Mild and medium tension | Forces and Newton's laws of motion | Physics | Khan Academy

You bought a huge canister aluminum can of super hot red peppers, 3 Kg worth, and you hung them from two strings from the ceiling 'cause you don't want anyone to get your super hot red peppers. You wanted to know what's the tension in both of these strings if this is the angle that the strings make with the ceiling.

What are these two tensions? This problem's hard! This problem's spicy! This is a spicy tension problem! Let's start with something a little more mild. We'll work our way up to this to see how this works. We're going to do that because even though this one seems very difficult and easier problems seem much different from it, the process that you use to figure out the answer is always the same.

So, I'm going to try to show you that you shouldn't get distracted. Even though the details are different, the process, the overall strategy, is the same. Let's start with something a little easier. Let's start with this: a nice red apple. Let's say this apple is 3 Kg so we can keep the same number. A 3 Kg apple hanging from a string. We want to know what's the tension in the rope.

Well, this one's easy! The way used to solve it is the way you solve all these problems. So even though it's easy, we're going to go through the entire process so you can see how it works and it won't take long.

We draw the forces. The force of gravity is exerted on this apple because it's always exerted on everything near the Earth, and it's mg. Then we've got a tension force, and the tension does not push. You can't push with a rope; you can pull with a rope! So this tension force points upward. I'll call it T. That's always the first step: you draw your force diagram.

Now you use Newton's second law for either the horizontal direction, or the vertical direction, or both if you have to. So, we're going to use Newton's second law, which looks like this: the acceleration in a certain direction equals the net force in that direction divided by the mass. Which direction do we pick? It's pretty obvious here. We'll pick the vertical direction because there are no forces in the horizontal direction.

I'll plug in the acceleration. If this apple is just hanging here and not moving, then it's not accelerating; it's just sitting here. Unless this was like an elevator or a rocket ship that had acceleration upward, you could do that. You'd plug in that acceleration, but if it's just hanging from the ceiling at rest, this acceleration is going to be zero.

That's going to equal the net force. We've got tension upward. To figure out what goes here, we draw this force diagram for a reason. This isn't just like happy fun painting time over here. This is a strategy; we draw these forces 'cause this lets us know what we plug into Newton's second law for the net force.

If it's not on here, I don't put it up here, and if it is on here, I have to put it in here if it's in that direction. So tension is vertical, and this is the vertical direction for the net force. So I put tension in here. I'm going to call upward positive. That means I make this tension positive.

How about mg over here? mg points down, so I do minus mg, and I divide by the mass. Well, now I just solve for the tension! So this is the process. You draw your force diagram, use Newton's second law, then you try to solve for the force that you want; in this case, it is the tension.

So I multiply both sides by m. I still have zero on the left-hand side, so 0 = T - mg. If I solve this for T, I get something that might not be all that surprising. I just get that the tension is equal to mg! You might be like, well duh, could have done that! That was like way more trouble than it had to be! I mean, you just knew it was mg; this tension just has to balance gravity, so it's just mg.

Why did we go through all this trouble? The reason is, I mean, it is equal to mg, but it won't always be equal to mg. So if you want to know what to do when it's not equal to mg, you have to know how to use this process.

Well, why would it not be equal to mg? Well, imagine this: imagine I had, say, two tensions up here. Or let's start with this: let's say I just pull down on the rope. So, I just come over to here and someone—or someone pulls down on the apple.

Let's say someone just pulls down the apple with an extra five Newtons, pulling this apple, which pulls this string, making it tighter. What would I do in that case? Well, I've got one more force here in my force diagram. I just add that, so that's 5 Newtons. It's pointing downward, so when I come over here to my net force, I've got to subtract five Newtons because that's pointing downward.

Now, I do my algebra just as before! I multiply both sides by m. I'll have another -5 Newtons here, and then when I solve for T, I'll add mg to both sides and I'll add 5 Newtons to both sides.

So, this would be my force. With numbers, what are we going to get? The mass was three, so we'll have 3 Kg times the acceleration due to gravity is 9.8. But, to make the numbers nice, let's just round to 10 for now—10 m/s². That way, we don't get lost in details and decimals.

So, plus 5 Newtons! So, it's mg when it's just hanging there, but if there's extra forces, it won't be mg. In this case, with a 5 Newton force downward, it's going to be 30 + 5, so it's going to be 35 Newtons.

So, it's equal to mg in the easiest possible case, but there—if there's extra forces, it won't be equal to that. Also, if there was another rope pulling up in the same spot as the first rope, now I'd have two tensions pulling up. Now, I've got another tension force pulling out this way.

So, what would I do now? I'd have to add another T. I'd have plus T. Well, T + T is just 2T, so I'd have a 2T here to solve for T. I'd just have to do 35 Newtons divided by 2.

