How to calculate tension in a multiple pulley system
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- เผยแพร่เมื่อ 17 ก.พ. 2014
- This engineering statics tutorial goes over how to calculate tension in a multiple pulley system that is in static equilibrium. There are 4 cables, 3 pulleys, and 2 masses in this system.
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Thanks for watching, I hope it helps!
This video was super helpful. I've been looking for a description of tension in a multiple pulley system all over the web. Thanks.
+Juliette Robinson Glad to hear it!! You can find the rest of the statics tutorials I made here: www.engineer4free.com/statics :)
hey teacher,.... i love u ....
I had trouble with pulleys and tensions, but I think this video might have given me a chance to get a passing grade on my physics exam in seven hours. Liked and subscribed. You also have a very soothing voice. I don't know why.
Thank you, watched this video to help understand a problem in the Fundamentals of Engineering exam.
Did you pass?
u r waaaaaaay amazing for a teacher
loved ur videos
Best and clearest ive seen yet. Minimal math. Great.
Dude that was some genuinely good explanation!! Thanks.
Thanks!!! =) =)
Better than my physics teacher and I'm really impressed😊💪
Thanks Mangala, means a lot!! (= (=
Man, thank you for helping me! It was concise and very understandable ☺️
Glad it helped! Thanks for watching 🙂
Explanation of the tensions in this pulley system via Newtons laws is excellent. It helped me along with other videos on mechanical advantage to understand better the motivation of definition of mechanical work
Thanks
Ni wtsp no msg chei problem photo pampista naku answer pampinchu
SIr,
This video was super helpful. I've been looking for a description of tension thanks for sharing this types of video for us please upload more & more videos
Glad you liked it!! Make sure you check out engineer4free.com/statics for the rest of my free statics videos :)
Excellent way of explaining the pulley system.!
+yogesh trakru thanks for the feedback, glad you liked it!
thank you very much, I was struggling to understand how can a pulley makes easier to lift something, if the tension on the cable was constant. well, happens that the tension in fact is constant along the cable, and yet, it can multiply the force applied. thank you for demonstrating it, so clearly and so short!
Glad I could help!! 🙂🙂
You explain that so well thank you!
You're welcome Jeremy! Thanks for watching 🙂
OMG !! It is really helpful. Thank you so much for making this video
That a great way of explaining it. Thanks!
amazingly clear!
simple and clear
very good explanation thanks!
Interesting, when reverse the mechanical advantage...in this case from question mark box to m, obtaining a quarter of initial tension for m...you obtain the mechanical dissadvantage useful for building fitness stations. Therefore, you don't have to have 20 plates (weights) or more at the beginning (100kg or more), you only need like half of that and according each cable and it's pulley system, you will have to push/pull/lift a higher amount than the set of weights available for the fitness station.
For example, if m is a set of weights with a total of 50kg and at the question mark you add another pulley with a cable, you will have to deadlift 200kg, which would be quite a good weight for a athletic guy.
THanks dude!
Thanks .This vid helped a lot :)
Thanks for letting me know!
Thanks for helping me.... U explained very well
Awesome! Thanks for letting me know =)
fantastico
really helpful
thankyou for sharing :)
thanks..... this video really helps
Very good explanation.
Thanks sir!! you really saved my butt from being eaten by my mechanical semester paper!! (ME:...tears of joy...)
Hahahaha glad I can help =)
very well explained
Awesome thanks!
Thanks it helped me a lot
Awesome glad to hear it, thanks for letting me know!
thanks so much for this!!
thank you :) this really helped
Glad to hear it Isabelle! Thanks for watching 🙂
nice work
Very helpful, thanks!
No worries!
YES! THANK YOU!
YES! YOU'RE WELCOME!
Amazing explanation 👍
Thank you 🙂
Very nice explanation .
Thanks Laraib! :0
Wow man thank you!
Hey you're welcome =)
what program are you using to draw and work this problem? i am interested in using it to work homework and other various problems
great job...
Thank you
Thank you so much
You're most welcome! There are more statics videos here too: engineer4free.com/statics =)
THANK YOU
This was awesome. New sub.
