You're very welcome! Statics can be really difficult if you don't understand the beginning stuff since it keeps building on it. I hope you understood every thing and I wish you the best with your studies! :)
I am so happy to hear that! Really well done on passing and getting through the course :) Also, thank you for taking the time to write your comment, it was really nice and I am glad these videos helped!
Some are also professors, and some are past students. For professors, they have to teach a lot of theory and proofs and have a lot of cover in a very short timespan. :) I am happy to hear these videos are helpful to you, keep up the awesome work!
i'll be having my midterms tomorrow and i've been cramming. I have the attention span of a potato so this video helps a lot because it is concise and straight to the point.
I’m glad you found the video helpful. I hope you do well on your midterms. I know cramming can be stressful, but you can do it. You have the potential to ace your exams. Best wishes!
Hey man, awesome video, it was extremely useful, and I tried to solve the problem on my own after viewing it to make sure I understood it. However, I believe there is a small error at 2:23 where F(ac) and F(ad) should be 58.4 instead of 54.8.
Glad to hear it was helpful. Also please see:www.cymath.com/answer?q=0.447C%3D0.447D%2C%20(-0.447*219.5)%2B0.895C%2B0.895D%3D0 for a breakup of solving the 2 simultaneous equations. Thanks!
If the question asked us to find the total length of the spring after the mass was hung, you would add the stretch to the 2 meters. Otherwise, for this question, it was just extra information that wasn't needed. It's the same as not caring about the length of the rope that's holding onto the crate. 👍
In the third example way it involved springs, don’t we add 2 m to 0.218 and 0.327 since both the springs at OA and OB starts with an unstretched distance of 2 m?
So the question is asking for the stretch, in other words, if our original length was x, how much more did it stretch from that point. The 2 m given is unnecessary info, probably to confuse students, or for another part that shares the same question.
Thanks! In 2-D, we have a vertical y-axis, and a horizontal x-axis. In 3-D, generally the vertical axis is the z-axis and the x-axis comes towards you, while the y-axis is horizontal. You can draw it however you like, it's up to you as long as you can follow through your own coordinates, you will get the same answers.
Hi in 1st question , wouldn't the position of W will slso be along y axis because its 4m left along Y and then in Z axis but you have only put the Z coordinate in position vector
Weight is not a position vector or a force along a line. It is simply weight expressed in cartesian form. It has just one component, that's a z-component (it's straight down). The other forces are all forces along ropes and have multiple components.
@@QuestionSolutions Thank you and one more thing, that means in such problems, I should take only the rope forces in three components, while writing the coordinates of the weight , only the vertically downward component are active, like along -Y( in 2D ) and along -Z( in 3D) , most of the time?
So first, I really need you to watch this video: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html It explains forces along lines/ ropes. When you need to express a force in cartesian form that's along a line, you need to follow the steps outlined. Weight is always straight down. It will have just a z-component. In 2D, the weight is along the y-axis. @@aashu_tosh
Please show your steps when solving for 2 unknown variables. Explain how you got Fca and Fda at 4:14. Show your steps if using elimination or substitution.
I am sorry, but that's something that won't happen on these videos. Solving for 2/3 unknowns is covered in high school mathematics and should be fundamental at his stage. If you are having trouble or you need a refresh, I highly encourage you to look up the methods needed to successfully do them. There are so many different videos on this subject on TH-cam :) It's like explaining and going through the steps of the Pythagorean theorem every time it's used. It's time consuming, extra work, and adds extra minutes to a video where that's not the main topic. I also believe many students already know how to do this, so to show these steps for every single problem in every single video is tedious and not efficient. The goal of these videos isn't to show how to solve equations, but rather, how to get those equations to begin with and how to apply the new material learned to solve a problem.
Hi, please from the first question, can you explain how you got FAC and FAD? If you could show a solution as to how you got it, I’d really appreciate. I’m having a hard time understanding it. Thank you
From the third equation, you can directly solve for FAB. Once you find FAB, plug it into the FY equation (2nd equation). Now you have 2 equations with 2 unknowns. You can solve them using the substitution method, or any method you're comfortable using. Here are some examples solved that way: th-cam.com/users/shortsHe7lrJEB04U?feature=share th-cam.com/users/shorts86uENomd53U?feature=share th-cam.com/users/shorts4euH1289_Kg?feature=share th-cam.com/users/shortsrAlhrq5hWFc?feature=share Let me know if you need further clarifications. Thanks!
