Super hot tension | Forces and Newton's laws of motion | Physics | Khan Academy

Oh, it's time! It's time for the super hot tension problem. We're about to do this right here. We've got our super hot can of red peppers hanging from these strings. We want to know what the tension is in these ropes. This is for real now; this is a real tension problem.

And here's the deal: you might look at this, you might get frightened. You might think, "I've got to come up with a completely new strategy to tackle this. I've got to throw away everything I've learned and just try something new." And that's a lie! You should not lie to yourself. Use the same process. We're going to use the same process we used for the easy tension problems because it's going to lead us to the answer.

Again, be careful—don't stray from the strategy here. The strategy works! So, we're going to draw our force diagram first; that's what we always do. We're going to say that the forces are the force of gravity on this can of red peppers, which is mg. If it's 3 kg, we know 3 kg times about 10. If we're going to say let's approximate g as 10 again to make the numbers come out nice, so instead of using 9.8, we'll say g is about 10.

And so, we'll say 3 kg times 10 m/s² is going to be 30 Newtons. So, the force of gravity downward is 30 Newtons. What other forces do we have? We've got this T1, and remember, tension does not push. Ropes can't push; ropes can only pull. So, T1 is going to pull that way. So, I'm going to draw T1 coming this way. So here's our T1.

And then we're going to have T2 pointing this way. So, this is T2. Again, T2 pulls just like all tension—tension can't push. So, I've got tension 2 going this way; that's it! That's our force diagram. There's no other forces. I don't draw a normal force because this can isn't in contact with another surface, so there's no normal force.

You've got these two tensions, the force of gravity, and now we do the same thing we always do after our force diagram. We use Newton's Second Law in one direction or another. So let's do it! Let's say that acceleration is the net force in a given direction divided by the mass. Which direction do we pick? Again, it's hard to say. We've got forces vertical; we've got forces horizontal. There's only two directions to pick: X or Y.

In this problem, we're going to pick the vertical direction, even though it doesn't really matter too much, but because we know one of the forces in the vertical direction—we know the force of gravity. The force of gravity is 30 Newtons. Usually, that's a good strategy—pick the direction that you know something about at least.

So, we're going to do that here. We're going to say that the acceleration vertically is equal to the net force vertically over the mass. And so now we plug in. If this can is just sitting here, if there's no acceleration, if this is not in an elevator transporting these peppers up or down, and it's not in a rocket, if it's just sitting here with no acceleration, our acceleration will be zero.

That's going to equal the net force in the vertical direction. So what are we going to have? So what are the forces in the vertical direction here? One force is this 30 Newton force of gravity; this points down. We're going to assume upward is positive; that means down is negative. So I'll just put -30 Newtons. I could have written negative mg, but we already knew it was 30 Newtons, so I'll write 30 Newtons.

Then we've got T1 and T2. Both of those point up, but they don't completely point up. They partially point up; only this vertical component we'll call T1y is going to get included into this calculation because this calculation only uses Y-directed forces. And the reason is only Y-directed forces affect the vertical acceleration.

So this T1y points upward; I'll do plus T1 in the Y direction. And similarly, this T2—it doesn't all point vertically; only part of it points vertically. So I'll write this as T2 in the Y direction, and that's also upward. So since that's up, I'll count it as plus T2 in the Y direction. And that's it! That's all our forces.

Notice we can't plug in the total amount T2 into this formula because only part of it points up. Similarly, we have to only plug in the vertical component of the T1 force because only part of it points vertically. And then we divide by the mass. The mass is 3 kg, but we're going to multiply both sides by 3 kg. We're going to get zero equals all of this right here.

So I'll just copy this right here; we'll use this over again. That comes down right there. But now there's nothing on the bottom here, so what do we do with this point? Now, you might think we're stuck. I mean, we got two unknowns in here; I can't solve for either one. I don't know either one of these. I know they have to add up to 30.

So what I'd find if I added 30 to both sides, I'd realize that these two vertical components of these tension forces have to add up to 30. And that makes sense—they have to balance the force downward. But I don't know either of them, so how do I solve here? Well, let's do this: if you ever get stuck on one of the force equations for a single direction, just go to the next equation.

Let's try A in the X direction. So for A in the X direction, we'd have the net force in the X direction over the mass. Again, the acceleration is going to be zero if these peppers are not accelerating horizontally. So unless this thing's in a train car or something and the whole thing's accelerating, then you might have horizontal acceleration.

