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

0 energy points

# Vectors word problem: pushing a box

Created by Sal Khan.

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 box is, right over here, and this is the target right over here. Let me write that. That is the target, that is where they are trying to get the box. And person A, because they can't for some reason, they can't push it exactly from behind the box, maybe there is not a good footing there, or I guess if they press there the box squeezes a little bit. So person A has to push it in a direction that is not exactly going in the direction of the target. So they push it in a direction that looks like this. Let me show you there, this vector I'm drawing essentially represents the force that they are exerting. This is their force vector. This is person's A force vector. And we know that the length of this vector, or another way of think about it, the magnitude of vector a is 330 N. And let's say that person B once again, because 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 right over here, so that right over there, that vector is the force of person b is pushing on to on the direction, and then the magnitude of that force, of that vector, of person b is pushing, is 300 N. And we know their angle that those make with the direction of the target. So this is the direction of the target right over here. We know that this is a 35 degree angle, and we know that 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 who is actually exerting more force in that direction? We see that person a is exerting a total, a higher magnitude force in this direction. And person b is doing in that direction. 330 N versus 300. But who is helping the box go more in that direction? And by how much more? And also, what is the total force now pushing the box in that direction? So I am assuming you have a 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 vector a. So vector a looks like this, and I will just draw it separately, so we just can clean it up a little bit, vector a looks like this. We know its magnitude; the magnitude of vector a is equal to 330 N. And if we say that the target is in this direction right over here, so that is the target, in some place out here, we have already been told that this is 35 degrees. So what we really wanna do is find the component, 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 a sub x, and we already know that the magnitude of a is 330 N. So the magnitude, let me just say, let me just write this way so we don't confuse, so the magnitude of the vector in the x direction, this vector right over here, we could write like this, the magnitude of that, we will just write it without the vector notation. So how can we think about that? Well, we know cosine is adjacent over hypotenuse. So we could write cosine of 35 degrees is equal to the length of the adjacent side, that would be a sub x, without the vector over it, we are just saying that it is the magnitude of vector a sub x, over the magnitude of vector a, over 330 N. Or we could say that a sub x is equal to 330 times the cosine of 35 degrees. And we can make the exact same argument for b. Vector b, so let me draw like this, vector b, maybe I will draw like this, let me do a little bit different. If this is the direction to the target, so once again I will just draw an horizontal line for that, than relative to that, vector b looks like something like this, so that is vector b. b sub x, in the direction of the target, so we would drop a perpendicular like that, 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? We could say that the magnitude of b sub x, we will just call that b sub x without the vector notation. Same exact logic. 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 N. So we could write that cosine of 15 degrees is equal to b 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 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 N, so that is a sub x, it is 270 N. And b sub x is 300 times cosine of 15 degrees, and we get 289, 777. 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 component of vector a going. So if we are rounding to the nearest Newton, this right over here, the magnitude of this vector right over here, b sub x, that is, if we round to the nearest Newton, 290 N. So this is approximately 290 N length, or I guess I can say magnitude. Well this one is a little bit shorter, we saw that if we round to the nearest Newton it is about 270 N. The length of this one is 270 N approximately. So if you would say how much more is person b pushing in the direction that we care about, it is about, well if we want to be a little bit more precise, we can subtract the two. So we can take 300 cosine of 15 minus 330 cosine of 35 and we get about 19,5 N of difference. That the blue, person b, is contributing 19,5 - 19,458- Newtons more in that direction towards the target than person a is. But if we want to talk about what is the total force going in that direction, than we would take the sum of these two things. So we would, the total force in that direction is going to be 560 N, if we round to the nearest Newton. So if you add this blue component to this pink component, you get this one right over here, which is 560 N. So this whole vector right over here, its magnitude, so I could write that as the magnitude of a sub x plus b sub x, which is the same thingas 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 560 N.