@@QuestionSolutions i am still confused. I just watched the videos you just recommended and i still don’t get it. I’m having a really hard time substituting.
I am not too sure how to help since it's hard to show these types of things over comments. Do you have a TA or a professor you can speak with during office hours to get help with algebra? Maybe this will help too? www.cuemath.com/algebra/substitution-method/ I'll try to do a simple example here. Let's say you have these two equations and you want to solve for x and y. 2x + y = 3 5x + 2y = 4 A way to solve it is to isolate for a variable in one of the equations. So let's isolate for "y" in the first equation. So all we did was subtract 2x from both sides and we get this: y = 3 - 2x Now we take this "y" value and plug it into the second equation. 5x + 2(3 - 2x) = 4 When we do that, notice we get an equation with just one variable. Here, that variable is "x". We can directly solve for "x" now. 5x + 6 - 4x = 4 x + 6 = 4 x = -2 Now that we figured out the "x" value, we can plug that value back into our first equation. 2x + y = 3 2(-2) + y = 3 Now we can solve for "y". y = 7 So our answers are x = -2 and y = 7. The process to solve using substitution is the same for all problems. They might seem more complicated, but the core of it is just isolating for one variable and plugging it into the other equation. @@Ijee12
So each point is with respect to the origin. In other words, we are looking for the distance to each point from the origin. Please see this video first: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
So you can solve them like you would any 3 equations with 3 unknowns. Using substitution or elimination. I think it's probably hard to see, but if you write these values with x and y instead, it might make it easier. For example, we can say F_CA = x, and F_DA = y. Then our first equation can be written as 15(4/14)+(6/14)x+(4/14)y=0, you can even convert the fractions to decimal form if it's easier. So the same equation can be written as 4.29+0.43x+0.28y=0. Writing the 3 equations like this will make it much easier to solve by isolating for one variable, plugging it into the next. Let me know if you need further help.
You can use the equilibrium equation for the z-components. It has just one variable and it's easily solvable for Fab. Pause the video at 2:16 and look at the very bottom equation. 👍 Let me know if you still need help.
So really, you are assuming a different direction for the force. That's fine, make sure to carry out the proper signs (positive/negative) until the very end, and if you get a negative answer, you know it's opposite to your assumption. 👍
Once you find FAB from the 3d equation, you can plug that value back into the other equations and solve. It's 2 equations with 2 unknowns, so use any technique you're comfortable with like elimination or substitution. Isolate for one variable and plug it back into the 2nd equation and solve. 👍
When you solve the three equations, you see that FAC won't carry a force. So you can interpret that as a cable that's there just for appearance, in other words, the two other cables carry all of the weight.
So you have 3 equations with 3 unknowns. You can use any method you learned in high school to solve them, like using elimination, substitution, etc. You can even graph them to see where they intersect. Also, you can double check your answers using wolfram alpha. :)
It depends on what you have to do. If it's a force along a line, it's a bit more tedious, but generally, you need a position vector, a unit vector, then using the unit vector, you need to express the force in cartesian form. I cover this in a different video before this, please check the description.
Actually, this is the usual method used in most North American curriculums to show 3D coordinate systems. If you're more familiar with another type, then use that, you'll arrive at the same answer.
This video is so so great!! I gotta be honest and admit that the explanation was a bit fast, simple, understandable and to the point. Thank you so much for publishing this video.
@@karellhyajervoso7979 For 99% of the problems, I use substitution. If it's more than 3 equations, I use a matrix. You can also use wolfram alpha to get a quick answer.