And if you did, it's not that big of a deal; you just plug it in there. But assuming this acceleration is zero because the peppers are just sitting there, not changing their velocity, we'll plug in zero. We'll plug in the forces in the X direction. These are going to be T1 in the X, so part of this T1 points in the X direction. Similarly, part of T2 points in the X direction.

So I'll call this T2x. We'll use these as the magnitudes. Let's say T2x is the magnitude of the force that T2 pulls with to the left, and T1x is the magnitude of the force that T1 pulls with to the right. So to plug these in, we've got to decide whether they should be positive or negative.

So this T1x, since it pulls to the right, T1x will be positive. We're going to consider rightward to be the positive direction because that's the typical convention that we're going to adopt. And T2x, since it pulls to the left, that's going to be a negative contribution. So, minus T2 in the X direction because leftward would be negative.

We divide by the mass. The mass was 3 kg, but again, we'll multiply both sides by three. We'll get 0 equals and then we just get T—same thing up here. So we'll just copy this thing here, put it down here, and again, you might be concerned. I can't solve this either! I mean, I can solve for T1x, but look at what I get if I just add T2x to both sides—I'm just going to get T1 in the X direction has to equal T2 in the X direction.

And that makes sense! These two forces have to be equal and opposite because they have to cancel, so that you get no acceleration in the X direction. And this was not drawn proportionally; sorry, this should be the exact same size as this force because they have to cancel since there's no horizontal acceleration. But what do we do?

We can't solve this equation we got from the X direction. We can't solve this equation we got for the Y direction. Whenever this happens, when you get two equations and you can't solve either because there's too many unknowns, you're going to have to end up plugging one into the other. But I can't even do that yet! I've got four different variables here: T1x, T2x, T1y, and T2y.

These are all four different variables; I've only got two equations! I can't solve this! So the trick—the trick we're going to use that a lot of people don't like doing because it's a little more sophisticated—now we've got to put these all in terms of T1 and T2 so that we can solve.

If I put T1y in terms of T1 and then sine of angles and cosine of angles, and I put T2y in terms of T2 and angles—let's do the same thing for T1x and T2x—I'll have two equations and the only two unknowns will be T1 and T2. Then we can finally solve! If that didn't make any sense, here's what I'm saying.

I'm saying figure out what T1y is in terms of T1. So I know this angle here—let's figure out these angles. So these angles here are—if this is 30, this angle down here has to be 30 because these are alternate interior angles. And if you don't believe me, imagine this big triangle right here where this is a right angle.

So this triangle from here to there, down to here, up to here—if this is 30, that's 90; this has got to be 60 because it all adds up to 180 for a triangle. And if this right angle is 90 and this side is 60, this side has got to be 30. Similarly, this side's a right angle. Look at this triangle—60, 90—that means this would have to be 30.

And so, if I come down here, this angle would have to be 60 just like this one. You see, it's an alternate interior angle, so that's 60! So this angle here is 60; this angle here is 30. We can figure out what these components are in terms of the total vectors. Once we find those, we're going to plug those expressions into here and that'll let us solve.

So in other words, T1y is going to be—once you do this for a while you realize this is the opposite side, so this component here is going to be the total T1 times sine of 30 because it's the opposite side. And if that didn't make sense, we'll derive it right here. So what we're saying is that sine of 30 is opposite over hypotenuse, and in this case, the opposite side is T1y.

So T1y over the total T1 is equal to sine of 30, and we can solve this for T1y now. We can get that T1y, if I multiply both sides by T1, I get that T1y is T1 times sine of 30. So that's what I said down here: T1 is just T1 times sine of 30.

Similarly, if you do the same thing with cosine 30, you'll get T1x is T1 cosine 30 by the exact same process. Similarly over here, T2 is going to be—or sorry, T2x is going to be T2, so T2 cosine 60 because this is the adjacent side. And T2y is going to be T2 sine of 60. And if any of that doesn't make sense, just go back to the definition of sine and cosine.

Write what the opposite side is over the total hypotenuse side, solve for your expression; you'll get these. So if you don't believe me on those, try those out yourselves! But those are what these components are in terms of T2 and the angles, T2, T1, and the angles.

