Weight and balance FAA test questions that can be quite confusing explained using our usual simple methods
thank you a lot for this video it was really useful
Dear Sir, I just wanted to say that I passed my instrument test the other day and wanted to say that I could not have done it without your help. I scored an 88% it was not exactly easy, but then nothing is when your learning. thanks again, Now off to get commercial..
Veni Vidi Amavi
thank you sir!! I’m taking my written in 2 weeks and weight and balance is something I need to be 100% good with before I do...(:
Couldn't you do for the second example just find "Z" by doing 150 from the weight of all 3, than multiple it by 100 from the arm to give you 5,000 then subtract from the known of 3750 giving you 1250 than dividing by 50 from the distance of 50? to give you the correct answer as shown below W |A | D x50|50|2500 y50|25|1250 ------------------- 100|37.5|3750 z50|100|5000 -------------------------- 150|100|5000-3750 ---------------------- (D)3750/(W)150=25??
Remember all A/C are designed to have a nose heavy cg in flight...why...because if the A/C is stalled for any reason, the nose will drop and if you have enough altitude then you can recover from a stall and fly away. If the nose did not automatically drop, you probably would not be able to recover! If your A/C weights for example 2,683 lbs on the ground, then in flight the wings would have to produce enough lift to support 2,683 lbs plus the balance force to keep the nose up force in flight...lets say that is 127 lbs...so in flight the wings would have to produce 2,683+127=2,810 lbs of lift to maintain level flight. Your A/C will always need to produce more lift than the A/C weighs to fly...that is the cost of the safety feature which would allow the A/C to recover from a stall. To minimize this effect...try to load you A/C to the Aft limits of your Wt & Balance envelope?
It was so good vedio tthanks please update more different size bars on both the side for example 4 inch left side and 3 inch in in right side which is supported in the middle
Bruce The Curmudgeon
These are HANDS DOWN the best teaching tools for aviation on the internet! I took another class and the tutorials are useless compared to these high quality clear and understandable videos!!! THANK YOU!!!!!!!!!!!!!!!!!!
Nnative English Tteacher
You forgot to add the arm's values(50 +25=75 0r 7500, that may confuse your students!!!
In the last problem what does the 50inches above the weight 500lbs represent?
So only the side on the right is now 200lb because the fulcrum was moved? Wouldn't the left side of the plank still have some weight take away from the 200lb on the right?
Hi - your demo for PC (VFR) link does not seem to work Regards, Philippe
How to solve for finding the fulcrum
Awesome explanations really in much simpler way. thanku. God bless u
how come the 250 weight can be 3 inches away from de fulcrum and balance the plank? wouldnt this weight need to be twice the distance between the 500 weight and the fulcrum??
Hello! Quick question :) if i were to put another "y" weight with it's center of mass exactly on the balance point of the fulcrum would it affect the center of mass of the assembly?
for some reason its more clear to understand the video than the Gelim book. Both great resources though.
Real useful simple & clear..thanks
This is exactly what I needed! Simple and clear explanations. Thank you.
Getting my PPL for helicopters and going to solo here pretty soon, and then take the test. So these questions were REMOVED from the Private Pilot test I'm going to take? It IS good info to know, but if I don't NEED to stress about it, then all the better...