Analytically find the magnitude of the resultant of the sum of the following vectors. Also, determine the angle it forms with respect to the positive x-axis. Okay, I'll tolerate it. So, we observe that we have 6 vectors that we are asked to add together, but how are we going to determine this sum? Well, for this I'm going to apply the rectangular component method, and in this case, remember that what we have to do is decompose each of our vectors precisely into its rectangular components. And for this, I'm going to do the following: I'm going to start by making a table with four columns, and in the first I'm going to write the magnitude of our vector, and in the second the direction, in the third the x-component, and in the fourth the y-component. And then I'm going to fill in this table, and I'm going to start by writing the magnitude of vector number one, which corresponds to this one here, and it's equal to three newtons. Now, what is the direction of this vector? Well, for that, remember that the direction of a vector is the angle it forms with respect to the positive x-axis. So, in this case, this horizontal line represents the x-axis, and it's positive to the right. So, what angle does this vector form with respect to the x-axis? The positive vector is this one here, but we don't know its exact value. However, we do know that the angle formed from this side to the other is 180 degrees. Therefore, we simply subtract this 45-degree angle from 180 to find its direction. 180 minus 45 degrees equals 135 degrees. Therefore, this is the direction of this vector, and I'm writing it in this column. Now that we have this, the x-component of this vector will be equal to the magnitude of our vector multiplied by the cosine of the angle, which is the direction. That is, by the cosine of 135 degrees. But 3 newtons multiplied by the cosine of 135 degrees equals -2.1 newtons, and I've rounded this result to one decimal place. On the other hand, the y-component is equal to the magnitude of our vector, which is 3 newtons, and this will be multiplied by the sine of the angle, which is the direction of the vector. That is, it multiplies the sine of 135 degrees and 3 newtons by the sine of the angle. 135 degrees is equal to 2.1 newtons, and I have rounded this result to one decimal place. I invited you to Eros, and here I want to make the following observation: only in the case where you are working with the direction of the vector, that is, with the angle it forms with respect to the positive x-axis, only in that case will the x-component always be related to the cosine and the y-component to the sine. Because if we work directly with the angles we are given, sometimes the x-component is related to the sine and the y- component to the cosine. This is because if we form a right triangle, we need to observe if the x-component corresponds to the adjacent side or the opposite side of the angle we are given. If it corresponds to the adjacent side, it will be related to the cosine, but if it corresponds to the opposite side, it will be related to the sine, and the same for the other component. So that is why we must be very careful in this, and I repeat, if we work directly with the direction of the vector, the x-component will always be related to the cosine and the y-component to the sine. We must be very careful with this, and having said that, I'm going to do exactly the same for the other vectors. Now, for vector number 2, which is this one we have here, it has a magnitude of 3.5 newtons, so we write it in this column. Now, what is the direction of this vector? Remember that it's the angle it forms with respect to the positive x-axis, that is, this angle we have here. But here we can directly observe that this angle is 90 degrees, so we write it in this column. And since we already have the direction of our vector, then the x-component will be equal to the magnitude, which is 3.5 newtons, multiplied by the cosine of the angle, which is the direction, that is, by the cosine of 90 degrees. Performing this multiplication, it equals 0 newtons. Now, for the sine component, in the same way, we have that it is equal to the magnitude of our vector, which is 3.5 newtons, and multiplied by the sine of the angle, which is the direction, that is, by the sine of 90 degrees. Performing this multiplication, it equals 3.5 newtons, and I'm going to continue with the next vector. This vector we have here has a magnitude of 2.5 newtons, and we write it in this column. Now, what is the direction of this vector? Well, in this case, it's the angle they're giving us. This is the angle our vector forms with respect to the positive x-axis. Therefore, the direction is 50 degrees. And again, performing the exact same procedure, we see that the x-component is equal to the magnitude of the vector multiplied by the cosine of this angle. Performing this operation, it equals 1.6 newtons, rounded to one decimal place. And for the y- component, which is equal to the magnitude of our vector, that is, 2.5 newtons, multiplied by the sine of the angle, which is the direction, meaning it multiplies by the sine of 50 degrees, performing this multiplication gives us 1.9 newtons, rounded to one decimal place. And I'm going to perform the same procedure for the other vector, which is number 4, and this one has a magnitude of 4 newtons. So I write it in this column. Now, what is the direction of this vector? Well, if we observe, it's entirely on the positive x-axis and also points towards The right side, therefore its direction is 0 degrees, and we write it in this other column. Now, the x-component will be equal to the magnitude of our vector x multiplied by the cosine of this angle, and performing this multiplication gives us 4 newtons. Now, for the z-component, we have that it will be equal to the magnitude of our vector multiplied by the sine of this angle, and performing this multiplication gives us 0 newtons. And I'm going to continue with the next vector, which is number 5, and this one has a magnitude of 3 newtons, so I write it in this column. Now, what is the direction of this vector? Well, it's the angle it forms with respect to the positive x-axis, that is, this angle we have here. But if we observe, if we make a complete turn, this corresponds to 360 degrees. Therefore, to