Introduction to Tension Video

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Introduction to Tension

Introduction to Tension I will now introduce you to the concept of tension so tension is really just the force that exist either within or applied by a string or wire when usually lifting something or pulling on something. So let’s say I had a weight here and let’s say it’s a 100 Newton’s and it’s suspended from this wire which is right here. Let’s say it’s attached to the ceiling right there. Well, we already know that the force if we’re on this planet that this weight is being pulled down by a gravity. So we already know that there’s a downward force on this weight which is a force of gravity and that equals a 100 Newton, right. But we also know that this weight isn’t accelerating. It is actually stationary and it has also has no velocity but the important thing is it’s not accelerating. So given that we know that the net force on it must be 0 by Newton’s Laws. So what is the kind of acting force? Well you don’t have to know about tension to say well the string is pulling on it, right? The string is what’s keeping the weight from falling. So the force that the string or this wire applies on this weight you can use a force of tension and that is also another way to think about it is that’s also the force that’s within the wire and that is going to exactly off set the force of gravity on this weight and that’s what keeps this point stationary and keeps it from accelerating. That’s pretty straightforward tension. It’s just you know the force of a string and just so you can conceptualize it on a guitar the more you pull on some of those higher the 10 strings that sound higher pitch. The more you pull on it, the higher the tension it actually creates a higher pitch note. So you’ve done a lot what tension lot actually when they sell wires or strings. The problem will tell you the tension that that wire and string can support which is important if you’re going to build a bridge or a swing or something. So tension is something that should be hopefully a little bit intuitive to you. So with that fairly simple example on, let’s create a slightly more complicated example. So let’s take the same weight instead of making the ceiling here let’s add two more strings. Let’s add the screen string and let’s make that green string there and it’s attached to the ceiling up here and that’s the ceiling now. Let’s see if this is the wall and let’s say there’s another string right here attached to the wall. So my question to you is what is the tension in these two strings? So let’s call this T1 and T2. Well, like the first problem this point right here, this red point stationary is not accelerating in either the left or right directions and it’s not accelerating in the up down directions. So we know that the net forces in both the x and y dimensions must be 0. My second question to you is what is going to be the offset because we know already that at this point right here there’s going to be a downward force which is the force of gravity again, the force the weight of this whole thing. We can assume that the wires have no weight for simplicity, so we know that there’s going to be a downward force here and this is the force of gravity, right the whole weight of this entire object of weight plus wire is pulling down. So what is going to be the upward force here? Well let’s look at each of the wires. This second wire T2 or we call it W2 I guess, the second wire is just pulling to the left. It has no y component, it’s not lifting up at all. So it’s just pulling to the left so all of the upward lifting, all of that is going to occur from this first wire from T1 so we know that the y component of T1, so if we say that this vector here. So we have this vector here which is T1 and we need to figure out what that is and then we have the other vector which is its y component and I’ll draw that like here. This is its y component. We can call this T1y and then of course it has x component too and this is once again, this is just breaking up a force into its component vectors like we’ve a vector force into its x and y component like we’ve been doing in the last several problems and these are just trigonometry problems, right? And so we could actually now visually see that this is T1x T1y now I forget to give you an important problem that you needed to know before solving it is that the angle that the first wire forms with the ceiling that this is 30°. So if that is 30°we also know that this is a parallel line to this so if this is 30 degrees this is also going to be 30°, right? So this angle right here is also going to be 30 degrees and we know about parallel lines and alternate, interior angles and we could have done it the other way. We could have said that if this angle is 30 degrees, this angle is 60° and these are right angles. So this is also 30 but that’s just review of Geometry that we already know. But anyway we know that this angle is 30°so what’s its y component? Well, the y component let’s see what involves the hypotenuse in the opposite side. Let me write SOH CAH TOA at the top because this is really just Trigonometry. SOH CAH TOA in blood red, so what involves the opposite in the hypotenuse? So that we know the sin 30° = T1y over the tension in the string going in this direction and so if we solve for T1y we get T1 sin 30° = T1y, right and what we want do we just say before we kind of dive into the Math, we said all of the lifting on this point, all of the lifting on that point is being done by the y component of T1, right because T2 is not doing any lifting up or down. It’s only pulling to the left, so the entire component that’s keeping this object up or keeping it from falling is the y component of this tension vector. So that has to equal the force of gravity pulling down so this has to equal the force of gravity. That has to equal this or this point so that’s a 100 Newton. I really want to heat this point home because it might be a little confusing to you. we just said this point is stationary. It’s not moving up or down. It’s not accelerating up or down and so we know that there is a downward force of 100 Newton so there must be an upward force that’s being provided by these two wires. This wire is providing no upward force so all of the upward force must be the y component or the upward component of this force vector on the first wire. So given that we can now solve for the tension in this first wire because we have T1, what is sine of 30, sin 30° in case you haven’t memorize it. Sin 30° = ½, so T1(1/2) = 100 Newton, divide both sides by ½ and you get T1= 200 Newton. So now we got to figure out what the tension in this second wire is and also there’s another clue here this point isn’t moving left or right. It’s stationary so we know that whatever the tension in this wire must be, it must be being offset by a tension or some other force in the opposite direction and that force in the opposite direction is the x component of the first wires tension so it’s this. So T2 = x component of the first wire’s tension and what’s the x component? Well, it is going to be the tension in the first wire 200 Newton x cos30°, it is adjacent over hypotenuse and that’s the √3/2 so it’s 200(√3/2)=100 √3. So tension in this wire is a 100√3 which completely offsets and it’s into the left and the x component of this wire is a 100√3 Newton’s to the right. Hopefully I didn’t confuse you. See you in the next video.

