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Determine the tension in cable CAD and the angle u which the cable makes at the pulley
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- čas přidán 18. 10. 2016
- The man attempts to pull down the tree using the
cable and small pulley arrangement shown. If the tension in
AB is 60 lb, determine the tension in cable CAD and the
angle u which the cable makes at the pulley. Get the book: amzn.to/2h3hcFq
I think this man single handedly saved every engineering undergrad ever
I wished so much somewhere was there for me thats why i do this on my free time
@@FinalAnswer hey man, youre doing great and i cannot thank you enough for helping me through my courses.
Is there any way to support your work?
my guy your videos are going to be timeless your helping out a 2020 engineering student and eventually will be helping out 2030 engineering students friggin legend
It’s 2024 and i feel blessed by the gods to have come across this legend
You are a hero for the MECH 1E Group from BCIT.
I followed the video to a T and checked my math 4 times and got it wrong
On FBD adjust the coordinate system so x' is along FB. Since FB is 30 degrees above horizontal, and FAD is 20 degrees above the horizontal, you know that FAD will be below x' 10 degrees. Clarifying, FAD and FAC are just T. Then we have FAC at u above x'. (u= theta - 10) Sum forces in x, 60-Tcosu-Tcos(10)=0. sum forces in y, Tsinu-Tsin(10)=0. We find u = 10 degrees, so theta must be 20 degrees. so 60=T(cos(10)+cos(10)), T=30.46 lb. Let me know if there is a flaw in this method as I made it myself.
this helped way more than the video thanks mate
way way better than the mess in that video lol. TY!
@@skyr6762 Glad I could help. Keep on pursuing engineering, I just graduated in May and have a job already. I'm assuming you just passed Statics. Get that Fluid Mechanics book early and fuck chegg.
This makes SO much more sense. Thank you so much!
where does he get .120 from?? 7:30 - 8:30
I would recommend restricting your line width so it doesn't get fat when you press down, assuming your program allows that.
It would really help improve the legibility of your process. (thankfully you have a nice, easy to understand voice so it doesn't matter too much but it would still be good).
is there a way you can solve this by using matrices?
what will happen if I divide formula 1 over 2;thanks
How did you get .694 sqrt(3) sin(phi) from squaring both sides of the equation? 7:19 - 8:08
He adds .347+.347 to get .694 and multiplied it by square root of 3 to give him 1.2
why does he want 1.2?
Edgar Moriel He wants 1.2 to use to quadratic formula to give him theta if you see its 1.2x so in the quadratic formula b= 1.2
@@Al41_42 where did he get an extra sin(phi) from in the middle term when squared both sides of the equation. that makes no sense to me.
@@HarisRazzaq I'm not sure how he got the extra sin either but an easy way I found to do this problem is draw the Free body diagram and split theta angle in half by drawing a line from point B through point A and to the tree. Then set 20 + (theta/2) = 30, theta would then equal 20 degrees and your phi angle would be 40 degrees. Hope this helps
You are a real one my guy
Thank YOU
Thanks
still couldnt get it so idk i guess i am not understanding. i cant see where i am messing up
Thanks a lot!!
such a fat problem this one. Thank you!
How did you get sqrt(3) by dividing 52/30 6:56/11:26
If you do on the calculator sqrt of 3, and compare with 52/30, will have very close numbers, so this is an approximation to don't keep writing the same small number (1.73~3).
Thank youuu.🤗🤗
How did you get -51.96 at the very start?
60(cos 30) = 51.96.... then subtract 51.96 from both side
Make sure you write big and clear things
thanks for saving me :)
What if my numbers dont work out as nicely!? =( I cant factor them
he didn't factor? he used quadratic formula
This is very well explained but it is also very ardous and that makes it easier to make mistakes. If you shift the axis, you can figure out a different angle and a clearer way to solve the problem without too much algebra. Here is the same problem solved a bit differently. Just scroll down until you get to that solution. openstudy.com/updates/57e48807e4b0df404bff6342
why did you set the equations equal to each other?
they were both equal to T
it is one rope.
i breezed through chapter 2, could do almost any problem. then instantly stuck on chapter 3. fuck my life.
But the Unit of “Tension” is Newton
Lbs force is also a valid unit
i love you
horrible explanation 7:12 - 9:12
MVP!
Why is your starting equation
60cos(30) - Tcos(20) - Tcos(theta) = 0
instead of
60cos(30) - Tcos(20) - Tcos(20 + theta) = 0
Shouldn't the angle of the upper rope be calculated from the ground, not the other rope?
He uses phi which is 20 +theta. He defined it in the first couple minutes
@@ExoTicHOAX I know this was 3 years ago but is it okay to use theta + 20 instead of "phi" to avoid confusion??