determine this angle, we must subtract 30 degrees from 360, and 360 minus 30 degrees equals 330 degrees, and we write this number in this column. And with this, we have that the x-component of this vector is equal to the magnitude of the vector multiplied by the cosine of this angle, and performing this operation gives us approximately 2.6 newtons, rounded to one decimal place. On the other hand, we have that the i-component will be equal to the magnitude of our vector multiplied by the sine of this angle, and this operation gives us a result of minus 1.5 newtons. Finally, for vector number 6, this has a magnitude of 2 newtons, so we write it in this column. And what is its direction? Well, remember that it is the angle it forms with respect to the positive x-axis, that is, the one we have in purple. And how are we going to determine it? Well, here we observe from this side to this other we have 180 degrees, and 180 degrees plus these 20 that we have here gives us a result of 200 degrees. This means that the direction of this vector is 200 degrees. Now, the x-component will be equal to the magnitude of our vector x multiplied by the cosine of this angle, and performing this operation gives us approximately minus 1.9 newtons. And on the other hand, the y-component It will be equal to the magnitude of our vector multiplied by the sine of this angle, and this is equal to negative 0.7 newtons, rounded to one decimal place. Now that we have obtained the rectangular components of all our vectors, what we are going to do next is obtain the sum of all the x-components and also the sum of all the y-components. I will start by determining the sum of all the x-components. So we have negative 2.1 + 0 + 1.6 + 4 + 2.6 - 1.9 newtons, which equals 4.2 newtons. On the other hand, I am going to calculate the sum of all the y-components, and we have 2.1 + 3.5 + 1.90 - 1.5 - 0.7 newtons, which equals 5.3 newtons. And with these last results that we have obtained, I am going to determine the magnitude of the vector and also the angle it forms with respect to the positive x-axis. Remember that the magnitude of the resultant vector is equal to the square root of the sum of the x-components squared plus The sum of the y-components is also squared, therefore, to determine the magnitude of the resultant vector, all I need to do is substitute the data into this expression. We then have that the magnitude of the resultant vector is equal to the square root of the sum of the squares of the x-components, that is, 4.2 newtons squared plus the sum of the squares of the y-components, and in this case, we have 5.3 newtons squared. This is again equal to the square root, and performing these operations, 4.2 newtons squared plus 5.3 newtons squared equals 45.7 newtons squared, and this result is rounded to one decimal place. The square root of 45.7 newtons squared equals 6.8 newtons, rounded to one decimal place. Therefore, what we have obtained is the magnitude or the resultant vector. Next, I will determine the angle that this vector forms with the horizontal, which sometimes will correspond to the direction of the vector and sometimes not, but I will explain this at the end. Well, this angle that it forms with the horizontal It will be equal to the inverse tangent of the absolute value of the sum of the y- components divided by the sum of the x-components. So all I have to do is substitute the data into this expression, and we have that the angle is equal to the inverse tangent of the absolute value of the sum of the components themselves, that is, 5.3 newtons divided by the sum of the x-components, which is 4.2 newtons. Now, if we observe, we have the unit newton multiplying in the numerator and in the denominator, therefore this will simplify, and this is equal to the inverse tangent of the absolute value of this division, and 5.3 divided by 4.2 is equal to 1.3, rounded to one decimal place. Now, the absolute value of 1.3 remains the same amount, so we have that this is equal to the inverse tangent of 1.3, and when we enter this into our calculator, it gives us a result of 52.4 degrees, and therefore this is the angle that the projectors form with respect to the horizontal. And now I'm going to explain what I mentioned a moment ago, that this angle sometimes goes Sometimes it corresponds to the direction of the vector, and other times it doesn't, because this angle gives us the angle that our resultant vector forms with respect to the horizontal. For example, here we have our horizontal, and if this were our resultant vector, then this angle would correspond to the one we have here, the one it forms with respect to the horizontal. And remember that the direction of the vector is the angle it forms with respect to the positive x-axis, that is, this one we have here. Now, how am I going to know if this angle corresponds to the direction of the vector or not? For that, I'm going to graph the components of the resultant vector and also its magnitude. And for that, I'm going to start by drawing our Cartesian plane. First, we observe that the x-component of the resultant vector is a positive quantity, therefore I'm going to represent it as an arrow coming out of the origin on the x-axis and pointing towards y. On the other hand, we observe that the y-component of the resultant vector is also a positive quantity, therefore I'm going to represent it as a vector that will start at the end of the x-component, and since it's a positive quantity, it will point upwards. And the resultant vector is the one that goes from the origin to the tip of the last vector, that is, this one we have here. And if We observe that this is the angle our vector forms with respect to the horizontal, but we can also see that it corresponds precisely to the direction of the vector, since it is the angle it forms with respect to the positive x-axis. Therefore, we can conclude that the magnitude of the resultant vector is 6.8 newtons and the direction of the vector, that is, the angle it forms with respect to the positive x-axis, is 52.4 degrees. Well done, Vito Aleros, and with this, I conclude this exercise and this video. I hope you liked it and, above all, that it was useful. And now I say goodbye. As you know, this is your virtual channel. I am Humberto, and I'll see you in the next video. Ah