Maaan, 1 hour ago I knew nothing about tension, I watched the first video on tension and now I`ve mannaged to solve this alone without watching the video. Yaaayy :))
I missed two classes that went over a friction and tension, and I just learned both topics in an hour! Such clear explanations, wish he was my professor.
Instead of solving for the cos and sin of whatever angle (and giving a radical answer), why not just explain this problem in simpler terms. You're unnecessarily complicating this problem by introducing square roots, and short cuts to algebra such as : multiplying both sides by two and then multiplying each side by the square root of three. Not everyone taking physics is fresh out of trigonometry, sir.
I signed in just to give this a thumbs up. Thanks so much Sal! You taught me in like 3 minutes (thank you RUclips for the ability to skip around) what I could not figure out after over 30 minutes of searching the internet. When I get into a good college thanks to you and get a lot of money, I will be sure to donate some back to Khan Academy! Haha
So I was sitting here doing physics homework, and I just couldn't get it. So I decided to look online for some help. As soon as I watched your video, something just clicked in my head and I got it! Thank you so much for everything you do, and have a great day.
My professor allowed us to make an "equation sheet" to use with our physics exam, this morning. I wrote this tension problem down, and sure enough, there was a problem exactly like this on the exam. Along with the rest of your videos, I can't thank you enough for helping me prepare; and if I wasn't broke, I'd try and donate. Maybe I can buy you a drink sometime! Thanks again Sal!
If you did the sum of the forces in the x and the sum of the forces in the y, it'd be a lot easier. x=T2 cos 30 - T1 cos 60. Then solve for T2. T2 = T1(cos 60/cos 30) y= T1 sin 60 + T2 sin 30 -10n = 0 Then plug in T2 into the second equation giving you: T1 (cos 30) + T1(cos 60/cos30)(sin 30) -10=0 Solve for T1 giving you T1=8.66N Since we already know T2 = T1(cos60/cos30), just plug in T1 making it 8.66 (or 5 square root of three) multiplied by (cos 60/cos 30). This gives 4.99N.
I'm sorry, I misunderstood the question because I rushed my answer. Even though you use the elimination method to find the point where a system of equations intersect, that's just a byproduct of what it's mainly meant to do. That is, it's a very useful way to eliminate one (or sometimes, more) variables from the equation. Because we had two variables, and we only wanted one, we used the elimination method to get us down to one. Note that we could have also used the substitution method.
Yes it is right, as T2 and T1 are not simply vertical but rather "diagonal" in a sense, as some of the tension is "wasted" horizontally. The only forces that must equal 10N are the vertical variables, which he demonstrates by setting both adding the vertical aspects and setting them equal to 10N. In my Physics class I thought the same thing. Now I have been enlightened and i don't need my senile grumpy old teacher anymore :)
In Singapore We learnt Design N Technology and will do questions like these. Thanks for helping me out for this^^LURVE U LOTS now I can let my Teacher now relax. HEY the homework involving this question, I am was the only student that bothered to tell the teacher that I will not be able to hand in the homework and no one even handed it in.
Tension is really just the cumulative effects of how matter is bonded "together". Each atom of the wire is secure in the wire itself, and the fact that I can't pull those atoms apart without sufficient force is proof that the tension exists. Of course, if I have enough force, the tension will disappear as the bonds will break, because I have applied enough force to break the bonds. So, I guess tension is just the cumulative affect of the chemical bonds. I'm not entirely sure, though.
At 8:58 you draw a box around the blue equation (sqrt 3)T1 - T2 = 0. Wouldn't it be easier to use this first to get T2 = (sqrt 3)T1, then plug this into the yellow equation above it to get T1 + (sqrt 3)(sqrt 3)T1 = 20N. That gives you T1 + 3T1 = 4T1 = 20N, so T1 = 5N. Not a critique, just a comment. Great video!
If you were to reverse the diagram and place the 10 N force on top the system would be in compression. It would be like a truss. Would the force in the diagonal member be the same, just in compression?
There is a proportion to check this problem or maybe lead to an easier way.( Never took physics) But I am taking trigonometry, so cos30/cos60= 5(rad3)/5 0.866025../0.5= 8.66025../5.0 radical 3 = radical 3
Could you have done sqrt(3)*T1 = T2, and then just plug it into the T2 for the Y vector formula? So you would have T1 + sqrt(3) * (sqrt(3)*T1) = 20? You get T1 = 5 N this way too, but I'm just wondering if your always better off using your method.
