I literally adore your techniques and explanation. You are one of the greatest out there. thank you for making this subject more fun and less stressfu,
this is the best i learned ma whole semester in few hours. tried to understand what lecturer saying but couldnt. this channel helped me to understand in seconds thankyou so much for ur hard work.
If I only found this channel before, I probably ace my statics last sem😭💖 Thank you so much sir for this. I'm reviewing all of my subjects today since I do not learn from my previous professors.
Excellent teaching skills! Thank you for breaking down the extremely basic reason behind how to get Fx and Fy are equal to the magnitude of the force x sin or cos!! Every other video I've seen skipped that first step and went right to what Fx and Fy are equal to. Now that I know that reason, it all clicked in my head.
Thank you very much. I am really happy to hear this video helped you out and made sense. Keep up the great work and I wish you the best in all your endeavors. Also, thanks for taking the time to write your nice comment :)
Thanks! You are helping me understand the basics and somewhat a bit complex topics in preparation for my board exam (licensure exam) for Civil Engineers this November 2024. I hope you upload lots of engineering-related topics soon.
Wish i found your channel back in college days. Now im a professor teaching Statics and plan to use your examples on my classes. May I? Examples and explanations are just so good.
Thank you very much and congratulations on your career! Yes, absolutely, you can use the videos but please don't remove the "question solutions" watermark and give proper citations. I wish you the absolute best with your teaching career.
Only your RUclips channel clear my concepts. heartly thanks. Your animation and explanation are great. again thanks. Kindly i request to not delete all videos. If you have any source of materials like animations for understanding strength of materials of mechanics of materials, therory of machiness subjects for gate and ese kindly tell me.
Thank you very much! I appreciate your comment, and I am really glad to hear that these help you out. Also, I don't think I've deleted any videos and probably never will. They are available for free, for anyone to learn from :)
you are absolutely correct, before I always think that sine is for y and x is for cosine, where infact sine must be used with respect to the opposite angle if you are going to get the side/length
In 3:22 I'm confused. It may be me or you might be wrong because you used 400 as the hypotenuse in the second component when it should be 800. But anyways, the result was correct, it's just the values that you input might induce misinterpretations.
Yes, it's a typo, because while I did mistakenly write 400 under the fraction, in the next step, I used 800. I am sure others can easily figure it out, but thanks for pointing it out :)
It's just solving 2 equations with 2 unknowns. You can graph them, use trigonometric identities, elimination, etc, whatever you like. You can also use wolfram alpha to check your answers. 👍
@@borgoniariocarlos.5699 It should have been covered in high school. Search "simultaneous equation solving." It's solving for 2 unknowns with 2 equations. www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-with-substitution/v/the-substitution-method
@@magalongmarkbrodettel.423 You must learn how to solve simultaneous equations, this should have been covered in high school. Eventually, you will need to solve 3 or 4 equations at the same time, so it might be a good idea to refresh your memory on how to do this now. www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-with-substitution/v/the-substitution-method
For problems that require a large amount of trigonometric identities to solve, I think it's faster to just graph it. Graph the 2 equations and see where they intersect, and you get your answer. :)
@Nick Jr In general, it's a ratio triangle. Think back to learning about similar triangles in math class, it's the same principle. The ratio triangle is a scaled down version of the big one. So we use the small triangle to find the sides of a bigger triangle. I also explain this with color coded sides at 5:27, I don't know how else to explain it 😅 If you have a specific question about the triangle, let me know and I'll do my best to help. :) Pleases see the following: www.mathsisfun.com/geometry/triangles-similar.html
When you break forces into components, there is no positives or negatives, it's just a component of a force. When you add components however, you need to consider which side is positive and which side is negative. In that case, if you pick right to the positive, any component facing left will be negative, and vice versa. Since we didn't assign positives and negatives for the components, you can even pick right to be negative, which means left facing forces will be positive. So you can see at 8:34, we have -4 sin 15.
I am not sure how to explain it more :( The 13 is the hypotenuse of the little triangle so that goes on the bottom. If we want the y-side, then we need to look for the value on the y-side of the little triangle. Here, it's 12, so that's what we use for the top of the fraction. If you need the x-side, then you look for the x-side on the little triangle. That's 5, so you write 5/13. If you were referring to the values, they were already given in the question diagram.
