Introduction to tension
An introduction to tension. Solving for the tension(s) in a set of wires when a weight is hanging from them. Created by Sal Khan.
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- If the block is not moving, and the forces are balanced, why doesn't the tension of both the ropes equal 100N? How can there be a tension of 200N in one of the ropes, if only 100N is pulling down on it?(66 votes)
- Imagine holding a something heavy (a physics textbook) in your hand with your arm straight down, by your side. Then raise your arm, still completely straight, up so it is at at an angle with you body. You will notice that it is much more difficult to hold the book in that position than if you let your arm hang straight down. It is the same principle.(139 votes)
- ls the force of gravity always equal the the weight of the object?(8 votes)
- Yes FirstLuminary, That happends to be the precise definition of weight. You see, mass is how much matter an object has, while weight on the other hand, is the extent to which gravity is pulling down on that matter.
Have a good day and study hard.(43 votes)
- the tension in the string at the beginning was 100 N. But when Sal added the two other strings to the diagram, what is the tension in that original string connected directly to the weight? for some reason this question is getting to me lol...(10 votes)
- It remains 100N. The weight is pulling on the string with a force of 100N, the string is pulling on the wires with a force of 100N.(9 votes)
- why is Ti equal to 100N(1 vote)
- The forces need to balance out somehow, because there is no acceleration in the system, implying a lack of net force. This means that the weight of the block needs to be balanced with the other vertical component, the vertical component of tension. This means that the two forces need to be equal and opposite, so they need to have the same magnitude, 100N.(21 votes)
- At2:32he says he's adding 2 more strings and then considers the tension in only those 2 strings but what about the first string?(6 votes)
- The tension in the original string must equal that of the two new strings because they are connected at the ends, the forces can then be projected onto the x and y axis.
where T0 is the original light blue string(5 votes)
- When we've attached the two new wires, does the original blue wire not provide any force to support the block anymore? I mean is it now just extending the downward force of the block to the two new wires and otherwise just hanging around? And if this is the case could you please explain why? I can set up and do a problem like this but I just don't understand why that original wire no longer supports the weight anymore in any way. Thanks.(4 votes)
- The original blue wire will still provide the same amount of force to the block, and it still has the same amount of tension on it, but is still transferring that force to the block.
To put it another way, if you were to cut the blue wire, the block would fall because you removed the force that the blue wire supplies to the block.(4 votes)
- Is the T1x = T2 an application of Newton's 3rd law? some thing liddat?(2 votes)
- No, this is an application of Newtons first law. Since there is no acceleration in the horizontal plane, we assume that those forces are equal in magnitude and opposite in direction(5 votes)
- What if T1 and T2 were symmetrical? (for example, the 100N object is in the middle, and the T1 and T2 wires form a symmetrical V)(3 votes)
- Then the horizontal components of T1 and T2 would cancel each other out, and the vertical components of T1+T2 would equal 100N.(4 votes)
- Why do we need to make the x and the y component of T1 wire?(4 votes)
- The fundamental reason behind this is that two mutually perpendicular vectors do not affect one another.Now, any vector can be resolved into two mutually perpendicular vectors, right? In this case T₁ᵪ and T₁ᵧ are those components.Now since T₁ᵪ is perpendicular to the 100N force acting downwards, it will not balance it, it will not add to it; it will not affect it in anyway.But the T₁ᵧ vector is parallel to the 100N force.Now, this T₁ᵧ MUST BE balancing the 100N downward force as the point, as Sal mentions, is not accelerating in any direction.So the 100N downward force is actually balanced by this T₁ᵧ=100N upwards force.