Well, that's goes into algebra when we're talking about solving a system of equations. As you may recall, it's possible to solve by either the elimination method or the substitution method. He used elimination. You and I like to use substitution. It all works out.
thats a right method but its a little confusing, y did we subtract the two equations?? i know to get T2 but how does this work? like when we subtract the two equations in math we get the point of intersection so how does it work here... finally is there is any other way to do this ?? thanx
omg! i think i get it :DDD .. Thanks bro. :) ... Your explanation of how its suppose to cancel the opposite forces and things really help my understanding.
It was a very good explanation and all, but while calculating the equation at the end, it was painful to watch. You chose a very complicated solution for such a simple equation.
Thanks for the video man but i dont really understand the part of the angles of the triangles its a little confusing. I hope you can help me out a bit.
I'm a little confused up with the rationale behind the y-component of the two tensions. I initially set the problem with a y-component, shared by both tensions, as equaled to the weight of the hanging mass, 10N. But, that's not the case here. Is it because the weight of the mass here is supported by two strings, so the tensions of their y components are divided between the two strings? I'm just thinking back to part 1 of the video where the y component of the tension of one of the strings was equaled to the weight of the mass... Hope for some clarification here. Thank you.
you're correct in thinking that it's divided between the two strings the tension just isn't divided equally because of the angles at which they hang one supports more tension in the vertical component(the steeper angle) and the other supports more tension in the horizontal component(the more shallow). Hope that helps.
Instead of subtracting the two equations wouldnt it be easier to solve for one variable in the first equation that you got and sub it in the second equation to solve for the other variable? I found that easier... but i guess its whatever feels more comfortable =|
at 5:55 how T1*Sin30 + T2*sin60 = 10 N instead of 10 * 9.8? Because W = m*g I thought it would be T1*Sin30 + T2*sin60 = 10 * 9.8 Is it because he already said it is 10N not 10 kg?
I had a problem.. what about the wire that connected the 10N block with T1 and T2? Shouldn't that also be taken into account when offsetting the force of gravity on that block?
are you talking about the weight of the wire? if so, the problem assumes it's weightless? If you are asking why we don't treat it the same as the other wires, I think it's just because we know that the wire is essentially just an extension of the weight of the block. If we created a free body diagram of that point where it intersects with the other two wires, it would look identical to the two wires interacting with the weight.
wasn't happy about the way I said it in my first comment ^^ I just want to ask why do you call it tension when there is no area involved for the dimension of tension [N/m^2]?
@ paulojunior201: first of all, you should be appreciative that this man is spending his time making videos to help others. Second, have you ever thought about the other people in the world, who don't understand certain concepts as quickly? Besides, his accomplishments in education will be better than you can ever achieve. So think before you speak you ignoramus.
This might sound really stupid, but why do we even need to consider the x-component, can't we just solve it the way we did the previous problem? I am really confused!
@rosyred158 There's a Kahn Academy app so you can have all his videos arranged on convenient playlists wherever you have internet connection. I've been using it all day to study for my physics final until my battery died haha
u can also say +T1 cos 30-T2 cos 60 = 0( addin all the x components) and equal to 0 because it is stationery ( T2 cos 60 is negative because it is in the negative x direction) also for adding the y components u have T1 sin 30 +T2 sin 60- 10= 0 (10 is negative cos it is in the negative y direction)... -10 will move to the opposite direction and become +10 when u collect like terms... correct me if am wrong
This is easy peasy. The part I'm having trouble with is a lorry pulling a car. Teacher never explained this to us. So flipping confused cuz I can't find ANY explanations anywhere
Srry, looks like angle y indeed IS 30. It looked like y in the first look but when I looked at it more deeply, it was clear. Well u still need to tell me how the net force at those two points are zero. Please its an urgent requirement.
+Anoushka Singh Well , you need to imagine a picture of that block of 10Kg attached to the ceiling with the help of two strings.Since the pulley is well in a state of equilibrium (i.e., it is not moving ) you can say that the Net Force acting from all the sides Is 0. Therefore ,after resolving the components of tension of strings on either side they are equalized to 0.
Honestly you made that way more confusing than it had to be all you need is two equations with two variables (t2)sin30=(t1)sin60 and (t2)cos30+(t1)cos60=10 from there it's simple algebra a lot Easier than what you just did
I get the way how to find the answer, but to tell you the truth, i dont understand why it gives the tension of the string/rope. So i wouldn't know what to do if there was 3 strings
Maaan, 1 hour ago I knew nothing about tension, I watched the first video on tension and now I`ve mannaged to solve this alone without watching the video. Yaaayy :))
I'm gonna fail this test tomorrow lol
lmfao howd it go man?
