Dynamics: Lesson 24 - Work and Energy Balance Hard Example
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- Опубликовано: 11 сен 2024
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PLEASE CONTINUE THE DYNAMICS VIDEOS, I HAVE AN EXAM IN 4 WEEKS!
Ibahesj Productions I have in 2 days
Alec Hex I have static’s and dynamics exam in an hour
Update: the dynamics exam crushed me, 40/100. Resit in 10 weeks. Wish me luck.
@@ibahesjproductions6009 damn bruh..
Update: took the resit and scored 58/100 which means I passed Dynamics :D
I'm taking dynamics right now. More videos please! They are so helpful!
For everyone wondering about the signs of PE for block a and b: The signs are for the work done by gravity. Not for actual PE. Block A moves down in the direction of gravity, hence the sign is positive. Block B however moves up against gravity, hence it's negative. It's confusing because PE increases when you move upwards (against gravity) normally. The signs are there to denote Work, not the actual PE increasing. It's a little tricky but he's doing it this way to keep from getting mixed up I think. Technically yes the PE of Block A goes down because it moves down, but the work DONE by the PE is positive. Hope that helps yall because I was lost af for a bit haha.
thanks bro
Sameee but before I read this comment I actually figured it out when he denoted it under the work applied area of the work energy equation!
right? for the PE values to be correct, the PE at state 1 and 2 for block b would need to be reversed. And that negative sign that suddenly appears at 13:49 ish is what saved it.
I'm kind of confused on why the potential energy "after release" is 0, when potential energy relies on height (mgh formula), knowing that block "B" gained a height of 1.73 ft in the vertical direction.
I had the exact same thought
@@bretthauser193 This is due to conservation of energy. The PE after release is zero because the lock moves. That motion is conversion from PE to KE. So that energy is conserved.
This is amazing. I bombed my first Dynamics test, and my second one is tomorrow. I feel so much more confident.
hi
Awesome ...please Dr Hanson continue with your wonderful work ..God bless you
"nothin sticks to Teflon... except pans" im weak
Thanks.
What about the work done by tension for each block?
good question! Work is done by the tensile force (T) on block B (=T*2ft) will be canceled out by the work done on Black A by the tensile force (2T); (=-2T*1ft). In other words, the work by tensile work on blocks is identical in magnitude but has a different sign. Therefore, writing work is done by tensile force does not make a difference in the final answer. This could be explained in the video to clarify but teaching is so tough and sometimes teachers forget explaining all details which is fine. Good job by the professor explaining how to solve this fairly complex problem.
@@MortezaNurcheshmeh Thanks 😊
Thank you for this video! There are so many steps involved and you do a great job explaining everything,
Why is Pe2 =0
DR. Jeff Hanson, thank you for solving this difficult Work and Energy Balance problem in classical Dynamics. I did not fully understand this problem from start to finish, however I will watch this video multiple times for a deep understanding and breakdown of this topic.
just a question why the kinetic energy before is 0...../ and why the position energy is 0 after release
Because before the blocks are released, their velocity is 0. Therefore, KE = 1/2 * m * (0)^2 = 0. The potential energy is 0 after they are released because all of their potential energy will be converted to kinetic/moving energy
thanks so much for your informative videos. in your video you mention 'your book' - could you tell us what that book is? regards
My dynamics class is passed this and has started fluids. 6 weeks to go. Do you teach a short dynamics class or something?
He prob teaches mechanics which covers both statics and dynamics. This is the second part of the class.
amazing stuff
you didnt explained why you suddenly put the negative sign on the PE1 of block B after pausing the video
I am pretty sure that since potential energy = mgh; you made a slight error. Main reason why is because you said potential energy= 60lbs*(0.1)sin(30), shouldn't it be PE=60/32.2*(32.2)(0.1)*(sin(30)) ?
Oh wait the gravities cancel out, my mistake haha.
how is the gravity canceled did he say cuz its not give in the question or did he make a mistake
Why do you not multiply by the distance for the work done by friction
ok you corrected it lol
can you quickly cover up till chapter 15? because we have done it in our university and its hard to understand it without your assistance.
Red marker out! 2:14
why did you say PE = rho (density) * g * h? 😅 shouldn't it be m*g*h? the one you mentioned is meant for pressure in liquid.
And where did you put gravity?
But in 10:38 aren’t the forces in the same direction with motion? i mean if the weights is our forces and the bokes are moving in the direction due of their weight and gravity there’s no outside force so we can detect why did we put on - for both WFa and WFb im confused 🥲
Regarding the negative sign in front of Block B's PE(1), I am wondering why we cannot put such negative sign in front of Block A's PE(1) instead? What's the reason; and when I actually put "- sign" in front of Block A's PE(1), I could not get the same result of two velocities at the end.
Bro hello, Normally the equation of work done by the mass is equal to U(w)=-mgh so we say block B goes up so it makes - since h will be positive . For example in the real life , if you want to pull something up , you have to give your energy (do work ) but while something goes down it goes itself , and release energy. There are a lot of machine creating energy thanks to this idea
But lets say if pe but negative sign in front of Block A's PE (1) , sign of the distance which we use fork work done by mass and friction will change as well , if you change them too , you will get the same answer.
2:05 rip red pen💔
i got different answer for acceleration
where come the 32.2
the value of gravity when the weight is in pound