David shows how to solve a super hot tension problem where a can of peppers hangs from two diagonal strings. Created by David SantoPietro.
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- This works out beautifully and I understand all the steps involved in getting the problem that YOU have.. Although I'm stuck on the same sort of problem in which it isn't a perfect 30,60,90 triangle... meaning the numbers are in decimals and it gets messy and confusing. The two angles given in my problem are 35 and 48. the mass of the object pulling is 65kg. I got all the way down to the point you reached around 11 minutes in.(39 votes)
- Yes, good call everyone. That was bad form on my part by prematurely turning trig functions into decimals. Leaving things as symbolic as possible is almost always the way to go.(13 votes)
- T1 and T2 happen to be mg*Sin30 and mg*Sin60 respectively. Is this just a coincidence?(7 votes)
- It works out this way because the angle between the two ropes (and thus the two tensions) is exactly 90 degrees (T1 and T2 are at right angles to one another, you can see this by noting that the other two angles in the "inside" triangle are 30 and 60, and the angles in a triangle add up to 180, so the angle between T1 and T2 is 90).
Because of this, you could rotate your reference frame. Suppose you pick the new y axis to point along T2, and the new x axis to point along T1. Now, the x equation and the y equation are each simple and have one unknown. However, you need to break gravity into the two components you mentioned, which are the components along these "new" axes. Hope that helps.(4 votes)
- what if two ropes pull the can with angle zero i.e. along x-axis so hard that they balance gravitational pull.how to find force here. how do they really affect cans accelaration in y direction(4 votes)
- one possibility is to froze the ropes then hold the can between the two ropes while making them horizontal. then it is not in the relm of tension anymore as there's no pulling force by them to the can
in other words, if the rope is ex"tensi"ble, it must make an angle to balance out gravity in the situation like above. otherwise, there's no "tensi"on you need to concern(1 vote)
- Earlier on, we learned that a vector can be moved and relocated within a vector diagram without changing its magnitude. Is it a shortcut in this problem to move the Gravitational Force vector head to tail with the Tension vector intersection? This would create 2 right triangles which would allow us to solve for T1 and T2 using trigonometry.
T1 = 30Ncos60 = 15N
T2 = 30Ncos30 = 15sqrt3N
This gives us the same answer, and eliminates an enormous amount of work. Is this a practical method? Or will the method shown in the video be of good use in more challenging problems thus the effort is worth it?(3 votes)
- Will there be any sort of problems with three tensions? How will you solve that then?(1 vote)
- Very minor annoyance I'm having, is that everytime you work with newton's second law, you tend to negate a force variable to imply that it goes in the opposite direction, but shouldn't you instead negate the magnitude of the force variable, because if it is truly negative, and you negate a negative, that's a positive.(2 votes)
- Can't we just write everything in terms of Ty by using tan(theta)?(1 vote)
- Why doesn’t T1(in Y direction)= T1*sin30 works for solving for T1 directly? After substituting it in Newton’s second law as follow: -30+T1sin30/3 = 0 solving this would bring T1 = 60N.(1 vote)
- -30+T1sin30/3 = 0 is incorrect because the weight of the can is not only supported by T1(in Y direction), it is also supported by T2(in Y direction). So it is -30 + T1sin30 + T2sin60 = 0(1 vote)
- When I tried to solve this problem by myself . I stopped when 0 = (-30N) + T1y + T2y . I noticed that T1y and T2y are equal because they are go out from the same point in the bottle to the roof so they have the same magnitude.
so I rewirte 0 = (-30N) + T1y + T2y to 30 = 2 T(1 or 2)y ... so T1y = 15 and T2y = 15 .... so sin30 = 15 / T1 ...so T1 = 15/sin30 = 30 , also i got T2 by the same way and both of them are wrong :D
I thought this way will work , why it dose not work ??
sorry for my bad E , I'm still learning .(1 vote)
- It doesn't work because T1 and T2 have different directions and magnitudes. The magnitude of their y component would only be the same if they had the same angle. I think...(1 vote)
- I managed to solve this on my own, I got magnitude of tension force 2 to be 27.7N though, I would assume inaccuracy & not any faults?(1 vote)
- [Voiceover] 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 gotta come up with a completely new strategy to tackle this. I've gotta 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 gonna use the same process we used for the easy tension problems, because it's gonna lead us to the answer again. Be careful. Don't stray from the strategy here. The strategy works. So we're gonna draw our force diagram first. That's what we always do. We're gonna say that the forces are force of gravity on this can of red peppers, which is MG, and if it's 3 kilograms, we know 3 kilograms times about 10, we're gonna 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 kilograms times 10 meters per second squared is gonna be 30 Newtons. And so the force of gravity downward is 30 Newtons. What other forces do we have? We've got this T1, remember tension does not push. Ropes can't push, ropes can only pull, so T1's gonna pull that way. So I'm gonna draw T1 coming this way. So here's our T1. And then we're gonna have T2 pointing this way, so this is T2. Again, T2 pulls, just like all tension. Tension pulls, 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, 'cause 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 did 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 gonna 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. Force of gravity is 30 Newtons. Usually that's a guod strategy, pick the direction that you know something about at least. So we're gonna do that here. We're gonna say that the acceleration vertically equals 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 in not 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 gonna equal the net force and the vertical direction. So what are we gonna 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 gonna assume upward is positive, that means down in a negative. So I'll just put -30 Newtons. I could have written -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. So part of them points to the right, part of them points upward. Only this vertical component, we'll call it T1Y, is gonna get included into this calculation, 'cause this calculation only uses Y directed forces. And the reason is only Y directed forces, vertical 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 in this formula, 'cause only part of it points up. Similarly, we have to plug in only 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 kilograms. But we're gonna multiply both sides by 3 kilograms, and we're gonna get zero equals all of this right here, so I'll just copy