Thanks for the sub!! Make sure to check out the full Statics playlist here: th-cam.com/play/PLOAuB8dR35oeXMk2C5fjHP2K306hGfk_w.html =)
Great thanks
You're welcome!
thanks for help
No problem
gr8 vid thank u
Ur welcome!
Excellent 👍
Thank you! More videos @ engineer4free/com/statics =)
Thank you. Was watching some pulley videos and kept thinking wouldn't a free body diagram make this much easier?
Yeaahhhhhh FBDs make analysis of these pulley problems so much easier!! It's worth checking out the rest of the statics vids that I did. This one is #23 out of 72 here: engineer4free.com/statics =)
sensational
Sir u are awesome_ can u upload a video of centre of mass
sachin dubey it is such an easy concept
simply means mass of the whole object is assumed to be confined in a point called CENTRE OF MASS
awesome thanks.
Thanks for watching!
amazing
Thanks Samuel 🙂
Jeez, you're good. +1 Sub
Thanks dawg. Happy to have you around =)
Amazing,!!, super
Thanks yo!
awesome
Thanks sir
You're welcome Sameeksha! 🙂
AMAZING
Thanks!
thank you!!!!!!!
You're welcome!!!!!
Thank you for video ,my question can I increase the pully to 6 pully so I can left heavy mass? Thank you
You can, but just be aware that with the more pulleys you add, the more crucial it becomes to have your information straight. This includes things like the mass of the pulleys, the length of the rope/system, and the maximum amount of tension the rope/pulleys can safely support, as well as the maximum load your anchors (in this case, the ones attaching the rope to the ceiling) can safely bear.
thanks bro
ur welcome bro
Sir how to assume direction of acceleration in case of pulley?? To solve by d'alembert's principle?
plz solve using line equations
thanks a lot
Happy to help! Thanks for commenting :)
very profound
Thanks!
thank u
ur welcome Asal
Thanks...
You're welcome 😊
upload more videos and we are gonna watch that too.
Nice
10000000000Thanks
I’m trying to use this to calculate the mechanical disadvantage so I know how much tension the weight stack on my home gym is producing per plate per exercise
Nice!! 💪
Thanks a lot sir It was really helpful May I know where are u from?
Glad to hear it!! I am from Canada.
Super dude
Thanks Prabin!!
What if the system's not in equilibrium?Will the tension in the string joining the centre of the pulley be equal to the sum of the tensions on either side of it on the opposite end?Like in your example, 2T=T+T
The books I refer to certainly say so but do not say why.Any help is appreciated
hey bro, I think its not logical for the system to not be in equilibrium, otherwise the whole pulley would fall apart, so solving this problem as if it were out of equilibrium is just too difficult and doesn't make sense anyway, hope this helps.
great
Thanks!
It would have been nice to denote the masses as m for the known mass and m_? or m' or something else for the unknown mass. That would've helped eliminate some ambiguity. Otherwise this was really good!
True, thanks for watching!
Awesome
Thanks :)
amazing explanation, college physics courses do such a poor job of explaining these fundamental concepts, which are actually simple if they are broken down into their elementary concepts of equilibrium. Thank you for the help!
helpful
+Golu Negi glad to hear it :)
Would this method work for all pulley problems?
Can you explain why the second tension on pulley A is also directed down? When I try to visualize it, I imagine the tension on the right of pulley A to essentially be pulled up. So I am confused on why its force is down.
Anthony Kupecz tension is always directed away from body
If you have 3t as described in your diagram, why is it not included in the equations?
nice video,more examples would be better.
Hey Pranjal, thanks for watching. I got one other example on tension: th-cam.com/video/JUd7f9iBdrY/w-d-xo.html but yea more would be better!
If the blue rope connects M directly to the ceiling does it still have tension T?
If the blue rope goes over only pulley A and the connects to the floor does it have tension T?
If the blue rope goes over only pulley A and then connects back to the box of M does it still have tension T or is it now T/2 ?