8:56 If I use Fcos(86.03) it results to 1.9687. If I use Fsin(86.03) it results to 1.9677 (as shown in the video). These 2 values are just approximately the same right? Nothing wrong with this? Sorry for this no brainer question.
Sir, I'm confused. Can you show how you get the angle 86.03. I actually tried to use the substitution method but it gives me 86.3. Perhaps, I'm not solving it right, so I want to see how did you do it. My exam is in 2 days, I hope you can enlighten me❤️
86.03 and 86.3 is very close, maybe it's a numerical or rounding error? See: www.wolframalpha.com/input?i=%28y%29sin%28x%29%3D1.963%2C+ycos%28x%29%3D0.1363 I assume you got the F value correct right? if so, you take the F = 1.9677 and plug it in, and then solve for θ, you will get 86.03. So you just take the inverse of sin. sin^-1=(1.963/1.9677) = 86.03 degrees. You can do the same with the cos value as well. Either works. I hope that helps!
Sir, can you enlighten me regarding the the value of theta and F from the last example? I got theta as 26.937 degree and for the F it's 4.333kN. Correct me if im wrong. Maybe I got it wrong.
You must look from the perspective of the angle. So opposite to the angle is the x-component, and sine = opposite over hypotenuse. Adjacent to the angle is the y-component, so that's cosine. See: ruclips.net/user/shortsvynnKlJD_Jo?feature=share
hi thank for your time , but how can be the angle that we get = 86.03 between F and y' and at the same time angle between x and x' = 30 ? if we add those two angles it would give us 116.03 which is more then 90 that should be between the x axis and the y axis ?
The force F actually goes past the x' axis, and is 3.97 degrees away from the x-axis. Remember that the diagram isn't always accurate, it's just a base point to start the problem. When we get new data, we have to reimagine the diagram in our heads. In this problem, we know that force F is actually 86.03 degrees away from the y-axis, which means it must go past the x-prime axis. Also, I am not sure why you are adding the 2 angles, since there is no relation between them. One will simply overlap the other 👍
So you can directly plug the cosine and sine values into your calculator. 8cos30 = 6.928 and -4sin15 = -1.035. After that, you're just adding the values together and isolating for Fsinθ.
I explain it from 5:39. I am unsure how else to go deeper. In essence, we are using ratio triangles to figure it out. The corresponding side is what gives the multiplying factor to figure out the component. I think the best way to understand it is to do a few examples with that specific method. I am sorry I couldn't be of more help :(
You might not be remembering trigonometric fundamentals. I highly encourage you to take a look at them and review the material. Remember that sin is opposite over hypotenuse whole cosine is adjacent over hypotenuse. You have to keep these in mind or you will have a hard time breaking forces into components 😅 Please see: www.mathsisfun.com/sine-cosine-tangent.html
@@QuestionSolutions apologies. Minute 8:57. I’d just like to know what was performed with these specific numbers. Already watched the recommended videos but I would still like to know what we would do with these specific numbers in our case. Thank you!
@@eduardobribiesca1980 You can use substitution to solve for the 2 equations since it's 2 equations with 2 unknowns. You will also need to use tan = sin/cos identity. If you don't want to go through the trouble, you can always graph the two equations.
@@dammy3985 So if you look at 7:48, you will see the components of force F. Notice how the x-component is opposite to the angle, which means it's sine, and notice how the y-component is adjacent to the angle, so it's cosine. After we get all the components, we need to figure out what value of force F would give us a resultant force of 8 kN along the positive x' line. So in essence, we are working backwards. I encourage you to try and solve the problem again, without watching the video and then when you get stuck, look to see what the next step is. It might be more helpful. 👍
So I think you might be a bit confused about how we draw the angles. The negative sign on the y component means the component is along the negative y-axis. That's why it was drawn downwards in the diagram. So when you find the angle, what you are finding is the angle with respect to the positive x-axis here. From FR to the positive x-axis. So an angle can't be negative. If you need further clarification, let me know.