The same exact way that was done in the previous 2 questions. Point O is at (0i+0j+0k) and point C is at (6i+4j+12k). Then you subtract O from C as shown at 5:29. Then find the magnitude, divide each term of r_OC by the magnitude and then multiply it by the force. Please kindly see this video if you need more background: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html I explain it step by step. Thanks! :)
I am not familiar with position vectors being called lambda vectors. Please see this video: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html In it, I explain what a position vector is, so if it matches with what your professor or book says is a lambda vector, then yes, if not, then no. 👍
@@sevgipnar5261 Point A is the origin since that's where everything starts from. The weight technically starts at point A. It just pulls straight down, so there are no other components.
@@sevgipnar5261 Okay, I see what's happening. You are mixing up forces expressed in cartesian form vs position vectors. The weight, which is 196.2 N, can be expressed in cartesian form as just -196.2K. When we need to express forces in the cables, we need to write position vectors from A to D, A to C, and A to B. We do NOT need any position vectors for the weight, because it just has a z-component. I am not sure how to explain this better, but one way to think about it is to realize that the cable holding the pot is straight up, it's not going to the left or right or anywhere else. It's just straight upwards, it does NOT have any other components. Again, you are mixing up expressing forces in cartesian form with position vectors. Please kindly watch this video first: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
So once you assume F_BA=15 kN, you now have 3 equations with 3 unknowns. Then you can solve them any way you like. For example, you can isolate for 1 variable in the first equation, plug that into the 2nd equation, then isolate for the another variable and plug that into the 3rd equation. See: www.symbolab.com/solver/simultaneous-equations-calculator/-b%5Cleft(%5Cfrac%7B4%7D%7B14%7D%5Cright)%2Bc%5Cleft(%5Cfrac%7B6%7D%7B14%7D%5Cright)%2Bd%5Cleft(%5Cfrac%7B4%7D%7B14%7D%5Cright)%3D0%2C%20b%5Cleft(%5Cfrac%7B6%7D%7B14%7D%5Cright)%2Bc%5Cleft(%5Cfrac%7B4%7D%7B14%7D%5Cright)-d%5Cleft(%5Cfrac%7B6%7D%7B14%7D%5Cright)%3D0%2C%20b%5Cleft(%5Cfrac%7B12%7D%7B14%7D%5Cright)%2Bc%5Cleft(%5Cfrac%7B12%7D%7B14%7D%5Cright)%2Bd%5Cleft(%5Cfrac%7B12%7D%7B14%7D%5Cright)-w%3D0%2C%20b%3D15?or=input
FAB is a very simple solve, it's just one variable with a single equation. The other 2 can be solved the same way, plug FAB into the other equation and solve. Use substitution or elimination. Isolate for one variable, plug that into the 2nd equation.
I use substitution, but you can use any method you're comfortable with; elimination, a matrix, etc. You can even graph the equations to see where they intersect. If you're pressed for time, use wolfram alpha or symbolab.
This is very vague, could you kindly send an example problem to contact @ questionsolutions.com. I can check it out and see if I cover any topics on that. 👍
So it's just a normal equation, you're most likely overthinking it. Let me re-write it with "x" as a variable. -196.2+0.894x=0 0.894x=196.2 x=(196.2/0.894) x=219.5 I hope that helps!
The weight is hanging straight down. The vector for weight is straight down, it's not slanted in any other direction. Which means, it has just a z-component.
So you can directly solve for F_AB from the 3rd equation at 2:16. After that, plug that value into the 2nd equation and isolate it for F_AC or F_AD. Then plug the isolated value into the first equation and solve. So easiest would be to use the substitution method. Imagine you have 2 equations like this: 2x+3y=5, 3x-6y=6. You can solve these two easily right? It's the same process.
@@QuestionSolutions i was to ask if your name was Jai , there is also another person on youtube channel named 'Jai apologetics' and both of you sound the same . Pardon me if asking name was inappropriate 🙏 .
You do not need to add the 2m since we are only looking for the stretch. So you can set the original stretch to 0, and then see how much more it stretched after the weight was hung. If the question asked us to find the total length of the spring after the mass was hung, you would add the stretch to the 2 meters. Otherwise, for this question, it was just extra information that wasn't needed. It's the same as not caring about the length of the rope that's holding onto the crate.
@@bigchungus208 Please watch: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html at time 1:10 I go over how to find points of interest. It's the exact same method used everywhere.