How’d it go
@Al-Mohsen Yahya Abbas I wouldn’t forget
You taught me more in 2 tension videos than my teacher has in 12 weeks lol..smh
I missed two classes that went over a friction and tension, and I just learned both topics in an hour! Such clear explanations, wish he was my professor.
Instead of solving for the cos and sin of whatever angle (and giving a radical answer), why not just explain this problem in simpler terms. You're unnecessarily complicating this problem by introducing square roots, and short cuts to algebra such as : multiplying both sides by two and then multiplying each side by the square root of three. Not everyone taking physics is fresh out of trigonometry, sir.
Leon Nicolas lol
Leon Nicolas remember its a older video :)
yeah :/
It's really easy for me this way. Some people WANT trigonometry because it's easier to them.
this is basic trig
I signed in just to give this a thumbs up. Thanks so much Sal! You taught me in like 3 minutes (thank you RUclips for the ability to skip around) what I could not figure out after over 30 minutes of searching the internet. When I get into a good college thanks to you and get a lot of money, I will be sure to donate some back to Khan Academy! Haha
So I was sitting here doing physics homework, and I just couldn't get it. So I decided to look online for some help. As soon as I watched your video, something just clicked in my head and I got it! Thank you so much for everything you do, and have a great day.
one of the equations can be made w.r.t one variable and then substituted in the other ...this is even simpler faster and takes less space :D
Thank you so much for everything you do for us! I finally understand forces. Once you understand it, it becomes quite easy.
I love these videos, but my problem is figuring the physics questions out in less than a minute
why are you trying to do it in less than a minute
@@woodychelton5590 that was when I was taking the MCAT. Fun times
@@cassie1790 what r u doing now
My professor allowed us to make an "equation sheet" to use with our physics exam, this morning. I wrote this tension problem down, and sure enough, there was a problem exactly like this on the exam. Along with the rest of your videos, I can't thank you enough for helping me prepare; and if I wasn't broke, I'd try and donate. Maybe I can buy you a drink sometime! Thanks again Sal!
tension was always something I got wrong but now I understand it completely thanks so much
when this was posted i was 2 years old, and now here i am watching this to help me in physics class. thank you dude
If you did the sum of the forces in the x and the sum of the forces in the y, it'd be a lot easier.
x=T2 cos 30 - T1 cos 60. Then solve for T2.
T2 = T1(cos 60/cos 30)
y= T1 sin 60 + T2 sin 30 -10n = 0
Then plug in T2 into the second equation giving you:
T1 (cos 30) + T1(cos 60/cos30)(sin 30) -10=0 Solve for T1 giving you T1=8.66N
Since we already know T2 = T1(cos60/cos30), just plug in T1 making it 8.66 (or 5 square root of three) multiplied by (cos 60/cos 30). This gives 4.99N.
damn, why this guy is so good at teaching?! Love him!
Hello
I'm sorry, I misunderstood the question because I rushed my answer.
Even though you use the elimination method to find the point where a system of equations intersect, that's just a byproduct of what it's mainly meant to do. That is, it's a very useful way to eliminate one (or sometimes, more) variables from the equation. Because we had two variables, and we only wanted one, we used the elimination method to get us down to one. Note that we could have also used the substitution method.
You are an awesome teacher. Thanks to the RUclips app on my Iphone you're always with me when I'm trying to figure out my physics homework.
Yes it is right, as T2 and T1 are not simply vertical but rather "diagonal" in a sense, as some of the tension is "wasted" horizontally.
The only forces that must equal 10N are the vertical variables, which he demonstrates by setting both adding the vertical aspects and setting them equal to 10N.
In my Physics class I thought the same thing. Now I have been enlightened and i don't need my senile grumpy old teacher anymore :)
YOU HAVE SAVED MY LIFE. well, my life. BUT my grades are my life at this point so thank you. I appreciate your videos very much
What are you doing now?
In Singapore We learnt Design N Technology and will do questions like these. Thanks for helping me out for this^^LURVE U LOTS now I can let my Teacher now relax. HEY the homework involving this question, I am was the only student that bothered to tell the teacher that I will not be able to hand in the homework and no one even handed it in.
Tension is really just the cumulative effects of how matter is bonded "together".
Each atom of the wire is secure in the wire itself, and the fact that I can't pull those atoms apart without sufficient force is proof that the tension exists.
Of course, if I have enough force, the tension will disappear as the bonds will break, because I have applied enough force to break the bonds.
So, I guess tension is just the cumulative affect of the chemical bonds.
I'm not entirely sure, though.
