Physics - Mechanics: Applications of Newton's Second Law (1 of 20) tension on horizontal blocks
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- Опубліковано 8 вер 2024
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In this video I will show you how to calculate tension 1 and tension of the rope connecting 2 of two masses being pulled by a 10N force.
I just wanna say that you are a great teacher, you helped me and a lot of students.
What I like about these lectures is that each lecture is ,generally ,short enough for the brain to take in.
This was the BEST explanation there could ever be for "problems of connected bodies"
I am watching this from India I was struggling with these types of questions in which we needed to find the tension but you made this concept easy fr. ♥️ From 🇮🇳
Btw I have one question. In India we study these topics in grade 11 what about in Usa?
If there is something called “Nobel prize for teaching” , you would definitely win it every time
R u joking us...😂🤌🤌
This is the best video on tension! Just wished there was one with friction as well
+AdrianColumbus
You may want to look at this playlist: PHYSICS 17 TENSION AND WEIGHT
I like it when you draw pictures it's very clear and easy to visualize, SO MUCH BETTER THAN MY SCHOOL TEACHER
The equations used in my textbook and taught looks so much more complicated and could never understand it, this is way better
So helpful. It took me like a week just to try to understand this by book and only 4 mins to get it right this time. Genius
I was so surprised this worked, saved me my many tears that weren’t worth shedding.
Glad it helped. 🙂
Your videos are the only reason why I am passing my physics class at college. Thank you for posting these--you're awesome!
I love this guy. I'm writing a physics exam tomorrow and he might have just saved my life.
thank you so much for this you're really a life saver!! I'm watching this 9 years later and it still halps so much
Glad it helped! Yes, the laws of physics will not age.
Sir i really cant express how happy I am. Ur teaching method is very simple and easy to understand. Ur channel is the 1 i was searching for. Ur gr8. U made my future
You're simply the best Physics teacher on UA-cam!
This is wonderful, really good job, much easier to understand than my lecturer who seems to enjoy jumping around the class randomly shouting with a incomprehensible accent a lot more than actually teaching .:/
Your videos have really helped me better understand physics, thank you!! Awesome bow ties btw haha
Denise,
It is great to get this kind of feedback.
Thanks
Wow this is pretty incredible, I was stuck on a problem like this for like 20 minutes, then I watched this video and just went like "obviously I am supposed to calculate my own problem like this" and solved the problem in 5 seconds. xD
Thank you!
Amazing we (me and my class mates) were struggling for more than a week to understand the concept but you in less than 4 min helped me a lot thanks🙏🏽🙏🏽🙏🏽
Glad it helped! There are thousands of videos on this channel about physics (every topic).
@@MichelvanBiezen Ok I'll see it thanks
Thank you from India🇮🇳
Welcome to the channel!
angel
Think about it this way. What if we removed the rope undergoing T2 and glue the two blocks together making it one piece. Would that change anything for T1? It should not.
Michel van Biezen I will not change anything. It will remain only t which will be equal to T=F= 100N and then will not talk about m1 and m2 but M= m1+ m2.
Thank you so much
seriously you made me cry
It is so simple !!!!
u r the universal best teacher.
Wonderful, very instructive. This series is helping me a great deal to understand the applications of Newton's 2nd law.
Your videos really help me solved my questions!! Thank you so much!
We are glad the the videos are helping. Thanks for sharing.
Amazing videos and great teacher!
Should of looked these up earlier, much easier to grasp than how my teacher does it. Excellent video.
Your videos are so informative. Thank you so much for taking the time to upload these lectures!
you are the best Sir!!
I salute you
Wow nice explanation sir
Thanks and welcome
Thank you so much you are so smart!!!
You're welcome!
Thank you so much i have just given up but when i saw this video it was exactly the same thing i could not do, exactly the same! Thank you so much now i understand!😁😀
you're simply the best
Thanks a lot. You really helped me so much.
Glad to hear that! And glad you found our videos.
i'm from algeria i found your video so much importan thank you teacher
Hi Tara, Welcome to the channel.
thanks you are great keep doing what you do :)
Excellent explain, many thanks!
very good explanation
very help similar to iB style
Glad it was helpful
You're even better than my professor
Thanks a lot sir
Most welcome
Very straightforward, thank you very much
Glad it helped
Hi Michael - your videos are extremely helpful! Thank you for all your help.
For a system of 3 blocks attached by 2 strings, what if the mass of the 3 blocks are unknown? You only know the applied force is F and the first tension is 1/2F. This is a question I'm stuck on.
Here's the diagram:
[C]--[B]--[A]---> F
The tension between B and A is 1/2 and the system is accelerated by magnitude F. What is the tension between blocks B and C. This one really stumped me. Any help would be appreciated. Thanks!
There appears to be some information missing. Since the tension between A and B is (1/2)F, mass A = mass B + mass C (but you don't know how the mass is distributed between B and C. The tension between B and C will be = (mass C) * a
I think you're right. There might be some info missing. I'll ask my professor about this. Thanks!
thanks a lot I was so confused with this :)
when finding the accelration we only take the external forces acting on the system wouldnt the 100 tension be force be internal? also thank you
The 100 N force is an external force acting on the system. (Something outside the system is applying the 100 N force).
