I watched your entire FE review course and it proved to be an excellent preparation for the exam. This morning I found out I passed! Thank you for all the thought and effort you put into your content! I found your examples to be harder than the actual FE questions. Cheers!
Wow, better than any course i've paid for. The books are hard to stick with and hard to stay motivated to do it yourself. Feels like I'm back in the classroom with a teacher who cares. Thanks for the great videos and thank you for giving me hope that I can pass the FE.
Thank you for all the videos!! I watched every video of the FE CIVIL and I passed the exam! The exam questions are similar and a bit easier than the questions in these videos. If you master the questions in the videos, you will pass the fe test. Thank you so much Mark, waiting for your PE videos.
Hey Mark, graduated in Fall 2021, and i started watching these a month ago. i am proud to say that because of your videos, i am now an EIT in the great state of Arizona
Thank you so much Mark. I passed the FE exam today on my first attempt. Your videos help a lot. I watched all your review session and bought an exam sample on NCEES site.
I was having SO much trouble with #12 when solving for P. I guess I’m so used to having the surfaces be horizontal that I completely disregarded P factoring into the normal force. Once I realized that, it made a lot more sense. Thanks, Mark!
*Appreciation Post* Started studying for FE since FEB 2022. It was a bit discouraging to start with Lindeburg's FE PRACTICE book, since while attempting the question I always felt the urge to check the solution in b/w. But your videos are helping me to revive my concepts. No I'm aiming to practice watching your videos and then try solving all the questions on the books later on to get my confidence back. P.S: Good Stuff! Hoping to finish all your reviews and pass FE CIVIL! Thanks!
Taking FE in mechanical in two weeks. Your videos are helpful. I am practicing your lectures. I am not going to give up- I will eventually pass the FE. Thanks, Matt.
Thank you so much for the video! FE in 32 days and this video series is fantastic and is keeping me motivated to study for the FE along with my course load :)
Thank you, Mark Matson, Your videos were instrumental in helping me pass the FE exam. Grateful for your work and excited about this achievement. You are definitely the best!
I can't thank you enough for the amazing videos you've shared on RU-vid. As I prepare for the FE exam in the US, your teachings have been very helpful for me and your explanations are clear,. I feel more confident than ever in pursuing my engineering career. Thank you for making a difference! Now I am moving to next video.
Thank you mark for your help! I got my results and pass first try because of your videos. I made a small donation but I couldnt write a message but I wanted to say thank you for all you do for the community.
For Question 6, I simply used a single moment equilibrium equation, still using method of joints, rather than the x and y equilibrium method you showed, which I found more difficult to understand and longer to execute. I got the same answer of 25.3 from it. To explain, I selected the same joint, Joint C, and then did moment equilibrium at Point B, which cancelled out the intersecting axial force BC, and just left AC unknown. So, essentially I treated it as a method of sections problem in a way, and I got the same answer of 25.3. Just another method that's possible for those that may find it easier.
Hey Mr. Mattson, hope you are doing well! Thanks for the great videos! Was curious on the last problem. I also got 1337 for P, but if we selected 1300 wouldn’t that be less than the force required to push the block up the ramp? I was conflicted between choosing between 1300 and 1400 due to the wording, but leaned to 1400 instead. If we got into this type of scenario on the FE, is it better to round to our closest answer instead of the logical answer? Either way - thanks so much for the help!
Having taken and failed the FE twice and now working towards passing it on my third go I would say statics is probably THE most useful section to be completely and utterly familiar with, knowing how free body diagrams and sum of moments work is useful on more than just the statics section and can really help you out on quite a few others.
Hi Matt, just want to clarify the answer on Q12 should be 1400N? Since calculated value should be atleast or greater than P=1338N? Or does rounding of numbers used on this problem?😅
Hello Mark, I gave the exam 2 weeks ago. Did not pass. I will try again soon. Your videos are very helpful. I should watch your videos before the exam. I got some similar question you have covered in this class. Thanks for uploading videos
I'm glad the content is relevant. Many of the basic concepts are things that will apply regardless of the specific question. If you can really master those first principles, like the equations of equilibrium, it will take you very far. Keep practicing and learning. Best!
Hello! For question 7, shouldn't the distance from point A to the center between B and C be 10.5 instead of 9? I say 10.5 because half of 9' is 4.5', not 3' like shown in the problem. So when you add 6' and 4.5', it gives you 10.5'. Thank you all your help Matt!
Hello Mark. I have a question about question 12. I am okey all with entire solution. I found P value 1337 like you. But Why did you choose 1300 N. Arent we supposed to choose bigger than 1337 N. I thought 1337 N is breaking point. If we want to cause motion to begin up Shouldn't we pick up bigger number from 1337? Bye the way thank you so much for your videos they are very helpful. I appreciate for your help. Thanks.
I can't argue with your logic... I probably could have written the question better, but "most nearly" is still closest to the 1300 answer. That's really the reason I picked it. You're absolutely right in that (in theory), you'd have to go to 1400 to get the block to actually move. So, hopefully, you don't get any questions that are poorly written on the exam :)
For question 7, how come you dont have to split the distributed load on the incline plane into x and y components? why can you just get the distance from A to the center of the distributed load?
Hi! I looked into your question. He somewhat did do this by calculating the distance using trig. He found the perpendicular distance from the force acting at that point to the moment point. Balancing moments about a point is different from balancing the forces in their x and y components. When calculating the moment, we factor in all the forces acting at some perpendicular distance rotating about that point, I believe they all don't necessarily need to be x and y acting forces, as long as there is a calculable perpendicular distance from the force. Hope this helps!
first thanks a lot for all ur effort and time, prof. Second, for the last problem, he asked for the force that will let the body start moving upward, I get 1338 N as u did solve it, but why we do not choose any force bigger than this number which is 1400 N as 1300 N is not enough to move it up. Plz clarify, thank you.