And why are we doing this? We're doing this so that when we plug in over here, we'll only have two variables. In other words, if I plug T1y—this expression here, T1 sine 30—in for T1y, similarly, if I plug in T2y as T2 sine 60 into this expression right there, look at what I'll get! I'll get 0 equals—so I'll get -30 Newtons, and then I'll get plus T1y was T1 sine 30.

So T1 and then sine of 30—we can clean this up a little bit. Sine of 30 is just a half, so I'll just write T1 over 2. And then, because sine of 30 is just 1/2, T2y is going to be T2 sine 60 and sine 60 is just √3/2. So I'll write this as plus T2 times √3/2. You might think this is no better! I mean this is still a horrible mess right here!

But look at this—this is in terms of T1 and T2! That's what I'm going to do over here. I'm going to put these in terms of T1 and T2, and then we can solve! So T1x is T1 cosine 30, so I'm going to write this as T1 √3/2. So this is T1 √3/2.

Now, T2x is right here. That's T2 sine 60; cosine 60 is a half, so T2x is going to be T2/2. So T2/2. So what I'm doing is if this doesn't make sense, I'm just substituting what these components are in terms of the total magnitude and the angle. I do this because look at what I have now!

I've got one equation with T1 and T2. I've got another equation with T1 and T2. So what I'm going to do to solve these—when you have two equations and two unknowns, you have to solve for one of these variables and then substitute it into the other equation. That way, you'll get one equation with one unknown, and you try to get the math right, and you'll get the problem!

So I'm going to solve this one; it's easier! So I'm going to solve this one for let's just say T2. So if we solve this for T2, I get that T2 equals—well, I can multiply both sides by 2, then I'll get T1√3. So T1√3, because the two here cancels with this two. So when I multiply both sides by two, it cancels out!

So you get T2 = T1√3; this is great! I can substitute T2 as T1√3 into here for T2. And the reason I do that is I'll get one equation with one unknown—I'll only have T1 in that equation now. So if I do this, I'll get 0 equal—you know what? Let's just move the -30 over; this is kind of annoying.

Here, just add 30 to both sides, so take this calculation here. We'll get +30 equals, and then we're going to have T1 over 2 from this T1. So T1 over 2 plus, I've got T2 = T1√3. So when I plug T1√3 in for T2, what I'm going to get is I'm going to get T1√3 times another √3 because T2 itself was T1√3.

So I'm taking this expression here, plugging it in for T2, but I still have to multiply that T2 by √3 and divide by 2. And so what do we get? √3 * √3 is just 3, so I get T1 * 3 plus T1/2.

We're almost there, I promise! T1/2 plus, and then this is going to be T1 * 3/2, so it's going to be 3T1/2. Well, what does that equal? T1/2 + 3T1/2 is just—it's just four halves, so that's just 2T1.

So this cleaned up beautifully! So this is just 2*T1, and now we can solve for T1. We get that T1 is simply 30/2. If I divide both sides, this left-hand side here by two and this side here, this right side by two, I get T1 is 30 over 2 Newtons, which is just—these should be Newtons; I should have units on these!

Which is just 15 Newtons! I did it; 15 Newtons! T1 is 15 Newtons. We got T1; that's one of them. How do we get the other? You start back over at the very beginning? No, not really; that would be terrible! You actually just take this T1, and you plug it right into here. Boop! There it goes!

So T2, we already got it. T2 is just T1√3. So all I have to do is multiply √3 by my T1, which I know now, and I get that T2 is just 15√3 Newtons. So once you get one of the forces, the next one's really easy. This is just T2, so T2 is 15√3, and T1 is just 15.

So in case you got lost in the details, the big picture recap is this: we drew a force diagram. We used Newton's Second Law in the vertical direction. We couldn't solve 'cause there were two unknowns. We used Newton's Second Law in the horizontal direction; we couldn't solve because there were two unknowns.

We put all four of these unknowns in terms of only two unknowns T1 and T2 by writing how those components depended on those total vectors. We substituted these expressions in for each component. Once we did that, we had two equations with only T1 and T2. We solved one of these equations for T2 in terms of T1, substituted that into the other equation, and we got a single equation with only one unknown.

We were able to solve for that unknown once we got that, which is our T1. Once we had that variable, we plugged it back into that first equation that we had solved for T2. We plugged this 15 in; we get what the second tension is. So even when it seems like Newton's Second Law won't get you there, if you have faith and you persevere, you will make it! Good job!