Introduction to tension (part 2)
A slightly more difficult tension problem. Created by Sal Khan.
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- Seems like the easiest way to do this problem was just putting the value 10N up the middle between them, then taking 10sin(60*)=T2 and 10sin(30*) = T1.(12 votes)
- But they come out correctly, is it just a fluke?(4 votes)
- Couldn't you have just done,
T2 = 10Sin60° = 5√3N = 8.66N
T1 = 10Sin30° = 5N
Or is it just luck that this happens to work in this situation? Having to go through the way in the video can be a bit tedious.(10 votes)
- What if I have more than 2 ropes, say 4. That would lead me to two equations with 4 unknowns. Or is it possible to derive two more equations with the increase of unknowns?(5 votes)
- Well, if you have 3 ropes, it could just be that 2 ropes are holding the weight, and the third is hanging slack, because it is too long. If you assume, that the ropes have the right length, that they are all under tension, or if you replace the ropes with bars (they support both tension and compression), it is solveable, but it gets complicated. btw this is called a "Statically Indeterminate Structure".
The way to do this is to calculate the deformation of the ropes/bars. Bars get a little longer if they are under tension and a little shorter under compression. I could make an example, but only if you care, it would be a bit of work.(2 votes)
- In the system of equations, how do you know which equation to subtract from the other? If I were doing this problem, I would have just subtracted the top equation from the bottom equation instead of the other way around, giving me 4T2 = 20√3, which basically gives me the same answer of T2 = 5√3.(3 votes)
- When solving a system of equations by elimination any of the two equations may be subtracted from another or added together. It does not matter if the top equation is subtracted from the bottom equation or vice versa and same for addition. The only thing that has to be seen is that a variable is eliminated.(6 votes)
- If i look at this problem i see that both y components must be equal because the vector has the same length. Using this you could solve the probelm much faster, couldn't you?(4 votes)
- Nothing would be wrong. The only thing you would have done is to change the question(1 vote)
- I was wondering on what contribution dose the rope on the bottom do to the overall tension supporting the block. I am talking about the rope that connects the mass and the point that attaches to t1 and t2. Dose the vertical wire contribute anything to the tension supporting the block or is t1 and t2 only responsible for pulling mass up against gravity. Is t1 and t2 divide the force of gravity that the bottom rope experinces?(3 votes)
- At5:17, Why does the tension of the combined y components not equal 10N*9.8(gravity) ?(1 vote)
- Why would you multiply 10 N times 9.8? The 9.8 is 9.8 N/kg. It tells you how many newtons there are per kilogram, if you are on the surface of the earth. If you multiply 10 N * 9.8 N/kg, you have 98 N^2/kg, which doesn't make much sense.
If the object is just hanging, and it is not accelerating, the sum of the upward tension forces has to equal the downward force, which is the weight. All forces should be in newtons. To get the downward force if you only know mass, you would multiply the mass by 9.8 N/kg.(3 votes)
- Visually, it appears that the y-components of both T1 and T2 are equal... but they are not. Does anyone have a good explanation for why we cannot judge these force vectors (or any force vectors, maybe?) by their appearance? When Sal said both y-vectors have to add to 10, and you can clearly see that they look the same, I would have stopped and said, well all right, both y-vectors are 5 N, and gone from there. Why would that be wrong?(2 votes)
- They look identical because they were drawn by hand as a representation. If it were a perfectly drawn diagram, the visual differences would be clearer. You can judge force vectors visually, but that would only give you a sort of "sense" of which rope may be exerting a greater force, or having a greater tension. This does not give you, however, the definite values we look for. These exact values we want to calculate (either y-components, x-components, or net rope tensions T and T2) are completely dependent on the ropes' inclinations with the ceiling.(1 vote)
- Hey Sal, at00:47you say that the point is not accelerating in any direction, how do you know that? like technically isnt it accelerating in the y because the point is not on the pink line parallel to it? like its below the pink line so isnt it accelerating downwards in the y direction.(1 vote)
- The weight is just hanging there. That's how the problem is given. If it is just hanging there, how can it be accelerating?(2 votes)
- Is it wrong if I just said:
10*Sin(30) = T1
10*Sin(60) = T2
By doing this I still got the correct answer. It doesn't seem right, but I don't really know whats wrong with it...