God bless you really worked after 12 years of my school finally got the point....thanks :)
At 8:58 you draw a box around the blue equation (sqrt 3)T1 - T2 = 0. Wouldn't it be easier to use this first to get T2 = (sqrt 3)T1, then plug this into the yellow equation above it to get T1 + (sqrt 3)(sqrt 3)T1 = 20N. That gives you T1 + 3T1 = 4T1 = 20N, so T1 = 5N.
Not a critique, just a comment. Great video!
If you were to reverse the diagram and place the 10 N force on top the system would be in compression. It would be like a truss. Would the force in the diagonal member be the same, just in compression?
There is a proportion to check this problem or maybe lead to an easier way.( Never took physics) But I am taking trigonometry, so cos30/cos60= 5(rad3)/5
0.866025../0.5= 8.66025../5.0
radical 3 = radical 3
thanks alot for the help, your an excellent teacher, easy to comprehend and good explanations
i have an A-level physics exam tomorrow morning. if i get an A, it'll be because of this dude :)
@fishysmell321 Looking at: y= T1 sin 60 + T2 sin 30 -10n = 0
Shouldn't this be y= T1 sin 30 + T2 sin 60 - 10 N = 0?
Could you have done sqrt(3)*T1 = T2, and then just plug it into the T2 for the Y vector formula? So you would have T1 + sqrt(3) * (sqrt(3)*T1) = 20? You get T1 = 5 N this way too, but I'm just wondering if your always better off using your method.
you are awesome keep up the good work
helped me out alot way better then my physics teacher!!
thank you!!!!! this is such a great tutorial :D im now ready to go to my physics class!!
Well, that's goes into algebra when we're talking about solving a system of equations. As you may recall, it's possible to solve by either the elimination method or the substitution method. He used elimination. You and I like to use substitution. It all works out.
Hi ! I am wondering what u r doing now? It would nice if u reply 😊
use the lami's theorem
What's that?
thats a right method but its a little confusing, y did we subtract the two equations?? i know to get T2 but how does this work? like when we subtract the two equations in math we get the point of intersection so how does it work here... finally is there is any other way to do this ?? thanx
How would you go about solving for T1 and T2 if the hanging weight is unknown?
Thank you khan saab,fantastic lecture as usual.
mind blown.... this really helps a lot! thank you very much!!! :D
Do you mean the equations to find T2 and T1? T2 and T1 must be equal, because they are equal forces. So subtracting one from the other must equal 0.
The point isn't accelerating in the x direction, so we must make sure the x components add to 0.
@tolsonw Yes. I didn't realize he chose the right wire as T1 and the left wire as T2.
Thanks so much
this was extremely helpful to me!!
Warm regards
thanks so much. that was so helpful
thanks!
MY god....compared to my physics teacher in school, you are so much better.
omg! i think i get it :DDD .. Thanks bro. :) ... Your explanation of how its suppose to cancel the opposite forces and things really help my understanding.
It was a very good explanation and all, but while calculating the equation at the end, it was painful to watch. You chose a very complicated solution for such a simple equation.
Thanks for the video man but i dont really understand the part of the angles of the triangles its a little confusing. I hope you can help me out a bit.
I'm a little confused up with the rationale behind the y-component of the two tensions. I initially set the problem with a y-component, shared by both tensions, as equaled to the weight of the hanging mass, 10N. But, that's not the case here. Is it because the weight of the mass here is supported by two strings, so the tensions of their y components are divided between the two strings? I'm just thinking back to part 1 of the video where the y component of the tension of one of the strings was equaled to the weight of the mass... Hope for some clarification here. Thank you.
you're correct in thinking that it's divided between the two strings the tension just isn't divided equally because of the angles at which they hang one supports more tension in the vertical component(the steeper angle) and the other supports more tension in the horizontal component(the more shallow). Hope that helps.
You really are my saviour :)
Phenominal! Thank you so much
Shouldn't the y-components of both wires be equal since they're on opposite sides of a rectangle?
thank you soooo, soooo much for this one.
thank you very much! this problem used to piss me off.......not anymore :D great videos!
keep up the good work, marvellous job this one
Thanks a lot Sir...
THANK YOU VERY MUCH!
Instead of subtracting the two equations wouldnt it be easier to solve for one variable in the first equation that you got and sub it in the second equation to solve for the other variable? I found that easier... but i guess its whatever feels more comfortable =|
At about 8:33 in the video, a value jumps from (10)(square root of 3) to (20)(square root of 3)
it became 20 because he multiplied it by 2 to remove the 1/2 from t1
Brilliant vid
u are great!
at 5:55 how T1*Sin30 + T2*sin60 = 10 N instead of 10 * 9.8? Because W = m*g
I thought it would be T1*Sin30 + T2*sin60 = 10 * 9.8
Is it because he already said it is 10N not 10 kg?