This is why we first MUST resolve the T₁ vector into two components, one perpendicular to the 100N force and one parallel to it.Hope I was clear enough :)(2 votes)
- At8:44he says 'sine of 30 degrees is one half' what does he mean by that? That kinda made me confused :((2 votes)
- What he is referring to is the trigonometric function sine. If you are confused, I suggest watching the videos on trig in one of the math playlists.(3 votes)
I will now introduce you to the concept of tension. So tension is really just the force that exists either within or applied by a string or wire. It's usually lifting something or pulling on something. So let's say I had a weight. Let's say I have a weight here. And let's say it's 100 Newtons. 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 pull down by gravity. So we already know that there's a downward force on this weight, which is a force of gravity. And that equals 100 Newtons. But we also know that this weight isn't accelerating, it's actually stationary. It also has no velocity. But the important thing is it's not accelerating. But given that, we know that the net force on it must be 0 by Newton's laws. So what is the counteracting force? You didn't have to know about tension to say well, the string's pulling on it. 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 view as the force of tension. Another way to think about it is that's also the force that's within the wire. And that is going to exactly offset the force of gravity on this weight. And that's what keeps this point right here stationery and keeps it from accelerating. That's pretty straightforward. Tension, it's just the force of a string. And just so you can conceptualize it, on a guitar, the more you pull on some of those higher-- what was it? The really thin strings that sound higher pitched. The more you pull on it, the higher the tension. It actually creates a higher pitched note. So you've dealt with tension a lot. I think actually when they sell wires or strings they'll probably tell you the tension that that wire or 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 let's, with that fairly simple example done, 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 this green string. Green string there. And it's attached to the ceiling up here. That's the ceiling now. And let's see. 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, is stationary. It's not accelerating in either the left/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 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, 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 could call it w2, I guess. The second wire is just pulling to the left. It has no y components. It's not lifting up at all. So it's just pulling to the left. So all of the upward lifting, all of that's going to occur from this first wire, from T1. So we know that the y component of T1, so let's call-- so if we say that this vector here. Let me do it in a different color. Because I know when I draw these diagrams it starts to get confusing. Let me actually use the line tool. So I have this. Let me make a thicker line. So we have this vector here, which is T1. And we would 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 could call this T1 sub y. And then of course, it has an x component too, and I'll do that in-- let's see. I'll do that in red. Once again, this is just breaking up a force into its component vectors like we've-- a vector force into its x and y components like we've been doing in the last several problems. And these are just trigonometry problems, right? We could actually now, visually see that this is T sub 1 x and this is T sub 1 sub y. Oh, and I forgot to give you an important property of this problem that you needed to know before solving it. Is that the angle that the first wire forms with the ceiling, this is 30 degrees. So if that is 30 degrees, we also know that this is a parallel line to this. So if this is 30 degrees, this is also going to be 30 degrees. So this angle right here is also going to be 30 degrees. And that's from our-- you know, we know about parallel lines and alternate interior angles. We could have done it the other way. We could have said that if this angle is 30 degrees, this angle is 60 degrees. This is a right angle, so this is also 30. But that's just review of geometry that you already know. But anyway, we know that this angle is 30 degrees, so what's its y component? Well the y component, let's see. What involves the hypotenuse and 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 and the hypotenuse? So opposite over hypotenuse. So that we know the sine-- let me switch to the sine of 30 degrees is equal to T1 sub y over the tension in the string going in this direction. So if we solve for T1 sub y we get T1 sine of 30 degrees is equal to T1 sub y. And what did we just say before we kind of dived into the math? We said all of the lifting on this point is being done by the y component of T1. 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, keeping it from falling is the y component of this tension vector. So that has to equal the force of gravity pulling down. This has to equal the force of gravity. That has to equal this or this point. So that's 100 Newtons. And I really want to hit this point home because it might be a little confusing to you. We just said, this point is stationery. It's not moving up or down. It's not accelerating up or down. And so we know that there's a downward force of 100 Newtons, 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's sine of 30? Sine of 30 degrees, in case you haven't memorized it, sine of 30 degrees is 1/2. So T1 times 1/2 is equal to 100 Newtons. Divide both sides by 1/2 and you get T1 is equal to 200 Newtons. So now we've got to figure out what the tension in this second wire is. And we 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 wire's tension. So it's this. So T2 is equal to the x component of the first wire's tension. And what's the x component? Well, it's going to be the tension in the first wire, 200 Newtons times the cosine of 30 degrees. It's adjacent over hypotenuse. And that's square root of 3 over 2. So it's 200 times the square root of 3 over 2, which equals 100 square root of 3. So the tension in this wire is 100 square root of 3, which completely offsets to the left and the x component of this wire is 100 square root of 3 Newtons to the right. Hopefully I didn't confuse you. See you in the next video.