What I want to know is:
1. Why is this wrong
2.Why did I get the correct answer
Thank you(2 votes)
- This program explains what is happening and why it is that when the sum of the angles equal 90 degrees you can get away with these seemingly simple "solutions." The derivation is in the comments as are the instructions for playing with the program to see other angles. You can edit the program and "break" the sums to 90 relationship to see what happens in those cases.
Welcome back. We'll now do another tension problem and this one is just a slight increment harder than the previous one just because we have to take out slightly more sophisticated algebra tools than we did in the last one. But it's not really any harder. But you should actually see this type of problem because you'll probably see it on an exam. So let's figure out the tension in the wire. So first of all, we know that this point right here isn't moving. So the tension in this little small wire right here is easy. It's trivial. The force of gravity is pulling down at this point with 10 Newtons because you have this weight here. And of course, since this point is stationary, the tension in this wire has to be 10 Newtons upward. That's an easy one. So let's just figure out the tension in these two slightly more difficult wires to figure out the tensions of. So once again, we know that this point right here, this point is not accelerating in any direction. It's not accelerating in the x direction, nor is it accelerating in the vertical direction or the y direction. So we know that the net forces in the x direction need to be 0 on it and we know the net forces in the y direction need to be 0. So what are the net forces in the x direction? Well they're going to be the x components of these two-- of the tension vectors of both of these wires. I guess let's draw the tension vectors of the two wires. So this T1, it's pulling. The tension vector pulls in the direction of the wire along the same line. So let's say that this is the tension vector of T1. If that's the tension vector, its x component will be this. Let me see how good I can draw this. It's intended to be a straight line, but that would be its x component. And its x component, let's see, this is 30 degrees. This is 30 degrees right here. And hopefully this is a bit second nature to you. If this value up here is T1, what is the value of the x component? It's T1 cosine of 30 degrees. And you could do your SOH-CAH-TOA. You know, cosine is adjacent over hypotenuse. So the cosine of 30 degrees is equal to-- This over T1 one is equal to the x component over T1. And if you multiply both sides by T1, you get this. This should be a little bit of second nature right now. That the x component is going to be the cosine of the angle between the hypotenuse and the x component times the hypotenuse. And similarly, the x component here-- Let me draw this force vector. So if this is T2, this would be its x component. And very similarly, this is 60 degrees, so this would be T2 cosine of 60. Now what do we know about these two vectors? We know that their net force is 0. Or that you also know that the magnitude of these two vectors should cancel each other out or that they're equal. I mean, they're pulling in opposite directions. That's pretty obvious. And so you know that their magnitudes need to be equal. So we know that T1 cosine of 30 is going to equal T2 cosine of 60. So let's write that down. T1 cosine of 30 degrees is equal to T2 cosine of 60. And then we could bring the T2 on to this side. And actually, let's also-- I'm trying to save as much space as possible because I'm guessing this is going to take up a lot of room, this problem. What's the cosine of 30 degrees? If you haven't memorized it already, it's square root of 3 over 2. So this becomes square root of 3 over 2 times T1. That's the cosine of 30 degrees. And then I'm going to bring this on to this side. So the cosine of 60 is actually 1/2. You could use your calculator if you forgot that. So this is 1/2 T2. Bring it on this side so it becomes minus 1/2. I'm skipping more steps than normal just because I don't want to waste too much space. And this equals 0. But if you seen the other videos, hopefully I'm not creating too many gaps. And this is relatively easy to follow. So we have the square root of 3 times T1 minus 1/2 T2 is equal to 0. So that gives us an equation. One equation with two unknowns, so it doesn't help us much so far. But let's square that away because I have a feeling this will be useful. Now what's going to be happening on the