So, you can start making these harder and harder and harder, and you could do that by adding extra ropes or adding forces down, and you might convince yourself erroneously you have to do something new, but you don't! You still just draw your force diagram, then you go to Newton's second law.

You put those forces in carefully, you solve for what you want to know. And if this is confusing, why is it 35 over 2? It kind of makes sense. The total downward force is 35 Newtons 'cause this is 30 Newtons of weight here. We've got 35 Newtons downward.

The total upward force has to be 35, and if these strings are attached at the same point, they're both going to bear the same amount of weight. They have to total up to 35, so you just get 35 over 2 amongst each of them because they each bear half the weight.

So, that was one of the easiest problems; we turned it into a little bit of a harder problem. It gets even harder if we add an angle. Remember those jalapenos were at an angle. We got to do this! We got to do one at an angle!

So, you got a chalkboard hanging from two strings. One string is horizontal, one string is up here at an angle. What do you do now? It's easy to convince yourself you got to try something new or go for a new strategy, but you don't! You solve this the same way.

We're going to draw our forces. We've still got a force of gravity down, so this force of gravity is just mg. For consistency's sake, let's say the mass of this chalkboard is also 3 Kg.

All right, what else do I have? I'm going to have a tension that points up to the right. It does not push. This is a rope; it can only pull. So, this tension's got to pull this way. We'll call this T1.

So, this is T1, and over on my force diagram, it would look like this. I'd have a T1 that points something like that. So, here's my T1. I'll put it right here—T1. And I've got one more force now; I've got this horizontal force here again.

It does not push; this is a rope; it can only pull. So, it pulls to the left. I'll call this T2. So, on my force diagram, I would have T2. That's it! Those are all my forces. I don't have a normal force; sometimes people want to draw. People are so used to there always being a normal force; they're like, normal force right?

And it's like, no! There's no surface touching this chalkboard; it's just hanging by strings! The only force keeping it up would be the vertical component of this T1. So, this is it. These are the only forces we've got.

What do you do after that? Same as the apple problem. We go to Newton's second law, and we say that the acceleration will be the net force divided by the mass, which direction do we pick? It's not quite as obvious here; we've got forces vertically and horizontally.

So, here's my advice: look for something that you know. In this case, I know the mass is 3 Kg and I know the acceleration due to gravity, or the magnitude of it, is 9.8, but in this case, 10! Since I know this force: remember this force is just 30 Newtons 'cause 3 * 9.8 or 3 * 10 m/s² is just 30 Newtons.

Since I know this force, it's a vertical force. I know something about that direction; I'm just going to start with that direction. See, I already know something about it! So, I'm going to do a in the y direction equals F in the y direction.

We'll start with the vertical direction. If you make a mistake and you pick the wrong direction, it's not the end of the world; just pick the other direction. I mean, there's only two to worry about, so if you screw one up, just go on to the next one! It's not that big of a deal.

All right! Acceleration vertically. Again, let's say this is at rest just hanging from these string rings, so we don't have to worry about any acceleration. Although, if there was, you would just plug that acceleration in here. It's not that much harder of a problem.

Zero equals... all right, what do we got? We need to put our vertical forces up top. I know this one: 30! Me: 30 Newtons downward. So, I'm going to have - 30 Newtons because those 30 Newtons point down, and I'm going to consider downward as negative, upward as positive.

This is really just mg; I could have wrote it as negative mg. What else do I have? I've got T1. T1 points up, but I can't add all of T1 here 'cause it doesn't all point up. I can only add all of T1 if T1 pointed straight upward, but it doesn't!

Part of it points up, so this part of T1 points to the right, pulls to the right. This part of T1 pulls up. It's this component right here that's going to be the component that actually causes this chalkboard to stay up. That keeps it from falling down because that's the part that's fighting gravity.

We'll call this T1 in the y direction; we'll call this T1 in the x direction. So, I need to add plus T1 in the y direction, and that's it! Those are the only two forces that are vertical: T1y and mg. Those are the only two forces that are vertical.

So, now I divide by the mass. I can multiply both sides by mass. m * 0 is still 0. I get 0 = 30 Newtons + T1y. Now, we have to figure out, okay, what T1y has to equal. I can solve this for T1y. I add 30 to both sides!

I'm going to get T1y equals 30 Newtons. And that makes sense! I mean, this T1y is the only component that's balancing out gravity. We know it has to balance because there's no acceleration vertically, so this T1y has to be the exact same size as the force of gravity.

I drew it; it's not proportional here; sorry about that! I should have drawn it with this component exactly the same length as this component because they have to be the same; they have to cancel. But that just tells me T1y!

