Four forces act on a particle as shown in the diagram. Determine 𝑅, the magnitude of their resultant, and find 𝜃, the angle between their resultant and the 𝑥-axis. Give your answer to the nearest minute if necessary.
We begin by simply labeling our four forces. Now these forces are acting in more than one direction. So they’re vectors. We have force one, 𝐅 one; force two, 𝐅 two; 𝐅 three; and 𝐅 four. The resultant of these forces is the vector sum of all four forces. It’s 𝐅 one plus 𝐅 two plus 𝐅 three plus 𝐅 four. But what are these four forces?
Well, we define 𝐢 as the unit vector acting in the horizontal direction, whereas 𝐣 is the unit vector that acts perpendicular to this, in the positive 𝑦-direction. And this means 𝐅 one is fairly straightforward to find. It’s simply eight 𝐢. The other three vectors though are going to require a little more work.
Let’s take our second force, 𝐅 two. It has a magnitude of five newtons. And it acts at an angle of 60 degrees to the positive horizontal axis. We need to split this into its horizontal and vertical components. And so we’re going to use right-angle trigonometry to do so. The hypotenuse we can say is five. Let’s call the adjacent side 𝑥 and the opposite side 𝑦. Then we can use the cos ratio to find 𝑥. We can say that cos of 60 degrees is adjacent over hypotenuse. It’s 𝑥 over five. And so 𝑥 equals five cos 60. Cos 60 is of course one-half though, so we find 𝑥 is equal to five over two.
We repeat this process for 𝑦. Sin 60 is equal to 𝑦 over five. And so 𝑦 is equal to five sin 60. This time though, sin of 60 degrees is root three over two. So we say that 𝑦 is equal to five root three over two. The vector 𝐅 two is therefore five root two 𝐢 plus five root three over two 𝐣.
Let’s repeat this process for force three. We’ll need to be a little bit careful with force three. We’ll find the sides of the triangle, but we need to notice that the horizontal component is going to be negative. It’s acting in the opposite direction. This time, the hypotenuse is three. We end up with an opposite side of three sin 60 and an adjacent of three cos 60. That gives us an opposite or a vertical component of three root three over two and a horizontal component of three over two. But remember, this is acting in the negative direction. So it’s negative root three over two. And we therefore say that vector three is negative three over two 𝐢 plus three root three over two 𝐣.
We repeat this for our fourth and final vector. This time, we need to notice that the horizontal and the vertical components are both going to be negative. The hypotenuse is six root three. Then we end up with a vertical component of six root three cos 60 — that’s the adjacent — and a horizontal component of six root three sin 60. Six root three cos 60 is three root three. So the vertical component is negative three root three. Then six root three sin 60 is nine. So the horizontal component is negative nine. And 𝐅 four is therefore negative nine 𝐢 minus three root three 𝐣.
Remember, the resultant is the sum of all of these forces. So let’s add together the horizontal and vertical components separately. The vertical components are eight, five over two, negative three over two, and negative nine. Whereas the vertical components are five root three over two, three root three over two, and negative three root three. This simplifies, and we end up with zero 𝐢 plus root three 𝐣, or simply root three 𝐣. So the resultant force acts purely in the vertical direction.
Now, of course, we’re looking to find the magnitude of this resultant. Normally, when finding the magnitude, we’d look to use the Pythagorean theorem. It’s the length of the vector. In this case though, since the vector is acting in only one direction, the magnitude is simply root three. It’s the length of the 𝐣 component. Similarly, since it’s acting in the vertical direction, it’s perpendicular to the positive 𝑥-axis. And so the angle 𝜃 between the resultant and the 𝑥-axis is 90 degrees. And we finished. 𝑅 is root three or root three newtons, and 𝜃 is 90 degrees.