Question on problem 12… i understand the method and how you get to your answer. But for this to work, do my assumptions on the direction of the components of force P need to be correct before entering into the system solver? On my first attempt assumed that the y component of force P goes up rather than down with respect to the plane. This yielded a force of ~1100 for P rather than the correct answer of 1337. I’m worried that I’ll make incorrect assumptions on the exam.
Hi! After looking into your question, I realized that when he switched the axis P became a force acting slightly downwards and to the right which indicates a negative Y (down) and positive X (right). I rotated my paper slightly to match his axis and I could see the way P points according to his defined axis. Hope this helps, good luck studying!
Hello Mark, love your videos! I have a question regarding question 7 (1:11:46) would the frame be stable and statically indeterminate? I got values of 3(2)+3 > 3(4)+0.
Hi Mark, Thanks for the videos. I'm taking my FE exam soon. Your Method of Joints (Part 1) video shows you using the sum of moments, but this video says not to use it. Would you please help clarify if/when to use the sum of moments with trusses?
For problem 6, why did we ignore the cable that the pulling force P is applied? when doing the summation of moments shouldn’t the length of the cable be part of the moment arms?
Hey Mark! I'm loving your videos! Thank you, for all of your content! I have a question for problem 6. I got -25.33kn indication that AC in compression. When taking the sum of forces using the method of joints, I didn't consider positive to be up/right and negative left/down. I considered any force acting towards a point(compression) as negative and positive going away(tension). Can you please clarify if I am doing this correctly? Thanks.
It sounds like your sign convention "any force acting towards a point(compression) as negative and positive going away(tension)" is correct. There are a lot of signs involved in this equation. Check that you added AC to the other side of the equation and that you had the double negative for BCx.
For Question 7: Why is it that Py and 200 are listed as separate forces in the Y direction? When solving this question before watching the video, I assumed that Py was 200lbs no?
Matt, I did it by hand as following: 1- Forces acting in the X direction: Fx= 507.8 N, Fy= N= 1,895.15, so therefore friction force (Ff) Ff=0.35x1895.15N= 663.30N Summary force acting in the X direction: Fx=507.80N & Ff=653.30N 2- Component Px= Fx+Ff= 507.80+653.30= 1161.10N 3- Problems states to find the magnitude of force 'P' (Acting Horizontally) that will cause motion. We can to calculate "P" having Px= 1161.10N as following: P= 1161.10N / Cos(15) = 1,202.06N we picked the same answer 'C' because P has to be more then 1,202.06N to be able to cause motion but we don't get the same answer of P= 1337N ; N 2241N and in this case the answer maybe should be "D" due to P has to be more then 1337N to be able to cause motion ? What am I missing ? Thank you Matt you are helping a lot !!
The normal force also has a component of P (not just the weight of the block). I think your friction force is low since I don't see Py in what you wrote above. Does that help?
@@MarkMattsonPE that is what I am missing. I see why you used the elimination method using the calculator due to we have 2 unknowns for 'N'. Thanks Matt Taking My FE very soon!
As for how 3' could be half the length of the triangle, it isn't! But the moment arm is indeed 9'. The resultant occurs at the centroid. So for the triangular distributed load (Rt), it occurs at b/3 = 9/3 = 3 (see FE Handbook 10.3 pg. 98). Then to get the moment arm, you have to make sure you are measuring the horizontal distance from A. This gives you 6 + 3 = 9. It would only occur at the halfway point for a symmetrically (often rectangularly) distributed load.
Hey Ruth! Do you mean question 4 at about 36:20? If so, I just made a guess based on the way applied forces (50 kN and 20 kN) are acting and what direction I think Ax should go. If you assume both Ax and Bx are to the right, when you solve the sum of moments equation at B, you'll get Ax is negative, meaning the arrow was shown in the wrong direction and the force really goes to the left. Does that help?
You can either use Ax and Ay or the tension in AB, but because I summed moments at A, neither causes a moment at A, because the line of action for all of those passes through A.
@@amberhall1119 I believe because, the force in the cable is the same if you analyze it at point B or at the end of the cable where it shows in the diagram. The moment distances are a ratio, and as one increases the other does as well, so mathematically they will cancel out because the force is the same throughout that section of the cable. Does that make sense? I know you posted this months ago, but at first glance I had the same question before I thought about it for a sec.
If you want to find the member force BC, you could use Bcy/Hyp or Bcx/Hyp. For this problem, I wanted to get back to sum forces in the X direction, so I used Bcx/Bcy to go from the Y-component to the X-component for the vector force. I hope that helps! These problems can stretch your mind for sure.
I may have messed you up with N for a variable and N for Newtons. The N on the left is a variable and on the right is Newtons. I just double checked it with Wolfram... see www.wolframalpha.com/input?i=systems+of+equations+calculator&assumption=%7B%22F%22%2C+%22SolveSystemOf2EquationsCalculator%22%2C+%22equation1%22%7D+-%3E%22-.35N%2B.9659P%3D508%22&assumption=%22FSelect%22+-%3E+%7B%7B%22SolveSystemOf2EquationsCalculator%22%7D%7D&assumption=%7B%22F%22%2C+%22SolveSystemOf2EquationsCalculator%22%2C+%22equation2%22%7D+-%3E%22N-.2588P%3D1895%22 for a solver of the two equations I wrote out.