I'm sorry, I don't solve individual problems because then I have to solve everyone's problems and I don't have the time to do it. But, please see this video: th-cam.com/video/AzZTyJtBHoE/w-d-xo.html I cover an example similar to the one you showed.
Such an underrated channel, definitely deserves more subs and views, thanks so much for explaining so well
Thank you so much for your kind comment :)
Needed a quick demonstration on how to complete problems like this, and this video was all I needed. Very concise and informative. Thanks, man!
Glad it helped! Best wishes with your studies.
I just found a great channel for teaching me physics. Thanks!
You're very welcome! Thanks for watching.
Wow wow wow, ur explanation is just incredible . So clear , so neat , it’s just perfect. You are a life saver ❤️🔥❤️🔥
Thank you so much for your kind comment. Best wishes with your studies!
Damn I love this guy the explanation is so simple and to the point
Thank you! 👍
thank you so much statics has been kicking my ass you explain this better than my professor thanks so much again
You're very welcome! Statics can be really difficult if you don't understand the beginning stuff since it keeps building on it. I hope you understood every thing and I wish you the best with your studies! :)
bro thank you soo much for yours vids, passed an exam i failed by watching ur stuff really helped a ton keep up the good work
I am so happy to hear that! Really well done on passing and getting through the course :) Also, thank you for taking the time to write your comment, it was really nice and I am glad these videos helped!
has anyone ever wondered why youtubers explain better than most of the professors😂😂
Some are also professors, and some are past students. For professors, they have to teach a lot of theory and proofs and have a lot of cover in a very short timespan. :)
I am happy to hear these videos are helpful to you, keep up the awesome work!
@@QuestionSolutions surely
Because of less salary for teachers
@@NebilGetachew-ob2gr I guess😂
They don't have these kind of animating visuals for better understanding but youtubers can edit
i'll be having my midterms tomorrow and i've been cramming. I have the attention span of a potato so this video helps a lot because it is concise and straight to the point.
I’m glad you found the video helpful. I hope you do well on your midterms. I know cramming can be stressful, but you can do it. You have the potential to ace your exams. Best wishes!
Hey man, awesome video, it was extremely useful, and I tried to solve the problem on my own after viewing it to make sure I understood it. However, I believe there is a small error at 2:23 where F(ac) and F(ad) should be 58.4 instead of 54.8.
Glad to hear it was helpful. Also please see:www.cymath.com/answer?q=0.447C%3D0.447D%2C%20(-0.447*219.5)%2B0.895C%2B0.895D%3D0 for a breakup of solving the 2 simultaneous equations. Thanks!
amazingly and very concisely explained ..
Thank you very much!
You are awesome, you are incredible, you are the saviour, you are the OG
Thank you so much for your kinds words, much appreciated :)
Brother very nice videos , tomorrow is my statics exams lets see what will happen , no practice but i will recommend this to my juniors
I wish you the absolute best on your exam! And thank you so much for the recommend. :)
for 3rd example, the spring we don't need to consider its length? for example r(OA) = -2j , r(OB)= -2i
If the question asked us to find the total length of the spring after the mass was hung, you would add the stretch to the 2 meters. Otherwise, for this question, it was just extra information that wasn't needed. It's the same as not caring about the length of the rope that's holding onto the crate. 👍
@@QuestionSolutions ohh that makes sense thank you very much !
@@yeohthianshenupm7885 You're very welcome!
In the third example way it involved springs, don’t we add 2 m to 0.218 and 0.327 since both the springs at OA and OB starts with an unstretched distance of 2 m?
So the question is asking for the stretch, in other words, if our original length was x, how much more did it stretch from that point. The 2 m given is unnecessary info, probably to confuse students, or for another part that shares the same question.
@@QuestionSolutions so, we can pretty much think of s as the change of displacement.
Wonderful explanation but I doesn't understand how to you draw the x,y and z axis. Y axis is horizontal but here it's vertical please tell.
Thanks! In 2-D, we have a vertical y-axis, and a horizontal x-axis. In 3-D, generally the vertical axis is the z-axis and the x-axis comes towards you, while the y-axis is horizontal. You can draw it however you like, it's up to you as long as you can follow through your own coordinates, you will get the same answers.
