I know this video was posted a while back but I still wanted to say THANK YOU! I had been struggling with a problem similar to this for a few days and you explained everything so easily. Thank you so so much!
Wow this video helped me out a HUGE amount! I have been searching for hours because the example my teacher did had the exact same angle for both sides so I couldn't figure out how to do it with different angles. She also skipped the substitution part which I think she should have at least noted while she was teaching so we could know. Thank you very much kind sir for posting this.
I want to thank you for your videos. It is a shame that I pay $2000 for a physics class and cant even comprehend as well as I just did watching this. Your a hero dude!!!
I love Khanacademy, however, I must say, this example puts Khan's example to shame. I am so impressed by how clear and concise you made this. Thanks a bunch!
There is a much quicker way to do this guys, using the sin rule. you need to find the adjacent angles of the wires first 90-35 = 55 90-42 = 48 make a triangle with those angles, and work out the missing angle by doing 180 minus the angles you have. 180 - (48+55) = 77 have one of the sides = 980 or whatever force value you are using and have this side either be called (a, b or c) any you want. (I usually go with c) Then use a/sin(A) = c/sin(C) (sin rule) re-arrange for the side your looking for. so lets say you want to workout T1 which is our "left" tension. you simply do 980*sin(48) (force * sin of (angle opposite to the tension) then divide by sin(77) (the missing angle/resultant force angle) all together you have 980*sin(48)/sin(77) = 747.4387353, which is a perfectly accurate value. So no substitution, no rounded numbers and no double equations.
Sorry to be so off topic but does any of you know a tool to get back into an Instagram account..? I somehow lost my password. I would love any tips you can offer me!
@Vivaan Howard I really appreciate your reply. I found the site thru google and Im waiting for the hacking stuff atm. Seems to take a while so I will get back to you later when my account password hopefully is recovered.
Thanks man. Re-starting college and haven't worked on this sort of deep problem solving in a long time. I couldn't quite figure out how to set up the formula(e). Symmetrical scenarios made sense, but I was lost on the asymmetrical problem.
Thank you this really helped me. Before receiving help from this i was going to try that method but i was not sure if that was right but now iam confident with my answer thank you
You are literally a life saver! Thank you so much for posting this video! It's super helpful and explanatory and it is going to be a great help for my impromptu physics quiz tomorrow. The only thing is, my physics teacher uses the sum of the forces in the x and y directions, which confuses me. Is there a way you can incorporate this? But thanks so much anyways this was such a good help. I could not find anything else as helpful as this video. :-)
What if there is an upward acceleration on the 100kg weight like 1.7 m/s squared upwards? Does that get multiplied directly to the tension of each cable or do you have to go back and add that 1.7 m/s to gravity so the actual force working on the object is 11.5m/s squared then take that, multiply it by the 100 kg to get weight and then start doing calculations?
LOL...I was standing on a stage that was creaking as I moved. You also hear a clicking sound as I change the stylus. I had to change the stylus in order to change color. I had to move them into and out of a tray as I was instructing. Sorry about that but I hoped the video helped!
A mass of 500 kg is suspended from a beam by two chains 1,5m and 2,7m long respectively. The distance between the suspension points is 3,746 m. Determine (a)The load in each chain (b) The vertical and horizontal reactions at the suspension points.
Hi, I am having trouble with a similar question. What if you have the same layout as above, but instead of having the weight of the suspended mass, you have the tension on one of the strings instead, such as T2. The question is to get the tension on T1 and then figure out the weight of the mass
Hi, what if the angles are very small because the cables are pretensioned to take up sag? How does that affect the tension in the cables? Say the rt angle is 1 degree and the left is 2 degree. That give me about 20 times the load in tensile force in each cable? That seems wrong. It implies that a taunt cable could easily snap while a loose cable would do fine? And what if there was some bounce on the cables? How to calculate the increased forces? I have a real life project and want to use cables to suspend a large hammock. I have seen some examples on parks. I could use some help in selecting cable size. Could i possibly contact you privately?
Hi I am struggling after 11:40 minutes. I am not quite sure how you got the answer T2 = 830.51 N. Could you please show me your working out for that answer. Many thanks.
This is probably the best explanation I've seen. No wonder you're a Doctor mate. Top effort
I know this video was posted a while back but I still wanted to say THANK YOU! I had been struggling with a problem similar to this for a few days and you explained everything so easily. Thank you so so much!
