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Current time:0:00Total duration:16:56

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 as 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 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 tangent problems because it's going to lead us to the answer again be careful don't 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 force of gravity on this can of red peppers which is mg and if it's three kilograms we know three kilograms times about ten if we're going to say let's approximate G is ten again to make the numbers come out nice so instead of using 9.8 we'll say G is about ten and so we'll say three kilograms times 10 meters per second squared is going to be 30 Newtons so the force of gravity downward is 30 Newtons what are the 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 gonna 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 poles just like all tension tension poles tension can't push so I've got tension two 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 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 force of gravity is 30 30 Newtons usually that's a good strategy pick the direction that you know something about at least so I'm 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 an elevator transporting these peppers up or down and it's not a no rocket if it's just seeing 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 negative 30 Newtons I could have written negative mg but we already knew it was 30 Newtons all right negative 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 t1 Y is going to get included into this calculation because this calculation only uses Y directed forces and the reason is only Y directed forces vertical forces affect the vertical acceleration so this t1 Y 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 is 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 three kilograms but we're going to multiply both sides by three kilograms we're going to 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 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 would find 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'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 these things in a train car something in 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 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 T one in the X so part of this T one points in the x direction similarly part of T two points in the X direction so I'll call this T two X will use these are the magnitudes let's say T two x is the magnitude of the force the T two pulls with to the left and T 1 X is the magnitude of the force that T 1 pulls with to the right so to plug these in we've got to decide whether they should be positive or negative so this T 1 X since it pulls to the right T 1 X 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 T 2 X since it pulls to the left that's going to be a negative contribution so minus T 2 in the X direction because leftward would be negative we divide by the mass the mass was three kilograms but again we'll multiply both sides by three 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 I can solve for T 1 X but look at what I get if I just multiply added T 2 X to both sides I'm just going to get t1 in the X Direction has to equal T 2 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 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 the X direction we can't solve this equation we got for the y direction whenever this happens when you get two equations 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 t1 X t2 X t1 Y and t2 y 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 t1 Y in terms of total t1 and then signs of angles and cosines of angles and I put t2 Y in terms of t2 and angles 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 t1 Y 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 an ID on this side 60 the sides got to be 30 similarly this sides 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 because 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 T 1 Y 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 sine of 30 is opposite over hypotenuse and in this case the opposite side is t1 y so t1 Y over the total t1 is equal to sine of 30 and we can solve this for T 1 Y now we can get the t1 Y 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 I forgot the 1 tu 1 times sine of 30 similarly if you do the same thing with cosine 30 you'll get the t1 X is t1 cosine 30 by the exact same process similarly over here t2 is going to be our sorry T 2 X is going to be t2 so t2 cosine 60 because this is the adjacent side and t2 Y 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 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 t1 why this expression here t1 sine 30 in 41 Y similarly if I plug in T 2 y is t2 sine 60 into this expression right there 42 Y look at what I'll get I'll get 0 equals so I'll get negative 30 Newtons and then I'll get plus t1 why was t1 30 so t1 and then sine 30 we can clean this up a little bit sine 30 is just 1/2 so I'll just write t1 over 2 and then because sine 30 is just 1/2 and then t2 Y is going to be t2 sine 60 and sine 60 is just root 3 over 2 so I'll write this as plus T 2 over 2 and then times root 3 and you might think this is no better I mean this is still an horrible mess right here but look at 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 t1 X is t1 cosine 30 so I'm going to write this as T 1 times cosine 30 and cosine 30 is root 3 over 2 so this is T 1 over 2 times root 3 and that should equal t2 x is right here that's t2 cosine 60 cosine 60 is 1/2 so t2 X is going to be T 2 over 2 so T 2 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 what I have now I've got one equation with t1 and t2 I've got another equation with t1 and t2 so 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 is easier so I'm going to solve this one for let's just say t2 so if we solve this for T 2 I get that T 2 equals well I can multiply both sides by 2 and I'll get T 1 times root 3 so t1 times the root 3 because the 2 here cancels with this 2 or when I multiply both sides by 2 it cancels out so get the 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 now if I do this I'll get zero equals negative you know let's just move the negative 30 over this is kind of annoying here just add 30 to both sides take this calculation here we'll get plus 30 equals and then we're going to have t1 over 2 from this t1 so T 1 over 2 plus I've got plus t2 is t1 root 3 so when I plug t1 root 3 in 42 what I'm going to get is I'm going to get t1 root 3 and then times another root 3 because t2 itself was t1 root 3 so I'm taking this expression here plugging it in 42 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 I get t1 times three halves plus t1 over 2 so I'll get 30 equals and then I get T 1 over 2 we're almost there I promise T 1 over 2 plus and then this is going to be t1 times 3 over 2 so it's going to be 3 T 1 over 2 well what does that equal T 1 over 2 plus 3 T 1 over 2 is just just 4 halves so that's just 2 T 1 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 here by 2 and this side here this right side by 2 I get t 1 is 30 over 2 Newton's which is just these should be Newton's I should have units on these which is just 15 Newton's whoo I did it 15 Newton's t 1 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 nah not really that would be terrible you actually just take this t1 and you plug it right into here boot there he 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 I get the t2 is just 15 times root 3 Newtons so once you get one of the forces the next one's really easy this is just T 2 so T 2 is root 15 root 3 and T 1 is just 15 so in case you got lost in the details the big picture recap is this we drew a force diagram we use Newton's second law in the vertical direction we couldn't solve because there were two unknowns we use 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