Great channel ever for statics course
Thank you very much!
At 2:51 how did you locate the points A,B,C,D ......I didn't understand....... please explain as I am unable to locate points in 3D
Please see this video first: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html where I show how to locate points.
man this channel is amazing
Thank you so much!
Hi in 1st question , wouldn't the position of W will slso be along y axis because its 4m left along Y and then in Z axis but you have only put the Z coordinate in position vector
Weight is not a position vector or a force along a line. It is simply weight expressed in cartesian form. It has just one component, that's a z-component (it's straight down). The other forces are all forces along ropes and have multiple components.
@@QuestionSolutions Thank you and one more thing, that means in such problems, I should take only the rope forces in three components, while writing the coordinates of the weight , only the vertically downward component are active, like along -Y( in 2D ) and along -Z( in 3D) , most of the time?
So first, I really need you to watch this video: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
It explains forces along lines/ ropes. When you need to express a force in cartesian form that's along a line, you need to follow the steps outlined. Weight is always straight down. It will have just a z-component. In 2D, the weight is along the y-axis.
@@aashu_tosh
Please show your steps when solving for 2 unknown variables. Explain how you got Fca and Fda at 4:14. Show your steps if using elimination or substitution.
I am sorry, but that's something that won't happen on these videos. Solving for 2/3 unknowns is covered in high school mathematics and should be fundamental at his stage. If you are having trouble or you need a refresh, I highly encourage you to look up the methods needed to successfully do them. There are so many different videos on this subject on TH-cam :) It's like explaining and going through the steps of the Pythagorean theorem every time it's used. It's time consuming, extra work, and adds extra minutes to a video where that's not the main topic. I also believe many students already know how to do this, so to show these steps for every single problem in every single video is tedious and not efficient. The goal of these videos isn't to show how to solve equations, but rather, how to get those equations to begin with and how to apply the new material learned to solve a problem.
You are 1000x better than my professor
Many thanks! Best wishes with your studies.
Thank you for these videos. You are truly amazing!
Thank you very much :)
More 3D examples please 🙏
😅
Hi, please from the first question, can you explain how you got FAC and FAD?
If you could show a solution as to how you got it, I’d really appreciate.
I’m having a hard time understanding it.
Thank you
From the third equation, you can directly solve for FAB. Once you find FAB, plug it into the FY equation (2nd equation). Now you have 2 equations with 2 unknowns. You can solve them using the substitution method, or any method you're comfortable using.
Here are some examples solved that way:
th-cam.com/users/shortsHe7lrJEB04U?feature=share
th-cam.com/users/shorts86uENomd53U?feature=share
th-cam.com/users/shorts4euH1289_Kg?feature=share
th-cam.com/users/shortsrAlhrq5hWFc?feature=share
Let me know if you need further clarifications. Thanks!
@@QuestionSolutions i am still confused.
I just watched the videos you just recommended and i still don’t get it.
I’m having a really hard time substituting.
I am not too sure how to help since it's hard to show these types of things over comments. Do you have a TA or a professor you can speak with during office hours to get help with algebra? Maybe this will help too? www.cuemath.com/algebra/substitution-method/
I'll try to do a simple example here. Let's say you have these two equations and you want to solve for x and y.
2x + y = 3
5x + 2y = 4
A way to solve it is to isolate for a variable in one of the equations. So let's isolate for "y" in the first equation. So all we did was subtract 2x from both sides and we get this:
y = 3 - 2x
Now we take this "y" value and plug it into the second equation.
5x + 2(3 - 2x) = 4
When we do that, notice we get an equation with just one variable. Here, that variable is "x". We can directly solve for "x" now.
5x + 6 - 4x = 4
x + 6 = 4
x = -2
Now that we figured out the "x" value, we can plug that value back into our first equation.
2x + y = 3
2(-2) + y = 3
Now we can solve for "y".
y = 7
So our answers are x = -2 and y = 7.
The process to solve using substitution is the same for all problems. They might seem more complicated, but the core of it is just isolating for one variable and plugging it into the other equation.