@@MichelvanBiezen thank you english is not my first langauge so i get pretty confused between internal and external
Isn't tension opposite to the motion? Or maybe because there is no friction you assume that T=F pulling the two boxes that's why the direction of tension also changes?
The direction of the tension between the 2 blocks depends on the reference point. Relative to the block on the left, the tension is to the right. Relative to the block on the right, the tension is to the left.
This is a very useful video... thank you..
Hello, thanks for the video! How would you solve a similar question but with three masses and 3 strings? Thanks!
We have examples of that. Keep watching.
@@MichelvanBiezen thank you! I was able to solve the question thanks to your videos. They are a lifesaver for sure!
If these boxes were on a surface with friction would box two be multiplied by the friction coefficient
If there was friction between the boxes and the surface then the friction forces acting on each of the boxes would be = mg x (coefficient of friction)
thank u so much sir..
Hi professor Van Biezen, I don't understand what you mean with: "the tension only applies to the mass that comes after it and not the mass that comes before it? Tension 2 of the 2 masses cancel each other out, right? And why is tension 1 equal to 100. Thanks
Tension 1 pulls on both m1 and m2, tension 2 only pulls on m2. This can be better understood if you draw free body diagrams. (one around both masses with tension 1 pulling on the box and a second one only around m2 with tension 2 pulling on the box).
Yes, I drew 2 free body diagrams, but I did it separately for box 1 and box 2 and not one around both masses. And when I drew the forces that are working on box 1, then I thought that T1 and T2 cancel each other out. So how do you know that you have to draw a free body diagram of both masses to find T1? and not separately for each box? Because I drew only a free diagram of both masses to find the acceleration. And for the calculation of the forces, I drew the free body diagram separately. (In case you don't understand what I mean: Ik teken eerst altijd een free body diagram van de 2 massa's om dan de versnelling te berekenen. Maar wanneer ik de versnelling heb gevonden teken ik de free body diagram voor elke massa afzonderlijk. En als ik dan naar de eerste massa kijk. Zie ik dan dat Fn en Fz elkaar compenseren en dat T2 gelijk is aan T1. Hoe moet ik bij deze oefening afleiden dat ik ook een free body diagram moet tekenen rond de 2 massa's om dan de krachten te kunnen vinden, want ik deed dit enkel om de versnelling te berekenen? Alvast bedankt.
What about questions with 2 tensions acting between m2 and m1
Your videos are helpful, thank you!
Can you please help me with this question:
A box is being pulled with a force of 410 kg at 49 degrees from the horizontal. If the box experiences a friction force of 85 N what is the acceleration of the box? After pulling it at 410 N for 10 seconds how fast is the box moving?
We need to know the mass of the box to find the acceleration. (did you mean to write that the box has a MASS of 410 kg?) Also, the assumption is that the box is pulled across a horizontal surface?
Michel van Biezen I meant 410 N and there's no mass in the question
In that case there may be some information missing.Since no mass is given we cannot use F = ma and find the acceleration. If we use the definition of W = change in energy = F x d we are not given the distance W = (410N x cos(49) - 85 N) x d = 184 N x d If we use impulse and momentum I = F x delta t = 184 N x 10 sec, but then momentum is equal to the change in momentum and then you'll need the mass.......
Excellent Sir. Regards
If m1 and m2 are different how come the acceleration both are the same? One force is exerted and is transmitted by tension in the strings to both masses but the accelerations produced are different! How?
Since they are connected, they act as one system and therefore every member of the system has no choice but to accelerate at the same rate.
@MichelvanBiezen First of all thank you so much for replying this fast. I asked this from my professor and aslo physics stack exchange and I've not found a satisfactory answer.
As we can see that if m2 is the second mass and it's lesser than m1 it means that the force transmitted to m2 is less than that of m1 right? Because acceleration is the same! Now the force on m2 should be equal to m2a as well because m1 and m2 are attached with a string. The string doesn't transmit the original force but a decreased one! My professor said it's because the force has to overcome m1's inertia first and that's why it drops. I still don't get it
THANK YOU!!!
Well explained, Choukran!
thx
Thanks alot
Most welcome
Thank you Sir, you helped me very much
You are most welcome
hi thank you so much for posting this video!!!
thank-you so much !!!
I didn't understand my teacher till I watch your video
The normal force on an object is not equal to its weight due to the 3rd law, because the 3rd law ties forces that are acting on different objects.
The reason N=mg is because the object is at rest on the horizontal axis and from newtons 1st law we deduce that N-mg=0.
right
Oh, so you think, that you are smarter than the teacher... Well listen here - The 3rd law says that if you act on something with a force, it will aplly the same force back on you. The 1st newtons law just says that there is no acceleration without a NET force (so the object is at rest or moving with a constant speed), it's not the reason that the other force exists thus you are wrong.
Everything you wrote is true except the conclusion.