I want to know what T1 is. How do I solve for what T1 is and what T2 is? These are what I want to figure out. What are the tensions? I don't just want the component; I want the tension!

And so now I say that this component T1y is going to be related to the total T1, and it's related through this angle here. So I can say that T1y, whatever this angle is right here—remember, we can use trigonometry!

We can say that sin(theta) is going to be the opposite side over the hypotenuse. In this case, the opposite side to this angle, opposite, is T1y. So, it's going to be T1y divided by the hypotenuse side, which is the total tension, in this case, we're calling that T1.

So, I want to solve for T1. So I multiply both sides by T1; I'll get T1 * sin(theta), and then I divide both sides by sin(theta). I'll end up with T1 equals T1 in the y direction divided by sin(theta).

I know T1 in the y direction—that was 30, or sorry, not 30—that was 30 Newtons. So, I've got 30 Newtons—that's my force upward. This vertical component right here had to be 30 Newtons 'cause it had to balance gravity divided by sin of the angle.

But what is this angle? We know this angle is 30, and you could probably convince yourself if I draw a triangle this way. Let's try to figure out—we want to figure out what this angle is right here because that's what this angle is here. So, if this is 30, and that's 90, then this has to be 60.

And if that's 60, and this is 90, then this has to be 30. So, this angle is 30° right here. So, that's 30°. So, this angle right here, which is this angle right here, has to be 30.

So, when I'm taking my sin, I'm taking my sin of 30°, and I get 30 Newtons divided by sin of 30. And sin of 30 is 1/2, so I get that this is 60 Newtons. And that might seem crazy! You might be like, wait a minute! T1 is 60 Newtons? 60 Newtons?!

The weight of this chalkboard is only 30! How in the world can the tension in this rope be 60 Newtons? I mean, if we just hung it by a single string, if we just hung this chalkboard by a single string over the center of mass, you'd just get a tension of 30 Newtons!

How can this be 60 Newtons? And the reason is this part has got to be 30 Newtons. We know that because it has to balance gravity, but that's only part of the total tension. So, if the total tension—if part of the total tension is 30, all of the tensions got to be more than 30.

In this case, it's 60 Newtons, so that's why it's larger in this case, 'cause it's at an angle. So, this component has to equal gravity, and this total amount has to be bigger than that so that its component is equal to gravity.

All right! How do we figure out T2? Well, you don't invent a new strategy. We keep going. We say that the acceleration in the horizontal direction is the net force in the horizontal direction divided by the mass.

So, we still stick with Newton's second law even when we want to find this other force. This force is horizontal, so it makes sense that we're going to use Newton's second law for the horizontal direction. Again, if this chalkboard is not accelerating, the acceleration is zero!

So, I'll draw a line here to keep my calculations separate. Equals net force in the x direction. Okay! Now I'm going to have T1 in the x direction, so this is going to be T1 in the x direction.

So, I'll have T1 in the x direction. That's positive because it points right, and I'm going to consider rightward positive. Minus T2—all of T2. I don't have to break T2 up; T2 points completely in the horizontal direction.

Now I divide that by the mass, and well, I can multiply both sides by mass. I'd get 0 equals T1 in the x direction minus... excuse me—minus T2. So if I solve this for T2, I'm going to get that T2, if I add T2 to both sides, I get that T2 just equals T1 in the x direction.

But how big is T1 in the x direction? We know T1 is 60 Newtons. We know T1 in the y direction—this piece here was 30 Newtons. How big is this piece? Well, we can use, instead of sin, now we can use cosine.

So if I use cosine, I can get the cosine theta, which is 30° here because this angle here is 30. The cosine of 30 would be the adjacent—that's this T1x. So, it's adjacent over the hypotenuse.

The hypotenuse is T1, and we know T1 was 60, so I can solve for T1x, and I get that T1 in the x direction, if I multiply both sides by 60 Newtons. This is 60 Newtons! I get the T1 in the x direction would be 60 Newtons * cos of 30, and the cosine of 30 is √3/2.

So I get 60 Newtons * √3/2, which means that T1 in the x direction is 30√3 Newtons. And that's what I can bring up here! This is T1x!

So since that's T1x, I can say T1x right here is 30√3 Newtons. And by Newton's second law in the horizontal direction, that's what T2 had to equal!

So, T2 equals 30√3 Newtons. And that shouldn't be surprising; this force here, in order to make it so that there's no acceleration horizontally, just has to equal this force here! Those are the only two horizontal forces. We knew T1x was 30√3. That's what we found.

So that means T2 also has to be 30√3 to make it so that these forces are balanced in the horizontal direction. All right! So, we did it! We figured out T1: 60 Newtons! We figured out T2: 30√3 Newtons! Now, we're ready.

Now we can figure out the super hot jalapeno problem! We'll do that in the next video!