Statics: Lesson 58 - Internal Forces M, N, V on a Frame Problem
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- čas přidán 26. 06. 2020
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Mate you're a very good teacher. It's a relief to listen to someone who's both fun and informative. I hope you're having an excellent day today!
Even if you are not aware of it, you are the static teacher of the Istanbul Technical University
And Clemson university lol
And for engineering students at the University of Toronto lol
Concordia university montreal as well
Yıldız technical also here.
Carleton University here in Ottawa aswell. Goes to show how weak engineering education at a post-secondary is in Canada....
great stuff! intuitive, explained in understandable terms, and fun to watch. thanks sir jeff!
Awesome video! Super helpful and easy to understand when you teach it! Thank you Jeff
Professor Hanson ,once again, thank you for another powerful analysis of Internal Forces on a Statics Frame Problem. This example is lengthy, however I understand it from top to bottom.
Great Video Jeff !!
Thank you for the awesome lecture. You are a natural born teacher.
I'm going to be so lost on my final tomorrow but the one thing I will remember will be "Don't get left at Walmart or you will feel down!" Thanks!!
Saving my life, one video at a time
Perfectly explained
Shouldn't the tension in pully D affect the beam at the center of the pully because it's the connection point between the pully and the beam And consequently , the vertical tansion on D have a 90# force, 5.4' away from E ?
the moment from point E still ends up the same either way
can anyone please explain why the pulley rope was cut between A and D?
Can i take moments at A. When looking for Moment at f?
can. it comes to the same answer
Why cant use the right side?
you can, but if you calculate the right side you need to know the force of Ex and Ey, which is not the most efficient way to solve.
In the first time that you take the momentum, in the E point momentum has to be 90x7,2 your answer 7,8 wrong because -> 1,8+3+2,4=7,2
no he is right you are forgetting the 0.6 radius on the E force 7.2+0.6=7.8
8:34 how is the distance 4.8? Shouldn't the distance be in the y-direction since the 90# force is in the x-direction?
He was talking abt the force in the y-drctn for the distance (4.8) in the x-drctn so it's correct.
left at walmart lolololol
i love you jeff hansen
shouldn't the total length when counting the third moment at 3:35 be 8.4 feet not 7.8 because the radius of the pulley is 0.6 so the diameter is 1.2
no dude . distance from A to E is 7.2 so you add the radius and it is 7.2+0.6=7.8 . keep in mind that E is in the center of the pulley and that's why you add the radius and not the diameter.
When taking moment at E there is a 90(0.6) part where the cable is below the bar. Why is it 0.6 and not 5.4 or 4.8?
Because the force is acting in the x direction, this means that the distance between the force and the axis point E resides on is 0.6m
wa7sh elkon yadactara
Sorry, but the way you resolved this problem, doesn't make sense at all. It didn't help :(
can anyone please explain why the pulley rope was cut between A and D?
can anyone please explain why the pulley rope was cut between A and D?
can anyone please explain why the pulley rope was cut between A and D?
can anyone please explain why the pulley rope was cut between A and D?
can anyone please explain why the pulley rope was cut between A and D?
To find the global equilibrium, you have to start by substituting all external supports (the wall in this case) with reaction forces. After removing the wall, you'd have to express the rope connected to the wall as a tension force. The same process is also why the supports at points A and B are substituted with reaction forces as x and y components.