+drorharush nah m8
Great teacher!!
i have to add that it took me three hours to do a question that had to do with exactly this. Very happy you could explain this to everybody in pedagogical manner, in a way where i could understand after watching the video from start to finnish only once.
you are special
I appreciate the comment, but I am far from "special". Just a simple man trying to navigate through life. 🙂
Hi mr , i tried to solve it by taking the sum of the forces (horizontally cuz there is no friction) of each mass and then with a system of equations solve for the acceleration of the whole system (i made it by using the large method) . but the problem is that the acceleration gave me 2.5 m/s^2. There is something else, in every exercise i do the formula always behaves like aT= sum(f)/(Number of masses * sum(m)). If there is any error, i would really like to know cuz im confused.
Your method is perfectly fine, but it is a bit more difficult and it takes more work. I prefer the F net = M total x a method as shown in the video. You may have made a small mistake somewhere.
@@MichelvanBiezen thanks for answering, i´ll check
Sir plz tell me the sequence of these videos. I want to hear all about Newton Laws,from which video i should start? There are cpations like Physics 15 Physics 5,what are these numbers?
These number represent the playlists. There are over 100 playlists on physics on this channel. Go to the homepage and you will see how they are organized. (or hit playlists)
Michel van Biezen Okay Thank you.
I wish i could listen all 100 playlists ^_^
My god thank u so much
You're welcome!
Dr can you shaw us how did you calculated the T1
show us in calculations please
+Michel van Biezen
I think that T1 should be equal to 130
Fnet=m1*a
100+T1=m1*a
T1=(6*5)+100
T1=130
am I correct?
Since an external force of 100 N is applied to the string, T1 must be equal to that external force.
Thank you so much!
Why we don y think about tension force when we calculate all the system
You can solve this problem in 2 ways, 1 ) Take the whole system as a single system and calculate the net force acting on the whole system (the tension is internal to the system). This is the method shown here. 2) Take each member of the system separately using free body diagrams and then solve the equations simultaneously to find the acceleration of the system.
thank you sir.
Can't we write it as T1+100=m(total ) a
Engine power generate a force which accelerate (and force you feel) a car because F = ma.
More power = more force pushing car forward.
Is that correct?
That is correct.
What if M2 is connected to the rope and pulled backwards with 50 newton.What will happen to the tensions and acceleration?
That is an interesting question. Since F = ma and the mass doesn't change, and now you have changed the force to half its original value, the acceleration will drop by half as well. (2.5 m/sec^2). The tension in between would now only have to accelerate the small mass. T = m1a = (6 kg ) (2.5 m/sec^2) = 15 N
What are the chances that you have/that you will show, a similar example with friction?
Btw, again, awesome videos!
I checked the playlist and I don't have an example like that with friction.
There are however a number of examples on an inclined plane and with a table and hanging weights very similar to that one you can use as examples.
Kkk thanks!
I was wondering what the tension would be if the blocks were experiencing a frictional force of 20N . Would the new tension be 70N+20N , or would it be 70N-20N ?
+Devious Minion Assuming that the total friction force for the 2 blocks combined is 20N then the net force would be 100N - 20N = 80N and the acceleration would be: a = 80N / (14kg + 6 kg). T1 would still be 100N T2 = m2*a + friction force on block 2
Thanks
+Michel van Biezen Hey, I was wondering sort of the same thing.
Block B is to the left (2kg) and block A is to the right (1kg). the system is being pulled to the right with a force of 13N. On block B there is friction R= 1N to the left.
What would the tension be on T1 and T2 and why?
I calculated it this way SumF= 13N-1N= 12N
Then a=F/Mt -> a=12N/3kg= 4m/s^s
That means T2 would be T2= Mb*a = 2kg* 4m/s^2= 8N
From what i understand from you i need to add friction to T2 but i dont really understand whats going on.
god bless you sir
T1 has to be 170N b/c T1-T2=F, T1=F+T2=100+70=170N
No, T1 = 100 N (It cannot be larger than the force pulling on the string).
How do you apply this situation to inclined planes?
There are lots of examples in the playlists on the inclined plane.
Is there a video tutorial on how to solve for connected bodies on inclined planes?
It is no different as having a single connected body. Draw a free body diagram around any of the objects in a string the and method will be exactly the same. There are a number of examples of that in the various playlists. Take a look here: PHYSICS 4.8 FREE BODY DIAGRAMS
What to do when friction is present
the force N2 is not the reaction force of m2g.
WHY DO ALL PHYSICS TEACHERS WEAR BOW TIES I SWEAR LOL
free swag
Bowties are cool
THOSE THAT CIRCLE/BALL HAVING MASS
what if the force has an angle?
than the force can be represented with 2 forces (1going up-down and one right/left) You ignore the up/down force (except if it's strong enough to lift the object up) and than you use the left-right force and ut becomes the same problem. You can devide forces in multiple ways (mathematicaly or geometrically) you can search that up - i cant draw it for you in the comments haha
T1 =30N
The results in the video are correct.
Still confused :(
Anthing in particular?
Tysm sir
You are welcome. Glad you found our videos. 🙂