10N refers to the weight of the object, not the mass.
thanks a lot! this really helped! btw, did you use paint to do this? and did you have a tablet laptop as well?
bless up
I had a problem.. what about the wire that connected the 10N block with T1 and T2? Shouldn't that also be taken into account when offsetting the force of gravity on that block?
are you talking about the weight of the wire? if so, the problem assumes it's weightless?
If you are asking why we don't treat it the same as the other wires, I think it's just because we know that the wire is essentially just an extension of the weight of the block.
If we created a free body diagram of that point where it intersects with the other two wires, it would look identical to the two wires interacting with the weight.
very nice
@zRemify yeah but if thats 30 degrees shouldnt angle y be 30 degrees as well? because its alternate angles??
I t is amazing thank u sooooooooooooooooooo much
wasn't happy about the way I said it in my first comment ^^ I just want to ask why do you call it tension when there is no area involved for the dimension of tension [N/m^2]?
Dude, thank you so much!
THANKS!!!!
Would the tensions change if the load was on a pulley (to make them equal regardless of angles)?
Convoluted. Also, it's hard to make out the numbers that you wrote.
Shouldn't the cos(60) be negative? Since the second quadrant on the xy plane contains negative cos values.
ButtholeLazer Cos 60 is in the first quadrant
wow never new simultaneous equations would be involved
@ paulojunior201:
first of all, you should be appreciative that this man is spending his time making videos to help others. Second, have you ever thought about the other people in the world, who don't understand certain concepts as quickly?
Besides, his accomplishments in education will be better than you can ever achieve. So think before you speak you ignoramus.
This might sound really stupid, but why do we even need to consider the x-component, can't we just solve it the way we did the previous problem? I am really confused!
@rosyred158 There's a Kahn Academy app so you can have all his videos arranged on convenient playlists wherever you have internet connection. I've been using it all day to study for my physics final until my battery died haha
Repeatedly press the time bar at 5:40.
I really wish you would do it in degrees and not radians
Im gonna come back to this vid after I pass the mechanical engineer licensure exam
replace my kpark physics honors teacher? cool, thanks. :D
Are these physics videos relevant to the new MCAT?
u can also say +T1 cos 30-T2 cos 60 = 0( addin all the x components) and equal to 0 because it is stationery ( T2 cos 60 is negative because it is in the negative x direction) also for adding the y components u have T1 sin 30 +T2 sin 60- 10= 0 (10 is negative cos it is in the negative y direction)... -10 will move to the opposite direction and become +10 when u collect like terms... correct me if am wrong
This is easy peasy. The part I'm having trouble with is a lorry pulling a car. Teacher never explained this to us. So flipping confused cuz I can't find ANY explanations anywhere
THANK YOU SO MUCH. Not to be a disrespectful student but my teacher literally has no idea what she's doing
great vid, are you a college professor?
When you set both X components together, wouldnt one have to be negative? since they cancel eachother out
charly aliaga they are in opposite directions so equal
bless you
is this for a level or as level
because i used it for as level questions - but some reason it did not work
your starrrrrrrrrrrrrrrr
is there a simpiler way to do this prob....
can we solve it by using trig functions since i knw the angle ??
can tension be negative because of gravity? Please comply. :)
Srry, looks like angle y indeed IS 30. It looked like y in the first look but when I looked at it more deeply, it was clear. Well u still need to tell me how the net force at those two points are zero. Please its an urgent requirement.
+Anoushka Singh Well , you need to imagine a picture of that block of 10Kg attached to the ceiling with the help of two strings.Since the pulley is well in a state of equilibrium (i.e., it is not moving ) you can say that the Net Force acting from all the sides Is 0. Therefore ,after resolving the components of tension of strings on either side they are equalized to 0.
FUCK DUDE i got lost when he started using radicals and stuff
Honestly you made that way more confusing than it had to be all you need is two equations with two variables (t2)sin30=(t1)sin60 and (t2)cos30+(t1)cos60=10 from there it's simple algebra a lot Easier than what you just did
@paulojunior201 apparently he needs to since you haven't grasped its soh cah toa not soh cah toah...
what do u mean...how do u do tht....
how would u do that...
I get the way how to find the answer, but to tell you the truth, i dont understand why it gives the tension of the string/rope. So i wouldn't know what to do if there was 3 strings
i feel stupid, i still cant do it x(