y components? So let's say that this is the y component of T1 and this is the y component of T2. What do we know? What what do we know about the two y components? I could've drawn them here too and then just shift them over to the left and the right. We know that their combined pull upwards, the combined pull of the two vertical tension components has to offset the force of gravity pulling down because this point is stationary. So we know these two y components, when you add them together, the combined tension in the vertical direction has to be 10 Newtons. Because it's offsetting this force of gravity. So what's this y component? Well, this was T1 of cosine of 30. This should start to become a little second nature to you that this is T1 sine of 30, this y component right here. So T1-- Let me write it here. T1 sine of 30 degrees plus this vector, which is T2 sine of 60 degrees. You could review your trigonometry and your SOH-CAH-TOA. Frankly, I think, just seeing what people get confused on is the trigonometry. But you can review the trig modules and maybe some of the earlier force vector modules that we did. And hopefully, these will make sense. I'm skipping a few steps. And these will equal 10 Newtons. And let's rewrite this up here where I substitute the values. So what's the sine of 30? Actually, let me do it right here. What's the sine of 30 degrees? The sine of 30 degrees is 1/2 so we get 1/2 T1 plus the sine of 60 degrees, which is square root of 3 over 2. Square root of 3 over 2 T2 is equal to 10. And then I don't like this, all these 2's and this 1/2 here. So let's multiply this whole equation by 2. So 2 times 1/2, that's 1. So you get T1 plus the square root of 3 T2 is equal to, 2 times 10 , is 20. Similarly, let's take this equation up here and let's multiply this equation by 2 and bring it down here. So this is the original one that we got. So if we multiply this whole thing by 2-- I'll do it in this color so that you know that it's a different equation. So if you multiply square root of 3 over 2 times 2-- I'm just doing this to get rid of the 2's in the denominator. So you get square root of 3 T1 minus T2 is equal to 0 because 0 times 2 is 0. And let's see what we could do. What if we take this top equation because we want to start canceling out some terms. Let's take this top equation and let's multiply it by-- oh, I don't know. Let's multiply it by the square root of 3. So you get the square root of 3 T1. I'm taking this top equation multiplied by the square root of 3. This is just a system of equations that I'm solving for. And the square root of 3 times this right here. Square root of 3 times square root of 3 is 3. So plus 3 T2 is equal to 20 square root of 3. And now what I want to do is let's-- I know I'm doing a lot of equation manipulation here. But this is just hopefully, a review of algebra for you. Let's subtract this equation from this equation. So you can also view it as multiplying it by negative 1 and then adding the 2. So when you subtract this from this, these two terms cancel out because they're the same. And so then you're left with minus T2 from here. Minus this, minus 3 T2 is equal to 0 minus 20 square roots of 3. And so this becomes minus 4 T2 is equal to minus 20 square roots of 3. And then, divide both sides by minus 4 and you get T2 is equal to 5 square roots of 3 Newtons. So that's the tension in this wire. And now we can substitute and figure out T1. Let's use this formula right here because it looks suitably simple. So we have the square root of 3 times T1 minus T2. Well T2 is 5 square roots of 3. 5 square roots of 3 is equal to 0. So we have the square root of 3 T1 is equal to five square roots of 3. Divide both sides by square root of 3 and you get the tension in the first wire is equal to 5 Newtons. So this is pulling with a force or tension of 5 Newtons. Or a force. And this is pulling-- the second wire --with a tension of 5 square roots of 3 Newtons. So this wire right here is actually doing more of the pulling. It's actually more of the force of gravity is ending up on this wire. That makes sense because it's steeper. So since it's steeper, it's contributing more to the y component. It's good whenever you do these problems to kind of do a reality check just to make sure your numbers make sense. And if you think about it, their combined tension is something more than 10 Newtons. And that makes sense because some of the force that they're pulling with is wasted against pulling each other in the horizontal direction. Anyway, I'll see you all in the next video.