Your first question: Yes
Your second question: Yes
Your third question: T/2
sir please tell how to calculate relation between velocities in pulley system
Hi sorry I don't have any videos on that yet, I expect to have some early in 2018
Directions of the tensions ?
Tension will be pulling away on both sides from any point in the rope/cable that you are observing.
Sir how to take out components...????
Provided the system is in equilibrium
How much weight would be pulling down on the ceiling
5W
Thanks god
Don't thank god, thank engineer4free :p ;)
What about if there's a bearing force on Ponit ,A, B,C??
Please answer
Hey Satish, if your problem involves bearing stress at those points, you can refresh on how to solve for it with this tutorial from my mechanics of materials series: www.engineer4free.com/4/bearing-stress
@@Engineer4Free thanks I'll go through it.
Great explanation! I try this on more complicated pulley-systems, but I always get something wrong.
I want to calculate these two systems: www11.pic-upload.de/12.04.15/fltzsqdn4c.jpg
At the left one, I got a contradiction. Because the Tension of the mass spilt into T/2 at the beginning. The free body diagram of A has then T/2 down and so need to have two times T/4 up. But C has two times T/2 up, an that is a contradiction to the rope, coming from A.
Can you help me with this problem?
Tim Pfeffer Hey this diagram bugs me because the angled cables wouldn't satisfy static equilibrium of the individual pulleys. For example, look at the weight, if there was a tension in the cable MA (which there needs to be to satisfy equilibrium on pulley A), it would apply a horizontal force of cos(angle) * tension to the weight, and it wouldn't be in static equilibrium because sum of force in x would not be zero)... Oh well. Its a bit sloppy but lets move on just pretend they are all straight up and down. Lets define some stuff for your left hand diagram: weight = mass*gravity=W. tension in cable MBC = T1. tension in cable MA=T2. tension in cable BAC=T3. We'll have 3 unknowns T1, T2, T3 that we will put in terms of W. free body diagram (fbd) of weight shows sum of forces in y = 0 so W=T1+T2. You are not allowed to assume that T1=T2=W/2. You have no grounds to do so, all you know is that one force down = sum of two forces up. So, from fbd M we get W=T1+T2. from fbd A we get T2=T3+T3=(2*T3). from fbd C we get T3=T1+T1=(2*T1). If you really want, you can find tension in the cable securing B to the ceiling and call it T4. from fbd B we get T4=T1+T1+T3. Now you just need to substitute like a mofo. W=T1+T2 ---> W=T1+(2*T3) ---> W=T1+2*(2*T1) ---> W=5*T1 ---> therefore T1=(1/5)W. Now that you know T1, you can solve for T3 because you know that T3 = T1+T1 from fbd C. T3=2*(T1) ---> T3 =2*(1/5)W --->T3=(2/5)W. Now you can finally solve for T2 from fbd A. T2=T3+T3. ---> T2=2*(2/5(W)). ---> T2=(4/5)W. For fun you can check the cable connecting it all to the ceiling and you get T4=T1+T1+T3=(4/5)W. Now you can draw the fbd for anything (the weight, pulley A, pulley, B, or pulley C) and you'll see that sum of forces in y direction will be zero for each one. I realize now I should have just drawn a picture and it would have been much clearer, but moral of the story is that you can't assume T1=T2=W/2 at the very beginning. Also don't forget to rage over the fact that the angled cables need to be drawn vertically to be technically correct. Hope that helps bruh.
@@Engineer4Free I did like your explanation actually much more better than my physics sir he explained me a lot of methods and one of them too didn't work due to overweight of complexity but it's simple to understand and find tensions for multiple pulleys when I go this way
We can trust this guy better than friendly neighborhood spiderman :-)
🕷❤️
FBD of C 4T acts in downward direction but in unknown mass block its acts in upward direction... Why??