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Introduction to Tension

I will now introduce you to the concept of tension so tension is really just the force that exist either within or applied by a string or wire when usually lifting something or pulling on something. So let’s say I had a weight here and let’s say it’s a 100 Newton’s and it’s suspended from this wire which is right here. Let’s say it’s attached to the ceiling right there.

Well, we already know that the force if we’re on this planet that this weight is being pulled down by a gravity. So we already know that there’s a downward force on this weight which is a force of gravity and that equals a 100 Newton, right. But we also know that this weight isn’t accelerating. It is actually stationary and it has also has no velocity but the important thing is it’s not accelerating. So given that we know that the net force on it must be 0 by Newton’s Laws.

So what is the kind of acting force? Well you don’t have to know about tension to say well the string is pulling on it, right? The string is what’s keeping the weight from falling. So the force that the string or this wire applies on this weight you can use a force of tension and that is also another way to think about it is that’s also the force that’s within the wire and that is going to exactly off set the force of gravity on this weight and that’s what keeps this point stationary and keeps it from accelerating.

That’s pretty straightforward tension. It’s just you know the force of a string and just so you can conceptualize it on a guitar the more you pull on some of those higher the 10 strings that sound higher pitch. The more you pull on it, the higher the tension it actually creates a higher pitch note. So you’ve done a lot what tension lot actually when they sell wires or strings. The problem will tell you the tension that that wire and string can support which is important if you’re going to build a bridge or a swing or something.

So tension is something that should be hopefully a little bit intuitive to you. So with that fairly simple example on, let’s create a slightly more complicated example. So let’s take the same weight instead of making the ceiling here let’s add two more strings. Let’s add the screen string and let’s make that green string there and it’s attached to the ceiling up here and that’s the ceiling now. Let’s see if this is the wall and let’s say there’s another string right here attached to the wall.

So my question to you is what is the tension in these two strings? So let’s call this T1 and T2. Well, like the first problem this point right here, this red point stationary is not accelerating in either the left or right directions and it’s not accelerating in the up down directions. So we know that the net forces in both the x and y dimensions must be 0.

My second question to you is what is going to be the offset because we know already that at this point right here there’s going to be a downward force which is the force of gravity again, the force the weight of this whole thing. We can assume that the wires have no weight for simplicity, so we know that there’s going to be a downward force here and this is the force of gravity, right the whole weight of this entire object of weight plus wire is pulling down.

So what is going to be the upward force here? Well let’s look at each of the wires. This second wire T2 or we call it W2 I guess, the second wire is just pulling to the left. It has no y component, it’s not lifting up at all. So it’s just pulling to the left so all of the upward lifting, all of that is going to occur from this first wire from T1 so we know that the y component of T1, so if we say that this vector here.

So we have this vector here which is T1 and we need to figure out what that is and then we have the other vector which is its y component and I’ll draw that like here. This is its y component. We can call this T1y and then of course it has x component too and this is once again, this is just breaking up a force into its component vectors like we’ve a vector force into its x and y component like we’ve been doing in the last several problems and these are just trigonometry problems, right?

And so we could actually now visually see that this is T1x T1y now I forget to give you an important problem that you needed to know before solving it is that the angle that the first wire forms with the ceiling that this is 30°. So if that is 30°we also know that this is a parallel line to this so if this is 30 degrees this is also going to be 30°, right? So this angle right here is also going to be 30 degrees and we know about parallel lines and alternate, interior angles and we could have done it the other way. We could have said that if this angle is 30 degrees, this angle is 60° and these are right angles. So this is also 30 but that’s just review of Geometry that we already know.

But anyway we know that this angle is 30°so what’s its y component? Well, the y component let’s see what involves the hypotenuse in the opposite side. Let me write SOH CAH TOA at the top because this is really just Trigonometry. SOH CAH TOA in blood red, so what involves the opposite in the hypotenuse?

So that we know the sin 30° = T1y over the tension in the string going in this direction and so if we solve for T1y we get T1 sin 30° = T1y, right and what we want do we just say before we kind of dive into the Math, we said all of the lifting on this point, all of the lifting on that point is being done by the y component of T1, right because T2 is not doing any lifting up or down. It’s only pulling to the left, so the entire component that’s keeping this object up or keeping it from falling is the y component of this tension vector.

So that has to equal the force of gravity pulling down so this has to equal the force of gravity. That has to equal this or this point so that’s a 100 Newton. I really want to heat this point home because it might be a little confusing to you. we just said this point is stationary. It’s not moving up or down. It’s not accelerating up or down and so we know that there is a downward force of 100 Newton so there must be an upward force that’s being provided by these two wires. This wire is providing no upward force so all of the upward force must be the y component or the upward component of this force vector on the first wire.

So given that we can now solve for the tension in this first wire because we have T1, what is sine of 30, sin 30° in case you haven’t memorize it. Sin 30° = ½, so T1(1/2) = 100 Newton, divide both sides by ½ and you get T1= 200 Newton.

So now we got to figure out what the tension in this second wire is and also there’s another clue here this point isn’t moving left or right. It’s stationary so we know that whatever the tension in this wire must be, it must be being offset by a tension or some other force in the opposite direction and that force in the opposite direction is the x component of the first wires tension so it’s this.

So T2 = x component of the first wire’s tension and what’s the x component? Well, it is going to be the tension in the first wire 200 Newton x cos30°, it is adjacent over hypotenuse and that’s the √3/2 so it’s 200(√3/2)=100 √3. So tension in this wire is a 100√3 which completely offsets and it’s into the left and the x component of this wire is a 100√3 Newton’s to the right. Hopefully I didn’t confuse you. See you in the next video.

Transcription by: Scribe4you Transcription Services