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Current time:0:00Total duration:8:56

Vector components from magnitude & direction: word problem

Video transcript

let's say that you have two folks that are trying to collectively push a box across the snow towards a target so this is where the boxes right over here and this is the target right over here let me write that that is the target that's where they're trying to get the box and person a person because they can't for some reason they can't push it exactly from behind the box maybe there's not a good footing there or I guess if they press there maybe the Box squeezes is a little bit so person a has to push in a direction that's not exactly going in the direction of the target so they push in a direction that looks like this and so let me show you there this this vector I'm drawing essentially represents the force that they are exerting this is their force vector this is person a's force vector so this is person a's force vector and we know that the length of this vector or one other way to think about the magnitude of vector a is 330 Newtons 330 Newtons and let's say person B once again because they can't they can't they can't push exactly in the direction of the target maybe the box is really soft right over here person B is pushing it this angle person B is pushing it this angle right over here so that right over there that vector is the force that person B is is pushing onto it in the direction and there the magnitude of that force of that vector of person B is pushing is 300 Newtons and we know their angles that those make with the direction of the target so if this is the direction of the target right over here we know that this is a thirty five degree angle thirty five degree angle and we know that this is a this is a 15 degree angle now what I want you to do is pause this video and think about how much of each of their force is going in the direction of getting the Box towards the target and then who is actually exerting more force in that direction we see that person a is exerting a total a higher magnitude of force in this direction than person B is doing in that direction 330 Newton's versus 300 but who's helping the Box go more in that direction and by how much more and also what's the total force now pushing the box in that direction so I'm assuming you how to go go at it and the key here is to find the component of each of these vectors the magnitude of each of these vectors in this direction in the direction towards the target and so let's look first at let's look first at vector a so vector a looks like this and I'll just draw them separately just so we can clean it up a little bit vector a looks like this we know it's a magnitude the magnitude of vector a is equal to 330 Newtons and if we say that the target is in this direction right over here so that's the target is someplace out here we've already been told that this is 35 degrees so what we really what we really want to do is find the component the the magnitude of the component going in that direction right over there and the way we can do that is just with our traditional trig functions this right over here is a right triangle we are looking for this side right over here we have let me just call that let me just call that a sub X and we already know that the magnitude of a is 330 Newtons so the magnitude let me just say that the magnet let me just write it this way just so we don't confuse so let's just say the magnitude of the vector in the in the X direction so this vector right over here we could write like this the magnitude of that will just write it without the vector notation so how could we think about that well we know cosine is adjacent over hypotenuse so we could write cosine cosine of 35 degrees is equal to the length of the adjacent side that would just be a sub X without the vector over it what you're saying that that's the magnitude of vector a sub X over the magnitude of vector a over 330 Newton's over 330 Newtons or we could say that a sub X is equal to 330 330 times the cosine of 35 degrees and we can make the exact same argument for B vector B so let me draw it like this vector B you could maybe I'll draw it like this just to make it a little bit let me do a little bit different if this is the direction to the target so once again I'll just draw a horizontal line for that then relative to that vector B looks something like this it looks something vector B looks something like that so that is vector V V sub X in the direction of the target so we would drop a perpendicular like that this would be this right over here would be the vector B sub X and so what is the magnitude of B sub X going to be equal to and we could say that the magnitude of B sub X we'll just call that V sub X without the vector notation same exact logic this is this is 15 degrees cosine of 15 degrees is going to be the length of the adjacent side over the length of the hypotenuse we already know that the length of the hypotenuse is 300 Newton's so we could write that cosine of 15 degrees is equal to V sub X length of the adjacent side over the length of the hypotenuse or that B sub X is equal to 300 times cosine of 15 15 degrees so let's get our calculator out and let's calculate what these things are so let's see we have 330 times cosine of 35 degrees gets us to 270 Newtons so that's a sub X is 270 Newtons and B sub X is 300 times cosine of 15 degrees we get two hundred and eighty nine point seven seven seven so what we see is even though B's magnitude is less than A's magnitude the component of vector B going in the direction of the target is actually larger than the come vector-a going so if we were rounding to the nearest to the nearest Newton if we were rounding to the nearest Newton this right over here the magnitude of this vector right over here B sub X that is 200 if we round to the nearest Newton 290 Newton's so this is approximately 290 Newton's length or I guess you say magnitude while this one this one is a little bit shorter is a little bit shorter we saw if we round to the nearest Newton it's about 270 Newton's the length of this one the length of this one is 270 Newton's approximately so if you were to say how much more is person B pushing in the direction that we care about it's about well if we want to be a little more precise we can subtract to the two so we could take 300 cosine of 15 minus 330 cosine of 35 and we get about 19 and a half Newton's difference that the blue person B is contributing 19 and a half 19 point four five eight Newtons more in that direction towards the target than person a is but if we wanted to talk about what is the total what is the total force going in that direction then we would take the sum of these two two things so we would the total force in that direction is going to be 560 Newtons if we round to the nearest Newton so if you add this blue component to this magenta component you get this one right over here which is 560 Newtons so this whole vector right over here its magnitude its magnitude so I could write that as the I could write that as the magnitude of a sub X plus B sub X which is the same thing as a sub X plus B sub X I already said that these are the equivalent of the magnitude of each of these vectors is equal to I could write approximately equal to 500 60 Newtons