@@Ijee12
Good afternoon, sir. Can you teach me how to define the value of each point. I'm still confused with the process (2:53)
So each point is with respect to the origin. In other words, we are looking for the distance to each point from the origin. Please see this video first: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
Is point A the origin?@@QuestionSolutions
No, the origin is always where the x,y,x axes meet.@@travoltavatileni7494
How come the point B is at 4i if 12 points m away from the origin. Should it be 12i instead of 4i?
love the content, surely made things easier....
Glad to hear! Thank you very much.
Time stamp 1:24, quick question, isn't it the C and D (i components, should be -2i for c and 2i for D?
ooppsss, it was correct, now im blind with the positive and negative direction of x axis
@@janecm6136 Glad it worked out :)
@@QuestionSolutions you don’t know how much your channel helped me. For real, hands down to you Sir 😭
@@janecm6136 I am happy it helped you out but it's also because of your hard work. So keep it up and do your best :)
In the first example, why isn't there a separate tension for the cable between the pot and A
Because the weight is translated onto the rope. In other words, the force in the rope is the weight of the pot.
4:14 how do we solve the equations after making assumptions that FBA =15kN ?
So you can solve them like you would any 3 equations with 3 unknowns. Using substitution or elimination. I think it's probably hard to see, but if you write these values with x and y instead, it might make it easier. For example, we can say F_CA = x, and F_DA = y. Then our first equation can be written as 15(4/14)+(6/14)x+(4/14)y=0, you can even convert the fractions to decimal form if it's easier. So the same equation can be written as 4.29+0.43x+0.28y=0. Writing the 3 equations like this will make it much easier to solve by isolating for one variable, plugging it into the next. Let me know if you need further help.
Please how did u get
A:(0i - 4j + 0k)
B:(0i - 7j+ 6k)
C:(2i + 0j + 0k)
D:(-2i + 0j - 0k)
See the first example of this video: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
how did you get the Fab @ 2:17
You can use the equilibrium equation for the z-components. It has just one variable and it's easily solvable for Fab. Pause the video at 2:16 and look at the very bottom equation. 👍 Let me know if you still need help.
Glad you are back. Will you ever cover especial relativity?
I get a lot of so many requests, I am hoping to cover them all. I will try my best but time is always the biggest factor. Will try :)
for the second question if we assumed the position vectors traveled from point A to the plate would we get wrong answers?
So really, you are assuming a different direction for the force. That's fine, make sure to carry out the proper signs (positive/negative) until the very end, and if you get a negative answer, you know it's opposite to your assumption. 👍
How do we know what’s the direction of the force
In the first problem in points C why is the 2 positive and in the D the 2 is negative
Please see this video first, especially the first example: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
In problem 1 how did you find Fac and Fad I’m still a little unsure of how you got 54.8
Once you find FAB from the 3d equation, you can plug that value back into the other equations and solve. It's 2 equations with 2 unknowns, so use any technique you're comfortable with like elimination or substitution. Isolate for one variable and plug it back into the 2nd equation and solve. 👍
Why fac is carry no force in sample problem 2
When you solve the three equations, you see that FAC won't carry a force. So you can interpret that as a cable that's there just for appearance, in other words, the two other cables carry all of the weight.
Hi sorry, but how did you get the FoA, FoB and FoC at 5:59?
So you have 3 equations with 3 unknowns. You can use any method you learned in high school to solve them, like using elimination, substitution, etc. You can even graph them to see where they intersect. Also, you can double check your answers using wolfram alpha. :)
What are the steps? Like 1. Find location 2. Position vector ... so on..?
It depends on what you have to do. If it's a force along a line, it's a bit more tedious, but generally, you need a position vector, a unit vector, then using the unit vector, you need to express the force in cartesian form. I cover this in a different video before this, please check the description.
tbh, the way you made the FBD is so confusing. commonly used coordinates will always label the vertical line as the "y" not "z".
Actually, this is the usual method used in most North American curriculums to show 3D coordinate systems. If you're more familiar with another type, then use that, you'll arrive at the same answer.
This video is so so great!! I gotta be honest and admit that the explanation was a bit fast, simple, understandable and to the point. Thank you so much for publishing this video.