@@johnwealth2205 I just want to make it clear that we aren't ignoring the negative sign. This negative sign is in relation to the coordinate system we used. So we got a negative value because the y-component is facing down, and that's only because we established up and to the right to be positive. In other words, our coordinate system had y and x at the top and on the right side (this is almost always the case anyways). So when we got a negative value, it indicates that our y-component faces down, along the negative y-axis. Think of it this way, the force itself can't be negative right? Like if we had a box, and there are 2 forces applied to it, one is pulling on it from the left, and another pulling from the right. The right will be considered positive if we assume right to be positive. Does that mean the left one in real life is negative? What is a negative force in our real world? There is no such thing right? So all we are saying is, a negative force is simply a force applied in the opposite direction to what we consider is positive. Once you draw your forces, when it's up or down, based on the positive and negative signs, we need an angle to show where it is with respect to our positive x and y axes. So we use the inverse of tan to find it. The angle isn't negative because it gives us a "magnitude" value of where our resultant force is, with respect to the x-axis. Going a bit further, what is a negative angle? In degrees, it just goes in a loop, from 0 to 360, and then back around again. Does that make sense?
@@QuestionSolutions Surprisingly you're really good at explaining..... I understand better now.... The negative sign is a representation that let's us know the position a value lies in the x or y axis , It simply shows whether or not it's in the positive or negative value of x or y..... But when it comes to angle ( as it concerns the meeting point of two lines) it can't be a negative value right? I'm just trying to make sure I understand what you're saying
@@johnwealth2205 That is correct, it is indeed just a way for us to know where the force lies, in the positive or negative axis. And yes, so angles aren't negative. Like imagine someone told you to draw an angle of -60 degrees? What would that be? Would it mean a counterclockwise turn? Instead, we do it differently. We just measure the angle with respect to an axis. It doesn't have to be just positive x-axis, it could be for any but most of the time, we use the positive x-axis as the reference line. That way, we don't have any negative angles. Please see: mathmonks.com/wp-content/uploads/2021/01/Reference-Angle.jpg
I am sorry to hear that. I hope you find other resources to help you out better. Don't worry, everyone learns at different paces, you just need to find your own groove. Best wishes with your studies!
@@alishamarroquin6301 you can slow down the playback speed, pause when you need to process something, and also follow along by writing the problem yourself. Pause to copy/comprehend when you need to. These are tactics I use. Something about trig has me always use the basics and it isn’t intuitive for me. I’ve been all the way through calculus, differential equations, physics, and I’m still slow when it comes to setting up sin/cos/tan problems.
On the first problem, when you are setting up your triangles, you moved Fy to the right to form your right triangle. If you moved Fx instead to form your right triangle, wouldn’t that give you Fx = 40sin30 and Fy = 40cos30? I’m trying to figure out why you moved Fy in the 400 N force, but Fx in the 800 N force.
No, your angle would not be 30 if you move the x-component up. You'd need the other side of the angle, which would be 90 - 30 = 60. So your x-component would be 400sin60 which is the same as 400cos30.
I guess if you want to think of it that way, there is no harm, but really, what we are trying to do is create a resultant force along the x' axis. So if we take all other other forces, we can replace it with our resultant force and it would have the same effect as all other forces effecting this bracket.
See how the 2 equations at 8:52 were solved: ruclips.net/user/shortsrAlhrq5hWFc?si=yiPoE3RxqFet_ib4
I cant believe I just found this channel, your explanation and visuals are so helpful!!! Thank you so much! God Bless you.
Really glad to hear it helps :) I wish you the best with your studies.
I literally adore your techniques and explanation. You are one of the greatest out there. thank you for making this subject more fun and less stressfu,
Wow, thank you very much! That was a really nice comment :)
I wish you the best with your studies, keep up the good work!
this is the best i learned ma whole semester in few hours. tried to understand what lecturer saying but couldnt. this channel helped me to understand in seconds
thankyou so much for ur hard work.
Glad to hear these videos are helpful :) Keep up the awesome work and do your best!
If I only found this channel before, I probably ace my statics last sem😭💖 Thank you so much sir for this. I'm reviewing all of my subjects today since I do not learn from my previous professors.
You're very welcome! Best wishes with your exams, I hope you do amazingly.
You're definitely the type of lecturer I need in my chemical engineering degree ❤❤❤😮
I hope you become an awesome chemical engineer! ❤️
Excellent teaching skills! Thank you for breaking down the extremely basic reason behind how to get Fx and Fy are equal to the magnitude of the force x sin or cos!! Every other video I've seen skipped that first step and went right to what Fx and Fy are equal to. Now that I know that reason, it all clicked in my head.