Wow this video helped me out a HUGE amount! I have been searching for hours because the example my teacher did had the exact same angle for both sides so I couldn't figure out how to do it with different angles. She also skipped the substitution part which I think she should have at least noted while she was teaching so we could know. Thank you very much kind sir for posting this.
Omg same my teacher does that and I've been searching hours to find help for our quiz tomorrow that she just told us about today -_-
@@InfinitySisters something about engineering teachers and skiping all the working.... odd
I want to thank you for your videos. It is a shame that I pay $2000 for a physics class and cant even comprehend as well as I just did watching this. Your a hero dude!!!
this vedio was probably the best what iv been searching since like hrs now !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I literally searched for 10 seconds
I love Khanacademy, however, I must say, this example puts Khan's example to shame. I am so impressed by how clear and concise you made this. Thanks a bunch!
There is a much quicker way to do this guys, using the sin rule.
you need to find the adjacent angles of the wires first
90-35 = 55
90-42 = 48
make a triangle with those angles, and work out the missing angle by doing 180 minus the angles you have.
180 - (48+55) = 77
have one of the sides = 980 or whatever force value you are using and have this side either be called (a, b or c) any you want. (I usually go with c)
Then use a/sin(A) = c/sin(C) (sin rule) re-arrange for the side your looking for.
so lets say you want to workout T1 which is our "left" tension.
you simply do 980*sin(48) (force * sin of (angle opposite to the tension)
then divide by sin(77) (the missing angle/resultant force angle)
all together you have 980*sin(48)/sin(77) = 747.4387353, which is a perfectly accurate value.
So no substitution, no rounded numbers and no double equations.
Sorry to be so off topic but does any of you know a tool to get back into an Instagram account..?
I somehow lost my password. I would love any tips you can offer me!
@Tucker Kai instablaster :)
@Vivaan Howard I really appreciate your reply. I found the site thru google and Im waiting for the hacking stuff atm.
Seems to take a while so I will get back to you later when my account password hopefully is recovered.
@Vivaan Howard it did the trick and I finally got access to my account again. I am so happy!
Thanks so much, you saved my ass!
@Tucker Kai glad I could help xD
Excellent video. You broke it down methodically where my textbook rushed to the answer.
I was confused for the last month how to solve it . Thanks to you it’s pretty easy now .
you deserve the nobel teaching prize
So clear explanation ..I will recommend your channel to others.
VERY efficient teaching! You have saved this college freshman's life!
thank you so much! I was struggling with the problem so much and forgot the crucial step of trig.... feels like that stuff never goes away!
Thank you a lot!!!!!! My final exam after 5 hours and you helped me save time. Thank you again!!
Thank you for taking the time to explain EVERYTHING. I think I actually finally understand how to set up these problems!
THANK YOU!! My professor teaches this course terribly and im so glad you can make it make sense for me.
THANK YOU! The only description of this problem that actually works!
Thank you so much! Your explanation is amazing, much better than any textbook. I think that I'm now pretty much sorted for my physics test tomorrow:D
Thanks man. Re-starting college and haven't worked on this sort of deep problem solving in a long time. I couldn't quite figure out how to set up the formula(e).
Symmetrical scenarios made sense, but I was lost on the asymmetrical problem.
Honestly, best explained lesson by far
Great Video, very detailed in everything. Best video I've came across so far! Thank you.
I've got a physics test tomorrow and I couldn't figure this out. Thanks for saving my ass! Great tutorial.
thank you! Finally learned what was going on in class through your video
This was a lifesaver. I understood it all perfectly.
best and detailed explanation so far on this topic. Many thanks to you!
Even in university, this helped a bunch
Thank you good sir
You are so flipping amazing! I finally understand these kinds of questions. THANK YOU!!!!
Thank you this really helped me. Before receiving help from this i was going to try that method but i was not sure if that was right but now iam confident with my answer thank you
Thank you this was extremely clear, thorough and most of all helpful!
thank you, great video. my professor makes no sense explaining this with all his free body diagrams.
So good i found your video, i have a test on this on monday and i need more practise
THANK YOU SO MUCH
THANK YOU SO MUCH! just spent ages trying to find some decent help
Great explanation, quite different from the method that I have learned. I'll try this, since it seems easier than the way I do it now :) cheers!
Thank you for this, I wish my physics teacher would teach things like you :)
Thank you! Thank you! A millions times, thank you!