Thank you so much for your comment. I am really glad, I try to make them as concise as possible :) Best wishes with your studies!
After finding Fab (@2:17), how did you find Fac and Fad when they were both two unknowns?
You have 2 equations with 2 unknowns, use any method you learned to solve them, substitution, elimination, graph them, etc.
What method did you use?
@@karellhyajervoso7979 For 99% of the problems, I use substitution. If it's more than 3 equations, I use a matrix. You can also use wolfram alpha to get a quick answer.
How did sir get force Fac and Fad
Please see: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
can you please explain to me how you got the components for force oc
The same exact way that was done in the previous 2 questions. Point O is at (0i+0j+0k) and point C is at (6i+4j+12k). Then you subtract O from C as shown at 5:29. Then find the magnitude, divide each term of r_OC by the magnitude and then multiply it by the force. Please kindly see this video if you need more background: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html I explain it step by step. Thanks! :)
The weight is acting on the Y axis why did you put it under K
Please give a timestamp so I know where you're referring to. Thanks!
Guys to get the locations of A,B,C,D at Question 2, assume the center of the plate is the origin.
The origin is always where the x,y,z axes meet.
Are position vectors the same as lambda vector?
I am not familiar with position vectors being called lambda vectors. Please see this video: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
In it, I explain what a position vector is, so if it matches with what your professor or book says is a lambda vector, then yes, if not, then no. 👍
In 1:17 why there is only z component ? Isn't there also has j component(+4j)
No, the weight is always straight down along the z-axis, so almost always, it will just have a z-component. Also, how did you get the +4j component?
@@QuestionSolutions because its starts from point A
@@sevgipnar5261 Point A is the origin since that's where everything starts from. The weight technically starts at point A. It just pulls straight down, so there are no other components.
@@QuestionSolutions how can point A origin ?didn’t we write A(0i -4j +0k )
@@sevgipnar5261 Okay, I see what's happening. You are mixing up forces expressed in cartesian form vs position vectors. The weight, which is 196.2 N, can be expressed in cartesian form as just -196.2K. When we need to express forces in the cables, we need to write position vectors from A to D, A to C, and A to B. We do NOT need any position vectors for the weight, because it just has a z-component. I am not sure how to explain this better, but one way to think about it is to realize that the cable holding the pot is straight up, it's not going to the left or right or anywhere else. It's just straight upwards, it does NOT have any other components. Again, you are mixing up expressing forces in cartesian form with position vectors. Please kindly watch this video first: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
excellent video
Thank you!
At 1:50 where did uou get the force magnitude
The force magnitude is represented by variables F_AB, F_AC, etc. We find those later on in the question.
(@ 4:18) how did you get the W= 25.714
So once you assume F_BA=15 kN, you now have 3 equations with 3 unknowns. Then you can solve them any way you like. For example, you can isolate for 1 variable in the first equation, plug that into the 2nd equation, then isolate for the another variable and plug that into the 3rd equation. See: www.symbolab.com/solver/simultaneous-equations-calculator/-b%5Cleft(%5Cfrac%7B4%7D%7B14%7D%5Cright)%2Bc%5Cleft(%5Cfrac%7B6%7D%7B14%7D%5Cright)%2Bd%5Cleft(%5Cfrac%7B4%7D%7B14%7D%5Cright)%3D0%2C%20b%5Cleft(%5Cfrac%7B6%7D%7B14%7D%5Cright)%2Bc%5Cleft(%5Cfrac%7B4%7D%7B14%7D%5Cright)-d%5Cleft(%5Cfrac%7B6%7D%7B14%7D%5Cright)%3D0%2C%20b%5Cleft(%5Cfrac%7B12%7D%7B14%7D%5Cright)%2Bc%5Cleft(%5Cfrac%7B12%7D%7B14%7D%5Cright)%2Bd%5Cleft(%5Cfrac%7B12%7D%7B14%7D%5Cright)-w%3D0%2C%20b%3D15?or=input
how did you solve Force ab, ac, ad at 2:21 ?