Thank you very much. I am really happy to hear this video helped you out and made sense. Keep up the great work and I wish you the best in all your endeavors. Also, thanks for taking the time to write your nice comment :)
It's very easy to understand
Tq so much
I never had this kind of explanation in my academics
Really glad to hear these help. Best wishes with your studies.
Why aren't you uploading more videos. Believe me your animated explanations are absolutely perfect
I am unable to upload during the month of July, will start again in August. :)
@@QuestionSolutions No issue, I wish my statics teacher could use 21st-century technologies to prepare his lectures but glad I found ur channel
@@talha3346 Thank you! I think everything needed for a first year statics course is covered in the statics A-Z playlist 👍
Thanks! You are helping me understand the basics and somewhat a bit complex topics in preparation for my board exam (licensure exam) for Civil Engineers this November 2024. I hope you upload lots of engineering-related topics soon.
Best of luck with your exam and all your future endeavors!
You’re an angel, this helped me actually understand why we do all of this
I am glad to hear that! Keep up the good work :)
Wish i found your channel back in college days. Now im a professor teaching Statics and plan to use your examples on my classes. May I? Examples and explanations are just so good.
Thank you very much and congratulations on your career!
Yes, absolutely, you can use the videos but please don't remove the "question solutions" watermark and give proper citations.
I wish you the absolute best with your teaching career.
Only your RUclips channel clear my concepts. heartly thanks. Your animation and explanation are great. again thanks. Kindly i request to not delete all videos. If you have any source of materials like animations for understanding strength of materials of mechanics of materials, therory of machiness subjects for gate and ese kindly tell me.
Thank you very much! I appreciate your comment, and I am really glad to hear that these help you out. Also, I don't think I've deleted any videos and probably never will. They are available for free, for anyone to learn from :)
really really appreciate this explanation, great work!
Thank you very much!
Extremely detailed explanation, better than a tutor
Thank you for your kind words! I’m glad that these videos were helpful to you.
This channel is amazing, thank you for the work you do. This video has helped me tremendously as a new mechanical engineering student
I am really glad to hear that. I wish you the best in your mechanical engineering degree.
you are absolutely correct, before I always think that sine is for y and x is for cosine, where infact sine must be used with respect to the opposite angle if you are going to get the side/length
It's a fairly common mistake for a lot of students 😅
Glad my professor recommending your channel
Wow, that's really nice to hear :) Keep up the great work and thank your professor too!
Thank you so much... I have always believed that the x and y components are cos and sin respectively... thank you so much
Don't worry, it's not just you, so many students tend to think this way and fall into the same pitfall again and again. 😅
u re the best my friend... im happy i found u
:) Many thanks!
In 3:22 I'm confused. It may be me or you might be wrong because you used 400 as the hypotenuse in the second component when it should be 800. But anyways, the result was correct, it's just the values that you input might induce misinterpretations.
Yes, it's a typo, because while I did mistakenly write 400 under the fraction, in the next step, I used 800. I am sure others can easily figure it out, but thanks for pointing it out :)
Hello there, how did you find the F = 1.9677kN and theta = 86.03? this was in the last question at 8:56
It's just solving 2 equations with 2 unknowns. You can graph them, use trigonometric identities, elimination, etc, whatever you like. You can also use wolfram alpha to check your answers. 👍
can you explain the procedure of solving that answer or equation?
Howww???
@@borgoniariocarlos.5699 It should have been covered in high school. Search "simultaneous equation solving." It's solving for 2 unknowns with 2 equations. www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-with-substitution/v/the-substitution-method
@@magalongmarkbrodettel.423 You must learn how to solve simultaneous equations, this should have been covered in high school. Eventually, you will need to solve 3 or 4 equations at the same time, so it might be a good idea to refresh your memory on how to do this now. www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-systems-topic/cc-8th-systems-with-substitution/v/the-substitution-method
2:54 that's what I used to think before watching this video 😭😭 thank you so much 🙏🏽
That's okay, now you know. :) Keep up the great work, best wishes with your studies.
thank u so much! your explanation is very clear. i learned a lot. keep up the good work! 😊
Happy to hear that! You as well, keep up the great work :)
at 8.56 you say 'you can solve them any way you like.', but i don't really know how to. Can you help me?
