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Current time:0:00Total duration:10:19

Video transcript

welcome back will now do another attention problem and this one is just a slight increment harder than the previous one just because we have to take out slightly more sophisticated algebra tools and reads in the last one but but it's not really any any harder but you should actually see this type of problem because you'll probably see it on an exam so let's figure out the tension in the wire so first of all we know that this point right here isn't moving so the tension in this little small wire right here is easy it's trivial it's the force of gravity is pulling down at this point it with 10 Newtons right because you have this weight here and of course since this point is stationary the tension in this wire has to be 10 Newtons upward that's an easy one so let's just figure out the tension in these two slightly more difficult wires to figure out the tensions of so once again we know that this point right here this point is not accelerating in any direction right it's not accelerating in the x-direction nor is it accelerating in the vertical direction or the Y direction so we know that the net force is in the X direction need to be zero on it and we know that the net force is in the Y direction need to be zero right so what are the net forces in the X direction well they're going to be the X components of these two of the of the of the tension vectors of both of these wires so let's I guess let's draw the tension vectors of the two wires okay so this t1 it's pulling the tension vector pulls in the direction of the wire along the same line so let's say that this is the tension vector of t1 if that's the tension vector its X component will be this let me see how good I can draw this it's intended to be a straight line but that would be its X component and its X component let's see this is 30 degrees this is 30 degrees right here and hopefully this is a bit second nature to you what is if this value up here is t1 what is the value of the X component its t1 cosine of 30 degrees t1 cosine of 30 degrees and you can do your sohcahtoa you know cosine is adjacent over hypotenuse right so the cosine of 30 degrees is equal to this over t1 is equal to the X component over t1 and if you multiply both sides by t1 you this this should be a little bit of second nature right now that the X component is going to be the cosine of the angle between the hypotenuse and the X component times the hypotenuse and similarly the X component here let me draw this force vector so if this is t2 if that's t2 this would be its X component and very similarly this is 60 degrees so this would be t2 cosine of 60 now what do we know about these two vectors we know that when you that their net force is zero or that you also know that the magnitude of these two vectors should cancel each other out or that they're equal I mean we they're pulling in opposite directions that's that's pretty obvious but and so you know that their magnitudes need to be equal so we know that t1 cosine of 30 is going to equal t2 cosine of 60 so let's write that down t1 cosine of 30 degrees is equal to t2 cosine of 60 and then we could we could bring the t2 onto this side and actually let's also I'm trying to save as much space as probable possible because I'm guessing this is going to take up a lot of room this problem what's the cosine of 30 degrees if you haven't memorized it already square root of 3 over 2 so this becomes square root of 3 over 2 times t1 right that's the cosine of 30 degrees and then I'm going to bring this onto this side so the cosine of 60 degrees the cosine of 60 degrees is actually 1/2 you could use your calculator if you forgot that so this is 1/2 T 2 I'm gonna bring it on this side so it becomes minus 1/2 I'm skipping more steps than normal just because I don't want to waste too much space and this equals zero but if you've seen the other videos hopefully I'm not I'm not creating too many gaps and that this is relatively easy to follow so we have the square root of 3 times T 1 minus 1/2 t2 is equal to 0 so that gives us an equation one equation with two unknowns so it doesn't help us much so far but let's let's where that away because I have a feeling this will be useful now what's going to be happening on the y-components so let's say that this is the Y component of t1 and this is the Y component of t2 what do we know what what do we know about the two why can I could have drawn them here too and then just shift them over to the left and the right we know that their combined pull upwards the combined pull of the two vertical tension bones has to offset the force of gravity right pulling down because this point is stationary so we know these two Y components when you add them together have the combined tension in the vertical direction has to be ten Newtons right because it's offsetting this force of gravity so what's this Y component well similar that this was t1 of cosine of 30 this should start to become a little second nature to you that this is t1 sine of 30 this y component right here so t1 let me write it here t1 sine of 30 degrees Plus this vector which is t2 sine of 60 degrees sine of 60 you can review your trigonometry in your circuit oh this is that that tends to be frankly I think what you know just seeing what people get confused on is the trigonometry but you can review the trig modules and maybe some of the earlier force vector modules that we did and hopefully this these will make sense I'm skipping a few steps and these will equal 10 Newton's 10 Newtons and let's rewrite this up here where I substitute the values so what's the sine of 30 actually let me let me just do it right here what's the sine of 30 degrees the sine of 30 degrees is 1/2 so we get 1/2 t1 plus the sine of 60 degrees which is square root of 3 over 2 square root of 3 over 2 t2 is equal to 10 and I don't like this all these twos and these one have this 1/2 here so let's multiply this whole equation by 2 so 2 times 1/2 that's 1 so you get t1 plus the square root of 3 t2 is equal to 2 times 10 is 20 so let's take this equation up here and let's multiply this equation by two and bring it down here so this original one we've got so if we multiply this whole thing by two I would do it in this color so that you know that it's the different equation so if you multiply square root of three over two times two I'm just doing this to get rid of the twos in the denominator so you get square root of three whoops square root of three t1 minus t2 is equal to 0 right because 0 times 2 is 0 and let's see what we could do what if we take this top equation because we want to start canceling out some terms let's take this top equation and multiply it let's multiply it by oh I don't know the - let's multiply it by let's multiply it by the square root of 3 so you get square root of 3 t1 right I'm taking this top equation multiply square root 3 this is just a system of equation so I'm solving for and so square root of 3 times this right here square root of 3 times square root 3 is 3 so + 3 t 2 is equal to 20 square roots of 3 and now what i want to do is let's let's I know I'm doing a lot of the equation manipulation here but this is just hopefully a review of algebra for you let's subtract this equation from this equation right so you can also view it as multiplying it by negative 1 and then adding the two so when you subtract this from this these two terms cancel out right because they're the same and so then you're left with minus t2 from here - this - 3 T 2 is equal to 0 - 20 square roots of 3 and so this becomes minus 4 T 2 is equal to minus 20 square roots of 3 and then divide both sides by minus 4 and you get t 2 is equal to 5 square roots of 3 so Newton's so that's the tension in this wire and now we can substitute and figure out t1 so with us let's do it let's use this formula right here because it looks suitably simple so we have the square root of three times t1 minus t2 well t2 is 5 square roots of 3 right five square roots of three is equal to zero so we have the square root of 3 T 1 is equal to 5 square roots of 3 divide both sides by square root of 3 and you get the tension and the first wire is equal to 5 Newtons so this is pulling with a force of tension of 5 Newtons or force and this is pulling the second wire with the tension of 5 square roots of 3 Newtons so this this wire right here is actually doing more of the pulling it's actually more of the force of gravity zuv is ending up on this wire and that that makes sense because it's deeper so since it's deeper it has its kind of it's contributing more to the the Y component it's good whenever you do these problems to kind of do a reality check just to make sure your your numbers make sense and if you think about their combined tension their combined tag tension is something more than 10 Newtons and that makes sense because some of their some of the force that they're pulling with is wasted against pulling each other in the horizontal direction anyway I'll see you all in the next video