem. However, it doesn't change the result because the negative is squared (-3.42) squared = 11.6964 (notice that this is positive, if you are getting a negative number make sure you have the number -3.42 in brackets before squaring) (10.24) squared = 104.8576 11.6964 + 104.8576 = 116.554 square root of (116.554) = 10.80 kN
In this case, just so that I could show that there are two ways to deal with finding components. One that treats the vector like the hypotenuse of a triangle. the second, that treats the vector like phasor (has a magnitude and direction)
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.
ani2000ify Hi there, this question shows a force system that is not balanced. That means the x component of the 5N force does not equal the x components of the 10N and 15N forces. One way to visualise this question is to replace the force vectors with distance vectors. Stand in a middle of a field, travel 10 metres at an angle of 45 degrees, then travel 5 metres at an angle of 150 degrees, then travel 15 metres at an angle of 300 degrees. The final location will not be the same as the starting location. Hope this answers your question.. if not let me know.
Hi there, make sure you put in a second set of brackets. The number you have looks like the answer to 3.42*3.42+10.24 = 21.9364 It should be √((10.24)²+ (3.42)²) or the answer to √ (3.42*3.42+10.24*10.24)
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.
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.
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.
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
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.
Sir could you please solve this question for me .I am not getting any idea to solve this Question What is the condition for equillibrium of concurrent force? 2 )A mass of 6 kg is suspended from ceiling by a rope of length 2 metre .A force of 50 N is applied at midpoint of rope in horizontal direction .What is the angle the rope making with vertical in equillibrium?(taking g = 10 m/s) 3 )what will be the angle made by the rope if its length is doubled?
2 questions 1) How can u use cos and sin if you have angles like 300 meaning its not even a triangle anymore? 2)Why can you go clockwise for the final resultant answer but before all angles were measured anti clockwise?
1) Trigonometry is used in triangles, 3 sided shapes. It can not be used in others unless you split the shape into triangles and work it out from there. 2) The final resultant was negative, subtract this answer from 360 it will give you the anti clockwise angle. Hope this helped :)
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.
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...
Hi Cephas Mwale. I'm assuming you mean the 'angle of the resultant of the net forces'. Answer: No, that is not correct. I have used the convention that a positive angle is measured anti-clockwise from the x-axis. In order for my answer to match that convention, I needed to calculate how big the angle is from the x-axis to the vector. In the video, I have found out the angle going clockwise. To get the anti-clockwise angle (from the positive x-axis), I used the idea that there is 360 degrees in a full rotation, and that the full 360 rotation is made up of the clockwise angle from the +ve x-axis and the anti-clockwise angle from the +ve x-axis. In short 360 = anti-clockwise angle from positive x-axis to the resultant vector + clockwise angle from positive x-axis to the resultant vector 360=anti-clockwise angle from positive x-axis to the resultant vector+18.5 anti-clockwise angle from positive x-axis to the resultant vector = 341.5
Why didn't you keep -3.42 as a negative number and instead used it as a positive number, when calculating the angle and R using trigonometry and Pythagoras. Even though, all the sin values added up to the negative 3.42.
When I was doing the calculation at that stage, I was treating the system of forces as a right triangle. So I was finding the angle as per the diagram drawn at 8:00 into the video. I was calculation the magnitude of the angle and resultants shown and letting the diagram indicate the direction.
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.
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
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
Why is the angle negative?? Isn't the angle supposed to be positive & is that our real angle or u have to either subtract or add to 180⁰ to find the final angle
I think you are talking about the angle at @11:10 for the resultant force. The angle is negative because there is a convention in Mathematics for angles. The angle is measured from the x-axis, and the positive direction is anti-clockwise. The angle could be written as 360 -18.5 = 341.5 degrees as a measurement from the x-axis... or could be written as -18.5 degrees as measured from the x-axis. Also, the angle symbol used (www.fileformat.info/info/unicode/char/2220/index.htm) means that the angle is measured from the x-axis with anti-clockwise positive here is an answer by someone else www.quora.com/Why-do-we-measure-the-angle-anticlockwise
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...
Not always. If you treat the vector as the hypotenuse of a right-triangle, you can find the the sides of the triangle using the angle with respect to the y axis. In this case you are likely to see that you use cosine for the y-component and sine for the x component. I have a comment further down describing that there are several ways to find the components.
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.
It is quite hard to answer your question without an image. Would you be able to post an image or provide a link to an image? It could be that the angle is given to you in the question and you haven't spotted this yet. It could be given by the dimensions in the question or by a small triangle that you can use to find the angle. It might be that the question is asking you to find an equilibrant. It could also be that the question has an error in it. I'll be able to help better with an image of the question that you are trying to solve
marianne pajatin .. thanks for your comment. I left out the negative sign because even with the negative sign I would have gotten the same answer. When you do your answer please be sure to put -3.42 into brackets... ie r^2=(10.24)^2+(-3.42)^2?
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.
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)
We could have ignored the first part. There are two approaches to breaking down the forces into components shown in this video. The first approach treats the vectors as triangles, where we find the size of the components, then add up the x components (if right, it would be added; and to the left, it would be negative). The second approach treats the vector as a phasor (try google this term and you should see a video of a vector that rotates, and that this rotating vector can be used to draw a sine wave (for y component) and cosine wave (for x component)
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.