For problems that require a large amount of trigonometric identities to solve, I think it's faster to just graph it. Graph the 2 equations and see where they intersect, and you get your answer. :)
I have watched a few vdeo and it is good and easy to understand ! Tq
Thank you! Glad to hear these were helpful to you :)
Your channel is golden, keep up the amazing work
Thank you very much, I really appreciate it.
Nice efforts ,and very helpful for me.
Glad to hear!
@Nick Jr In general, it's a ratio triangle. Think back to learning about similar triangles in math class, it's the same principle. The ratio triangle is a scaled down version of the big one. So we use the small triangle to find the sides of a bigger triangle. I also explain this with color coded sides at 5:27, I don't know how else to explain it 😅 If you have a specific question about the triangle, let me know and I'll do my best to help. :) Pleases see the following: www.mathsisfun.com/geometry/triangles-similar.html
I do have a question. Is it possible to use the parallelogram rule on Question 2, for example, instead of vector analysis?
You can, but it's more effort.
Wow he is just good as the organic chemistry tutor
Thank you, best wishes with your studies!
Sir I just want to clarify in your example no.3 is Fx=4 sin15 or Fx=-4sin15?
When you break forces into components, there is no positives or negatives, it's just a component of a force. When you add components however, you need to consider which side is positive and which side is negative. In that case, if you pick right to the positive, any component facing left will be negative, and vice versa. Since we didn't assign positives and negatives for the components, you can even pick right to be negative, which means left facing forces will be positive. So you can see at 8:34, we have -4 sin 15.
5:58 how did you get 12/13 ? I cant understand that quick sorry
I am not sure how to explain it more :( The 13 is the hypotenuse of the little triangle so that goes on the bottom. If we want the y-side, then we need to look for the value on the y-side of the little triangle. Here, it's 12, so that's what we use for the top of the fraction. If you need the x-side, then you look for the x-side on the little triangle. That's 5, so you write 5/13. If you were referring to the values, they were already given in the question diagram.
8:50 how did you get Fcostheta = 0.1363? I did 8 sin 30 = 4 and 4 cos 15 = 3.863 then minus 6 but when i add them together it says 1.863
So just focusing on the equation for y-axis forces, all you do is 8sin30 - 4cos15. That gives you 0.1363. So you're just isolating for the Fcosθ term.
8:56
If I use Fcos(86.03) it results to 1.9687.
If I use Fsin(86.03) it results to 1.9677 (as shown in the video).
These 2 values are just approximately the same right? Nothing wrong with this?
Sorry for this no brainer question.
Yes, it's just errors in rounding. 👍
@@QuestionSolutions Thanks!
And the easy way to understand to all concept.
Awesome :)
Sir, I'm confused. Can you show how you get the angle 86.03. I actually tried to use the substitution method but it gives me 86.3. Perhaps, I'm not solving it right, so I want to see how did you do it. My exam is in 2 days, I hope you can enlighten me❤️
86.03 and 86.3 is very close, maybe it's a numerical or rounding error?
See: www.wolframalpha.com/input?i=%28y%29sin%28x%29%3D1.963%2C+ycos%28x%29%3D0.1363
I assume you got the F value correct right? if so, you take the F = 1.9677 and plug it in, and then solve for θ, you will get 86.03. So you just take the inverse of sin. sin^-1=(1.963/1.9677) = 86.03 degrees. You can do the same with the cos value as well. Either works. I hope that helps!
Sir, can you enlighten me regarding the the value of theta and F from the last example? I got theta as 26.937 degree and for the F it's 4.333kN. Correct me if im wrong. Maybe I got it wrong.
ahh disregard Sir, my computations were wrong. I got it now! Thank you
@@janecm6136 Glad to hear you got it :)
hey! may i ask what method did u used to solve the last part?
Fsintheta=1.963
Fcostheta=0.1363
Please see: ruclips.net/user/shortsrAlhrq5hWFc?si=yiPoE3RxqFet_ib4
Sir my only question is that why did you use Sine to find the x components?
Please see: www.mathsisfun.com/sine-cosine-tangent.html
I wish I had found this channel on my firsts year on college
Never too late, I think :)
how did you find the final answers in the last question? F=1.9677 kN and theta=86.03
Apologies for the late reply. I've solved it for you here: ruclips.net/user/shortsrAlhrq5hWFc
bro is my savior
Thank you!
how did the F1x on the second example become sin. Should it be cosine?