FAB is a very simple solve, it's just one variable with a single equation. The other 2 can be solved the same way, plug FAB into the other equation and solve. Use substitution or elimination. Isolate for one variable, plug that into the 2nd equation.
how did you find Fca , Fda and W @4.48
I use substitution, but you can use any method you're comfortable with; elimination, a matrix, etc. You can even graph the equations to see where they intersect. If you're pressed for time, use wolfram alpha or symbolab.
Brooo i need help on solving problems that needs the use of lambda
This is very vague, could you kindly send an example problem to contact @ questionsolutions.com. I can check it out and see if I cover any topics on that. 👍
2:13 how did you get that answer?
So it's just a normal equation, you're most likely overthinking it. Let me re-write it with "x" as a variable.
-196.2+0.894x=0
0.894x=196.2
x=(196.2/0.894)
x=219.5
I hope that helps!
can u explain how u got the answer in more detail rather than skipping it plz .
Please let me know where I skipped steps and I can help you out better. Thanks!
I was wonder how u got your answer ?
You are saving me
Literally gonna use your channel to help through mechanics 1😅✊🏻
Awesome! Best wishes on your course :)
Ican understand why w components are in z axe only
The weight is hanging straight down. The vector for weight is straight down, it's not slanted in any other direction. Which means, it has just a z-component.
@@QuestionSolutions thak you bro M really understand now , that's mean the w like Thread facing down
how did you get Fad and Fac at 2:22?
So you can directly solve for F_AB from the 3rd equation at 2:16. After that, plug that value into the 2nd equation and isolate it for F_AC or F_AD. Then plug the isolated value into the first equation and solve. So easiest would be to use the substitution method. Imagine you have 2 equations like this: 2x+3y=5, 3x-6y=6. You can solve these two easily right? It's the same process.
It's so fast.
Thank you for your feedback!
Srry i got lost at 2:15
So you're solving for 1 variable. Please see: www.cymath.com/answer?q=-196.2%2B0.894F%3D0
Jai?
Sorry, I don't understand?
@@QuestionSolutions i was to ask if your name was Jai , there is also another person on youtube channel named 'Jai apologetics' and both of you sound the same . Pardon me if asking name was inappropriate 🙏 .
@@firaol2517 Oh, no, I just didn't know what you meant. I don't know who that is 😅
Thank you so much for helping usss..I wish your channel reache to millions subs
Thank you so much!
In problem 3, Soa and Sob are solved. But is that the total stretch? Do we add the given 2m of unstretched that was given in the problem?
You do not need to add the 2m since we are only looking for the stretch. So you can set the original stretch to 0, and then see how much more it stretched after the weight was hung. If the question asked us to find the total length of the spring after the mass was hung, you would add the stretch to the 2 meters. Otherwise, for this question, it was just extra information that wasn't needed. It's the same as not caring about the length of the rope that's holding onto the crate.
Does it not matter from where you take the position vector from?
I am not sure of your question, sorry! Do you think you can give me a timestamp that's referring to a specific spot?
can u pls tell us how to write three dimensional coordinates
Please see these 2 videos:
- th-cam.com/video/mz7gPpIL0Gk/w-d-xo.html
- th-cam.com/video/CCeWy1kmxMs/w-d-xo.html
thats great you mention what one needs to be able to do before tackling this. right on.
Thank you :)
How'd you get the points of interest?
At 1:24
@@bigchungus208 Please watch: th-cam.com/video/CCeWy1kmxMs/w-d-xo.html at time 1:10 I go over how to find points of interest. It's the exact same method used everywhere.
@@QuestionSolutions thank you very much appreciate the help
@@bigchungus208 You're very welcome!
Hi sir can you help me with this problem (P-3-6.23) Thanks
drive.google.com/file/d/1-mU-QprQLPxMu0YXlNB7KJkdVXpfhXC1/view?usp=drivesdk
I'm sorry, I don't solve individual problems because then I have to solve everyone's problems and I don't have the time to do it. But, please see this video: th-cam.com/video/AzZTyJtBHoE/w-d-xo.html I cover an example similar to the one you showed.
Sir thank you very much for your reply and recommendation. I highly appreciate your advise.