this right here. We use this over again, that comes down right there. But now there's nothing on the bottom here. So what do we do at this point? Now you might think we're stuck. I mean, we've 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 I'd do fine, if I added 30 to both sides, I'd realize that these two vertical components of these tension forces added up 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 have the net force in the X direction, over the mass, again, the acceleration is gonna 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 it did, it's not that big of a deal, you just plug it in there. But assuming it's acceleration 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 gonna be T1 in the X. So part of this T1 points in the X direction. Similarly, part of T2 points in the X direction. We'll call this T2X. We use these as the magnitude. Let's say T2X is the magnitude of the force that T2 pulls with to the left, and T1 is the magnitude 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 gonna consider rightward to be the positive direction, 'cause that's the typical convention that we're gonna adopt. And T2X pulls to the left. That's gonna be a negative contribution, so minus T2 in the X direction. 'Cause leftward would be negative. We divided by the mass, the mass was 3 kilograms, but again, we'll multiply both sides by 3, we'll get zero equals, and then we just get T, the 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 multi, or if I added T2X to both sides, I'm just gonna 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 have no acceleration in the X direction. And this was not drawn proportionately, 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 X direction. We can't solve this equation we got from 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 gonna 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 gonna use that a lot of people don't like doing because it's a little more sophisticated, now we've gotta put these all in terms of T1 and T2 so that we can solve. If I put T1Y in terms of the total T1, and then sines of angles, and cosines of angles, and I put T2Y in terms of T2 and angles, and I do the same thing for 1X and 2X, 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 over 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 gotta be 60, 'cause it all adds up to 180 for a triangle. And if this right angle is 90, and this side's 60, this side's gotta 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, 'cause 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 gonna plug those expressions into here, and that will let us solve. In other words, T1Y is gonna be, once you do this for awhile you realize, this is the opposite side. So this component here is going to be 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, 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 the T1Y if I multiply both sides by T1. I get that that's T1 times sine of 30. So that's what I said down here. T1 is just T, oh sorry, forgot the one. T1 times sine of 30. Similarly, if you do the same thing with cosine 30, you'll get that T1X is T1 cosine 30, by the exact same process. Similarly over here, T2 is going to be, I'm sorry, T2X is gonna be 2. So T2 cosine 60, because this is the adjacent side. And T2Y is gonna 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, the total hypotenuse side, solve for your expression, you'll get these. 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 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 is T2 sine 60 into this expression right there for T2Y, look at what I'll get. I'll get zero equals. So I'll get negative 30 Newtons, and then I'll get plus T1Y was T1 sine 30, so T1, and then sine 30, we can clean this up a little bit. Sine 30 is just a half. So I'll just write T1 over 2, and then 'cause sine 30 is just one half. And then T2Y is gonna be T2 sine 60, and sine 60 is just root 3 over 2. So I'll write this as plus T2 over 2, and then times root 3. And you might think this is no better. I mean this is still a horrible mess right here. But, look at. This is in terms of T1 and T2. That's what I'm gonna do over here. I'm gonna put these in terms of T1 and T2, and then we can solve. So T1X is T1 over cosine 30, so I'm gonna write this as T1 times cosine 30, and cosine 30 is root 3 over 2, so this is T1 over 2 times root 3. And that should equal T2X is right here, That's T2 cosine 60, cosine 60 is a half. So T2X is gonna be T2 over 2. So T2 over 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 in the angle. And I do this, because look at what I have now, I have got one equation with T1 and T2. I've got another equation with T1 and T2. So what I'm gonna do to solve these, when we 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 gonna solve this one is easier, so I'm gonna 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, and I'll get T1 times root 3. So T1 times root 3, because the 2 here cancels with this 2, or when I multiply both sides by 2 it cancels out. So we get that T2 equals T1 root 3. This is great. I can substitute T2 as T1 root 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 zero equals negative, you know what, let's just move the -30 over. This is kind of annoying here. Just add 30 to both sides, then take this calculation here. We get plus 30 equals, and then we're gonna have T1 over 2, from this T1, so T1 over 2 plus, I've got plus, T2 is T1 root 3. So when I plug T1 root 3 in for T2, what I'm gonna get is, I'm gonna get T1 root 3, and then times another route 3, because T2 itself was T1 root 3. So I'm taking this expression here, plugging it in for T2, but I still have to multiply that T2 by a root 3 and divide by 2. And so, what do we get? Root 3 times root 3 is just 3. So we have T2 times 3 halves, plus T1 over 2. So I'll get 30 equals, and then I get T1 over 2, we're almost there, I promise. T1 over 2, plus, and this is gonna be T1 times 3 over 2, so it's gonna be 3 T1 over 2, or what does that equal? T1 over 2 plus 3 T1 over 2 is just 4 halves. So that's just 2 T1. So this cleaned up beautifully. So this is just 2 times T1, and now we can solve for T1. We get that T1 is simply 30 divided by 2. If I divide both sides, this left hand side by 2, and this side here, this right side by 2, 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. Whoo, 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 root 3. So all I have to do is multiply root 3 by my T1, which I know now. And I get that T2 is just 15 times root 3 Newton. So once you get one of the forces, the next one is really easy. This is just T2. So T2 is 15 root 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, because there were too many 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, T2, and T1 and T2 in them. We solved one of these equations for T2 in terms of T1, substituted that into the other equation. 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 have that variable, we plug it back into that first equation that we had solved for T2. We plug 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.