You must look from the perspective of the angle. So opposite to the angle is the x-component, and sine = opposite over hypotenuse. Adjacent to the angle is the y-component, so that's cosine. See: ruclips.net/user/shortsvynnKlJD_Jo?feature=share
Thank you for this video...it is really useful for us
I am glad to hear that!
Thank you so much sir.. your explanation is superb 🌠🤓
You're most welcome! Best wishes with your studies :)
hi thank for your time , but how can be the angle that we get = 86.03 between F and y' and at the same time angle between x and x' = 30 ? if we add those two angles it would give us 116.03 which is more then 90 that should be between the x axis and the y axis ?
The force F actually goes past the x' axis, and is 3.97 degrees away from the x-axis. Remember that the diagram isn't always accurate, it's just a base point to start the problem. When we get new data, we have to reimagine the diagram in our heads. In this problem, we know that force F is actually 86.03 degrees away from the y-axis, which means it must go past the x-prime axis. Also, I am not sure why you are adding the 2 angles, since there is no relation between them. One will simply overlap the other 👍
I hope you notice me, how did you simplify in the part of 8:43?
So you can directly plug the cosine and sine values into your calculator. 8cos30 = 6.928 and -4sin15 = -1.035. After that, you're just adding the values together and isolating for Fsinθ.
@@QuestionSolutions My calculator shows that the answer is 5.89 and not 1.963. ☹️
@@ronsking4083 So you have 6.928 +1.035 - 6 = 1.96
on 5:10 there should be Fx = 30c0s30 and Fy=30sin30 but you have done opposite ?? are you wrong there or am i !???
Please see: ruclips.net/user/shortsvynnKlJD_Jo?si=N1D9pVhLm1IIRKt7
Which book do you follow sir
Any book used is always listed in the description below.
Can we donate to you because you teach better than our professors that we actually pay for
If you would like, and only if you want to, you can donate here: ko-fi.com/questionsolutions
I am glad these videos were helpful to you :)
How F2x=26(5/13) and F2y=26(12/13)??? I can't understand please explain
I explain it from 5:39. I am unsure how else to go deeper. In essence, we are using ratio triangles to figure it out. The corresponding side is what gives the multiplying factor to figure out the component. I think the best way to understand it is to do a few examples with that specific method. I am sorry I couldn't be of more help :(
8.56 did you use simultaneous equation to solve that?
I don't remember 😅 if the trigonometric identities are complicated, you can always graph them.
Which app you are using to show this
No app, just animated using after effects, equations written by hand and then typed using LaTeX.
Why did you use sin at f1x while cos at f1y..?
You might not be remembering trigonometric fundamentals. I highly encourage you to take a look at them and review the material. Remember that sin is opposite over hypotenuse whole cosine is adjacent over hypotenuse. You have to keep these in mind or you will have a hard time breaking forces into components 😅 Please see: www.mathsisfun.com/sine-cosine-tangent.html
How’d you solve for F and how’d you get the angle? Thanks.
Please let me know of a timestamp so I know where to look.
@@QuestionSolutions apologies. Minute 8:57. I’d just like to know what was performed with these specific numbers. Already watched the recommended videos but I would still like to know what we would do with these specific numbers in our case. Thank you!
@@eduardobribiesca1980 You can use substitution to solve for the 2 equations since it's 2 equations with 2 unknowns. You will also need to use tan = sin/cos identity. If you don't want to go through the trouble, you can always graph the two equations.
Srry i got lost how did u get F i know how u got F sin0 and F cos0 but after that i got a bit lost
Please give me a timestamp so I know where to look. Then I can explain it better.
@@QuestionSolutions srry its 8:47
srry from 8:45 i started getting lost also isnt Fx = Cos0 and Fy = Sin0 because that us wat i was thought
@@dammy3985 So if you look at 7:48, you will see the components of force F. Notice how the x-component is opposite to the angle, which means it's sine, and notice how the y-component is adjacent to the angle, so it's cosine. After we get all the components, we need to figure out what value of force F would give us a resultant force of 8 kN along the positive x' line. So in essence, we are working backwards. I encourage you to try and solve the problem again, without watching the video and then when you get stuck, look to see what the next step is. It might be more helpful. 👍
thanks so much for this
You're very welcome! Best wishes with your studies.
