Vector word problem: resultant force | Vectors | Precalculus | Khan Academy

We're told that a metal ball lies on a flat horizontal surface. It is attracted by two magnets placed around it. We're told that the first magnet's force on the ball is five newtons. We're then told the second magnet's force on the ball is three newtons in a direction that is 100 degrees rotation from the first magnet's force, and we can see that drawn here. This is the first magnet's force, it's five newtons, and then the second magnet's force is three newtons at a 100-degree angle, 100-degree rotation from the first magnet's force.

Now they're asking us a few interesting questions. What is the combined strength of the magnets' pulls? And then they also say, what is the direction of the magnets' combined pulls relative to the direction of the first magnet's pull? So I encourage you to pause this video and have a go at this on your own before we work through this together.

All right, now let's work through this together. So they're really saying, if I take the sum of these two vectors, what is going to be the resultant force vector? What is going to be the magnitude of that resultant force vector, and what is its direction going to be?

There are two ways we could approach this. We could break down each of these vectors into their respective components and then add the respective components, and then from that figure out what the magnitude and direction is. We do that in other videos. Or we could take the geometric approach, so that's what we're going to do here. To help us with that, we're going to use what we've called the parallelogram rule, which is really the same idea as the head-to-tail addition of vectors.

I could take the 3-newton vector, I can shift it over so its tail is at the head of the 5-newton vector. It would look something like this. So this is 3 newtons right over there. And then I could also go the other way around; I could take the 3-newton vector first and take the tail of the 5-newton vector at the head of the 3-newton vector and shift it like this. You can add in either direction, and either way you look at it, when you start at the tails and you get to the head of the second vector, you're going to have a resultant force that looks like this, which is the diagonal of this parallelogram.

So there we go, and let me just call that our force vector right over there. So if we can figure out the length of this line, of this diagonal right over here, that would be the magnitude of this force vector. Now how can we do that? Well, let's just think geometrically about what else we can figure out about what's going on over here.

This is a parallelogram, so this is a hundred-degree angle right over here. This angle right over here is also going to be 100 degrees. We also know that these two opposite angles are also going to have the same measure right over here, and we also know that the sum of all of the angles in a quadrilateral are going to be 360 degrees. So if these two make up 200 degrees, we have 160 degrees left that have to be split between that one and that one. So we know that this is 80 degrees and we know that this is 80 degrees.

Well, how does that help us? So we know the length of this brown side; we know the length of this side right over here. We know the angle between them, and what we're trying to do is figure out the length of the side opposite this angle, opposite this 80-degree angle. Some of you might remember the law of cosines here, and the law of cosines I always imagine as an adaptation of the Pythagorean theorem so that we can deal with non-right triangles.

The law of cosines will tell us that the magnitude, I'll just write it over here, the magnitude of this vector, which is the length of this diagonal, is going to be equal to the square root of— we're going to have this side squared, so let me write 3 squared plus this side squared, plus 5 squared minus 2 times this side—so times 3 times that side, so times 5 times the cosine of 80 degrees.

And so let's get our calculator out to calculate that. I'll start with taking the cosine of 80 degrees. Then I'm just going to multiply that times looks like 30. So times 30 is equal to that. Let's put a little negative there, and then to that I'm going to add 25 and 9, which is 34. So plus 34 is equal to that.

And now I just take the square root of all of that, and they tell us to round our answer to the nearest tenth, so I can round this to approximately 5.4. So this is approximately 5.4 newtons. Now they say, what is the direction of the magnets' combined poles relative to the direction of the first magnet's pole?

So really, what we want to do is figure out this angle right over here; let's call that theta. Well, we know what the length of the side opposite is, so maybe we could use the law of sines. The law of sines would tell us that the sine of theta over the length of the side opposite to it is going to be equal to—let's pick another angle we know—sine of this angle, sine of 80 degrees, over the length of the side opposite to it.

And so this is approximately 5.4. And so if we want to solve for theta, we can multiply both sides by three. So we're going to get sine of theta, I'll just stay in this purple color for simplicity, is equal to three times sine of 80 degrees divided by 5.4.

And then we could say that theta is equal to the inverse sine of all of this business: 3 sine of 80 degrees over 5.4. So we're going to take 80 degrees, take the sine of it, we're going to multiply that by 3, divide that by 5.4. That equals that, and then I'm going to take the inverse sine of all of that, and they want us to round to the nearest integer. So that's approximately 33 degrees.

When you do the law of sines, it's possible that you're also dealing with an obtuse angle, and when you do all of this you get the acute one, and then you would have to make an adjustment; but that's not what we're dealing here. So we know that this theta is approximately equal to 33 degrees, so we know the magnitude of the force, and we know that it forms an angle of approximately 33 degrees with the direction of the force of that first magnet.