Every point A, B, C , and the mass centre of each block has forces acting up and forces acting down. For each of A, B, C, and the mass centre of each block, the forces acting up at that point are equal and opposite to the forces acting down. This is because we are told that the system is in static equilibrium, so every part must also be in static equilibrium. Point C has tension from each side of the green cable pulling up, and tension from the pink cable pulling down. The mass centre of the unknown block has tension from the pink cable pulling up, and the force of it's weight "pulling" down. The tension force inside the pink cable is equal to the weight of the unknown mass block (which is 40N). So the pink cable pulls up on the block with a force of 40N, and also pulls down on pulley C with a force of 40N. I recommend watching this other video that I made introducing tension in cables: th-cam.com/video/nVfAKHMeUxg/w-d-xo.html and then rewatching this video. Cheers, hope that helps!
If I equate 2T=mg , => T=(mg)/2 and then come to the conclusion that the unknown mass= 4{(mg)}/2 . would it be correct??
Yeah but I don't know why you'd do that. The unknown mass is 4 times greater than the known in this prob, but T=mg, not T/2=mg
@@Engineer4Free I've no idea why I asked this question in the first place!!
What is the force acting on the body m??
F = ma , so the force is equal to its mass * acceleration due to gravity = 10 kg * 9.81m/s^2 = 98.1 kgm/s^2 = 98.1 N
can we out tension t2 on pulley c
is there is any change in answer ?
+Chetan Sharma can we take tension t2 on pulley c
is there is any change in answer ?
+Chetan Sharma If this particular set up is to remain in static equilibrium, the mass hanging off pulley C MUST be 4 times greater than the mass hanging off pulley A. T can be any value, but it depends on m. in fact, T = m*g. So if for example we picked a value for m that was twice as big, T would then just be twice as big. At first when I did the problem in terms of m and T, that was the general case, then at the end, I just picked a random value for m, to give a more tangible example. I could have easily picked m=20kg, then the unknown mass hanging off pulley C would have needed to be 80kg to satisfy static equilibrium.
How did you get 98.1 N? Sorry Im studying for a test for a new job
Alex Tuioti The force or tension is calculated by F (or T in this case) = mass x g ( acceleration due to gravity)
So T = 10kg x 9.81 = 98.1N
William Grady Thanks for helping out community members!
How can you use pulleys to ADD weight?
You could set them up in a way that a cable will pull on the bottom of an object, adding more force in the downward direction. Is that what you mean?
Engineer4Free
I was at a fitness store where the owner fixed abd sold weight lifting equipment. There was a cable machine there, and the leg attachment on this machine had about 4 pullies near the leg press. I commented on this, and the owner of the shop said that the pulley design doubles the weight of the weight stack, because legs are typically stronger than upper body pulling.
I can't really comment on it too much w/o seeing the machine in detail. However, if you imagine that we remove pulley C from the set up in this video, and replace the green cable coming off of pulley B with your legs, then your legs must exert a force of 196.2 N to hold the 98.1 N weight in place. Effectively doubling the weight that your legs see compared to the actual weight that is being held up somewhere in the machine. So they probably did design the machine on a principal similar to having your legs replace the green cable and your arms coming off the blue cable to make your legs see a greater load than say the chin up bar or whatever other things are also connected to the same weight without having to adjust the pin as often.
Just looking at that picture of pulleys lost me. You are obviously not showing how the green rope of pulley C is attached on pulley B. Is it tied to the center of pulley B where they pulley B rotates around?
In 2D statics problems, ropes will either be wrapped around the outside of a pulley, or attached to a pin in the centre. That's how you will see everyone draw them. So yeah, one end of the green rope is attached to the centre of B, pulling straight down on B. Similar to the red rope which is attached to the centre of pulley A, pulling straight up on A. It is also standard to assume in 2D statics problems that there is no friction in 2D statics problems unless clearly indicated in the problem.
at A why is there tension pulling downwards?
The blue rope has the same tension everywhere in the rope, so it more of less has the effect of pulling on anything that it is attached to, and if it's wrapped around something it pulls with each end that's coming off that thing, effectively doubling the pulling force on the object if the ropes are parallel to each other. Watch this other video I made, introducing the idea, hopefully it clears it up: engineer4free.com/4/why-is-tension-the-same-everywhere-in-a-rope