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Current time:0:00Total duration:17:30

Video transcript

you bought a huge canister aluminum can of super hot red peppers three kilograms worth and you hung them from two strings from the ceiling because you don't want anyone to get your super hot red peppers and 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 is hard this problem spicy this is a spicy tension problem let's start with something a little more mild and 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 three kilograms so we can keep the same number three kilogram apple hanging from a string we want to know what's the tension in the rope well this one's easy the way you 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 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 the rope so this tension force points upward I'll call it T that's always the first step you draw your force diagram and 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 we pick it's pretty obvious here we'll pick the vertical direction because there are no forces in the horizontal direction and I'll plug in the acceleration if this apples 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 they 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 celebration is going to be zero that's going to equal the net force we've got tension upward to figure out what goes here like we draw this force diagram for a reason this isn't just like happy fun painting time over here where you see this is a strategy we draw these forces because 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 and how about mg over here mg points down so if 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 you use Newton's second law and then you try to solve for the force that you want in this case is the tension so I multiply both sides by M I'd still 0 on the left-hand side so 0 equals t minus mg and if I solve this for T I get something that might not be all that surprising I just get the tension is equal to mg and you might be like well duh could have done that with 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 and 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 we'll imagine that it's imagine I had say two tensions up here let's start with this let's say I just pull down on the rope I just come over here and someone someone pulls down on the Apple let's say someone just pulls down the apple with an extra 5 Newtons all right pulling this Apple which pulls this string which makes it tighter what I do in that case well I've got one more force here in my force diagram I just add that so that's five Newtons it's pointing downward so when I come over here to my net force I've got to subtract 5 Newtons because that's pointing downward and now I do my algebra just as before I multiply both sides by M I'll have another minus 5 Newtons here and then when I solve for T I'll add mg to both sides and I'll add 5 Newtons to sides and so this would be my four so with numbers what are we going to get the mass was three so we'll have three kilograms times the acceleration due to gravity is 9.8 but to make the numbers nice let's just round to 10 for now 10 meters per second squared that way we don't get lost in details and decimals so +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 plus 5 so it's going to be 35 Newtons so it's equal to mg in the easiest possible case but 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 have two tensions pulling up now I've got another tension force pulling up this way so what would I do now I'd have to add another T I'd have plus T well T plus T is just 2t so I'd have a 2t here I'd have a 2t here to solve for T I just have to do 35 Newton's divided by 2 so you can start making these harder and harder and harder and you can 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 and 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 because this is 30 Newton's of weight here we've got 35 Newtons of 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've got a chalkboard hanging from two strings one strings 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 and for consistency's sake let's say the mass of this chalkboard is also three kilograms all right what else do I have I'm going to have a tension that points up into the right it does not it doesn't point this way tension doesn't push tension pulls so this tensions got a 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 they're always being 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 be the net force divided by the mass which direction do we pick is 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 three kilograms and I know the acceleration due to gravity or the magnitude of it 9.8 but in this case 10 since I know this force remember this force is just 30 Newtons because 3 times 9.8 or 3 times 10 meters per second squared 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 because I already know something about it so I'm going to do a in the y-direction F in the y-direction we'll start with the vertical direction if you make a mistake in you picked 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 she's going 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 strings 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 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 feet 30 Newtons downward I'm gonna have negative 30 Newtons because those 30 Newtons point down and I'm going to consider downward as a negative upward is positive this is really just mg I could 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 because 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 pulled 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 t1 X is not vertical that's horizontal and this t2 is horizontal so I've included t1 Y and mg those are the only two forces that are vertical so now I divide by the my mass I can multiply both sides by mass M times 0 is still 0 I get 0 equals negative 30 Newtons plus t1 Y and now we have to figure out okay like t1 y t1 Y has to equal what I can solve this for t1 why I add 30 to both sides I'm going to get t1 y equals positive 30 Newtons and that makes sense I mean this t1 Y is the only component that's balancing out gravity we don't have to balance because there's no acceleration vertically so this t1 Y 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 t1 why 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 t1 Y is going to be related to the total t1 and it's related through this angle here so I can say that t1 Y whatever this angle is right here remember we can use trigonometry and we can say that sine theta is going to be the opposite side over the hypotenuse and in this case the opposite side to this angle opposite is t1 Y this is going to be t1 Y divided by the hypotenuse side is the total tension so that's always the total magnitude of the force in this case we're calling that t1 so I want to solve for t1 so I'm going to if I multiply both sides by t1 I'll get t1 times sine theta and I divide both sides by sine theta I'll end up with t1 equals t1 in the Y Direction divided by sine theta I know t1 in the Y direction that was 30 degrees or sorry not 30 degrees out of 30 Newton's so I've got 30 Newtons that's my force upward this vertical component right here had to be 30 Newtons because it had to balance gravity divided by sine 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 degrees right here so it's 30 degrees so this angle right here which is this angle right here has to be 30 so when I'm taking my side I'm taking my sine of 30 degrees and I get 30 Newton's divided by sine of 30 and sine of 30 is 1/2 so 0.5 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 Newton's 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 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 and in this case it's 60 Newtons so that's why it's a larger in this case because 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 T - well you don't invent a new strategy we keep going we just going to 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 we'll draw a line here to keep my calculations separate equals net force in the x-direction okay now imma have t 1 in the x-direction so this is going to be T 1 in the X so I'll have T 1 in the x-direction that's positive because it points right I'm going to consider rightward positive minus T 2 all of T 2 I don't have to brake T 2 of T 2 points completely in the horizontal direction and I divide that by the mass and well I can multiply both sides by mass I'd get 0 equals T 1 in the x-direction - scuse me - t 2 so if I solve this for T 2 I'm going to get the T 2 if I add T 2 to both sides I get the T 2 just equals T 1 in the x-direction but how big is T 1 in the x-direction we know T 1 is 16 we know t 1 in the Y direction this piece here was 30 Newtons how big is this piece well we can use instead of sine now we can use cosine so if I use cosine I can get the cosine theta which is 30 degrees here because this angle here is 30 cosine of 30 would be the adjacent that's this t1 X so it's adjacent over the hypotenuse the hypotenuse is t1 and we know T 1 t1 was 60 so I can solve for t1 X and I get that t1 in the X direction if I multiply both sides by 60 Newton's this is 60 Newton's I get the t1 in the X direction would be 60 Newton's times cosine of 30 and the cosine of 30 is cosine of 30 is root 3 over 2 so I get 60 Newton's times a root 3 over 2 which means that T 1 and the X Direction is 60 over 2 would be 30 so this is 30 root 3 Newtons and that's what I can bring up here this is T 1 X so since that's t1 X I can say t1 X right here is 30 root 3 Newtons and by Newton's second law in the horizontal direction that's what t2 had to equal so t2 equals 30 root 3 Newtons we figured it out t2 equals 30 root 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 t1 X was 30 root 3 that's what we found that means t2 also has to be 30 root 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 root 3 Newtons now we're ready now we could figure out the super hot jalapeno problem we'll do that in the next video