Thank you!!
You're very welcome!
thank you
You're very welcome!
In 4:54 your angle is supposed to be in minus, u ignored the minus sign in the "365.68"
So I think you might be a bit confused about how we draw the angles. The negative sign on the y component means the component is along the negative y-axis. That's why it was drawn downwards in the diagram. So when you find the angle, what you are finding is the angle with respect to the positive x-axis here. From FR to the positive x-axis. So an angle can't be negative. If you need further clarification, let me know.
@@QuestionSolutions So since the angle is negative we tend to ignore the negative sign, That's to say an angle can only be positive?
@@johnwealth2205 I just want to make it clear that we aren't ignoring the negative sign. This negative sign is in relation to the coordinate system we used. So we got a negative value because the y-component is facing down, and that's only because we established up and to the right to be positive. In other words, our coordinate system had y and x at the top and on the right side (this is almost always the case anyways). So when we got a negative value, it indicates that our y-component faces down, along the negative y-axis. Think of it this way, the force itself can't be negative right? Like if we had a box, and there are 2 forces applied to it, one is pulling on it from the left, and another pulling from the right. The right will be considered positive if we assume right to be positive. Does that mean the left one in real life is negative? What is a negative force in our real world? There is no such thing right? So all we are saying is, a negative force is simply a force applied in the opposite direction to what we consider is positive. Once you draw your forces, when it's up or down, based on the positive and negative signs, we need an angle to show where it is with respect to our positive x and y axes. So we use the inverse of tan to find it. The angle isn't negative because it gives us a "magnitude" value of where our resultant force is, with respect to the x-axis. Going a bit further, what is a negative angle? In degrees, it just goes in a loop, from 0 to 360, and then back around again. Does that make sense?
@@QuestionSolutions Surprisingly you're really good at explaining..... I understand better now.... The negative sign is a representation that let's us know the position a value lies in the x or y axis , It simply shows whether or not it's in the positive or negative value of x or y..... But when it comes to angle ( as it concerns the meeting point of two lines) it can't be a negative value right? I'm just trying to make sure I understand what you're saying
@@johnwealth2205 That is correct, it is indeed just a way for us to know where the force lies, in the positive or negative axis. And yes, so angles aren't negative. Like imagine someone told you to draw an angle of -60 degrees? What would that be? Would it mean a counterclockwise turn? Instead, we do it differently. We just measure the angle with respect to an axis. It doesn't have to be just positive x-axis, it could be for any but most of the time, we use the positive x-axis as the reference line. That way, we don't have any negative angles. Please see: mathmonks.com/wp-content/uploads/2021/01/Reference-Angle.jpg
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I honestly still don't get it .I feel like he goes too fast and doesn't explain why he does what he does . I am just a slow learner .😥😞😞😞😥
I am sorry to hear that. I hope you find other resources to help you out better. Don't worry, everyone learns at different paces, you just need to find your own groove. Best wishes with your studies!
@@alishamarroquin6301 you can slow down the playback speed, pause when you need to process something, and also follow along by writing the problem yourself. Pause to copy/comprehend when you need to. These are tactics I use. Something about trig has me always use the basics and it isn’t intuitive for me. I’ve been all the way through calculus, differential equations, physics, and I’m still slow when it comes to setting up sin/cos/tan problems.
It is not 365 that is 765
I don't know where you're referring to. Please use timestamps. Thanks!
On the first problem, when you are setting up your triangles, you moved Fy to the right to form your right triangle. If you moved Fx instead to form your right triangle, wouldn’t that give you Fx = 40sin30 and Fy = 40cos30?
I’m trying to figure out why you moved Fy in the 400 N force, but Fx in the 800 N force.
No, your angle would not be 30 if you move the x-component up. You'd need the other side of the angle, which would be 90 - 30 = 60. So your x-component would be 400sin60 which is the same as 400cos30.
How do you simply those functions. I am having trouble with that.
Please provide a timestamp so I know which functions you're referring to. Thanks!
so you can think of x' as a 4th force? (problem 4, at 7:15)
I guess if you want to think of it that way, there is no harm, but really, what we are trying to do is create a resultant force along the x' axis. So if we take all other other forces, we can replace it with our resultant force and it would have the same effect as all other forces effecting this bracket.