You are literally a life saver! Thank you so much for posting this video! It's super helpful and explanatory and it is going to be a great help for my impromptu physics quiz tomorrow. The only thing is, my physics teacher uses the sum of the forces in the x and y directions, which confuses me. Is there a way you can incorporate this? But thanks so much anyways this was such a good help. I could not find anything else as helpful as this video. :-)
Am surely going to make it in physics. thanks sir for the wonderful lesson
Great Video on this subject. Thank you so much for explaining it very well
What if there is an upward acceleration on the 100kg weight like 1.7 m/s squared upwards? Does that get multiplied directly to the tension of each cable or do you have to go back and add that 1.7 m/s to gravity so the actual force working on the object is 11.5m/s squared then take that, multiply it by the 100 kg to get weight and then start doing calculations?
You're a life saver!!!
I appreciate this video, it helped me a lot. Thank you.
Thank you!! Amazing video.
LOL...I was standing on a stage that was creaking as I moved. You also hear a clicking sound as I change the stylus. I had to change the stylus in order to change color. I had to move them into and out of a tray as I was instructing. Sorry about that but I hoped the video helped!
thank you SO MUCH YOU SAVED MY LIFE ♥
Good work. Thank you for a clear explanation!
this was extremely helpful. thank you
wow dope video mahn..... I want more of your videos.
Brilliant method ❤️💡
Thanks now im ready for my Physics Test :D
great explanation understood everything.thank you
thank you, you really helped me , may god bless you
You saved me, thank you.
Praise the engineering gods.
Thanks for the great video!!
Thank you so much! Helped so much!
SO helpful! Thank you so much ☺
Thank you! Very helpful.
A mass of 500 kg is suspended from a beam by two chains 1,5m and 2,7m long respectively. The distance between the suspension points is 3,746 m. Determine (a)The load in each chain
(b) The vertical and horizontal reactions at the suspension points.
Thank you so much! Helped me a ton
Great!!! Helped so much!!!
Great explanation, Thank you
Big Thumbs up , great help
What a champ :)
Thank you sir
Well Payed Sir
Brilliant! Thank you!
you have literally saved my ass
thanks man, brilliant
You sir, rock.
Thank you so much.
THANK YOU SO MUCH!
thank you for this video
What if the other tension is pulling the object on the lower part of that imaginary horizontal component?
THX VERY MUCH!!!
you're a god
Would, or how would, the tensions change if the load was hanging from a pulley on the T1/T2 line?
What are those sounds of stretching
thank you. made it very easy to follow
god bless you sir
Thank you, sometimes you need someone better than your physics teacher to explain these things.
Thank you thank you thank you thank you
Why isnt the Ti(0.82) a negative value?
Thank you!
how do you know to calculate for t1 in terms of t2 instead of t2 in terms of t1?
You can do both
Personal preference
@SciencePi's channel Hello Sir ive one question in tension force in rope beam triangle can i send u pic for that ?
Is the adjacent always t1x?
you rock bro
THANK YOU SO MUCH
Why do you use Cos and not Sin like in your last video?? please explain. Im having trouble differentiating how and when to use it
Hi, I am having trouble with a similar question. What if you have the same layout as above, but instead of having the weight of the suspended mass, you have the tension on one of the strings instead, such as T2. The question is to get the tension on T1 and then figure out the weight of the mass
just to clarify something, he rounded his answer when finding .9 right? Because I got 827.4299
Chris Fang Mine is .90
Use a fx-991ES PLUS calcu
This is how he do it (.74)/(.82)
helps sooo much
why dont the total N add up? 980N down ....but over 1500n of force on the cables
Thank you so much :)
Hi, what if the angles are very small because the cables are pretensioned to take up sag? How does that affect the tension in the cables? Say the rt angle is 1 degree and the left is 2 degree. That give me about 20 times the load in tensile force in each cable? That seems wrong. It implies that a taunt cable could easily snap while a loose cable would do fine? And what if there was some bounce on the cables? How to calculate the increased forces? I have a real life project and want to use cables to suspend a large hammock. I have seen some examples on parks. I could use some help in selecting cable size. Could i possibly contact you privately?
thank you you beaut
Hi I am struggling after 11:40 minutes. I am not quite sure how you got the answer T2 = 830.51 N. Could you please show me your working out for that answer. Many thanks.
why do we not calculate the tension of the vertical cable?
what if it was at a constant speed ?
Hey, the sum of the opposites add up to 984.45N. A difference of 4.45N upwards. Is there a mistake cause sum of the vertical forces add up to zero?
why isn't the equation t2-t1=0 when t1 is in the negative x direction and t2 is in the positive direction? which would make it cos(35)/cos(42)