(II) Three vectors are shown in Fig. 3–41. Their magnitudes ar...

Question

(II) Three vectors are shown in Fig. 3–41. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with the +𝓍 axis.

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Accepted Answer

Welcome back everyone. In this problem. The figure below shows three vectors and Q in the xy plane with their length specified in arbitrary units. We want to find the sum of these three vectors and express the result. And in first its individual components and second its magnitude and angle relative to the positive Y axis. For our answer choices A says the X component is negative 21 its Y component is 35 magnitude is 41 and the direction is 32 degrees. B says they are 66 70 96 and 46 degrees respectively. C says it's 30 35 46 and 48 degrees respectively. And D says negative 21 70 73 and 58 degrees respectively. Now, if we're going to find the individual components of our vectors, first, we need to ask ourselves, what do we know about the components of a vector and the vector itself? Well, let's take vector P. For example, the X component of P would be right here along the X axis. Let's call that PX and the Y component of P would be parallel to the Y axis as called at Py. Now notice that the vertical and horizontal components form a right triangle form a right angle with each other. And because of that, we can apply what we know about vectors. Because recall that for a Victor A, OK, add 30 degrees to the horizontal axis, then the X component of vector A will equal a class theta and the Y component of vector A is going to be equal to a sine theta where A is the length or magnitude of a vector. Now, if we look at our three vectors here, we know their lengths, we know what angle they are with the horizontal. So that means we can figure out the X and Y components for all of them, add them all up and that will give us the individual components for the result. And so let's see if we can do that. So I'm going to make a column here here with the X and Y components for each vector. And now remember we have to pay attention to what direction the vectors are in. So we are going to find the X and Y components for Victor P four, Victor Q and then we're gonna find it for vector R. So let's start with Victor P notice that both components for vector P are going in a positive direction, the X component is going eastward and the Y component is going north. So that means their components will be positive. So since this magnitude of P is 42 and it's at 58 degrees to the horizontal. The X component of P is going to be equal to 4258 degrees. And the Y component of P is going to be 42 sine 58 degrees. Next, for vector Q notice that it's going in the negative X direction. So the X value is going to be negative but it's going in the positive Y direction. So the Y component will be positive since its magnitude is 31 units and it's at 30 degrees to 33 degrees to the horizontal. The X component of Q is going to be 31 plus 33 degrees or negative 31 costs 33 degrees and its Y component is going to be 31 sine 33 degrees where that value is positive. Finally, for our component O I don't know why I put R here. My, my, my apologies, you know what it is. I'm saying the alphabet, Pqr, I'm not saying OPQ, I think that is but for Victor, O for Victor O, OK, we notice that for its components, its X component is going left in the negative direction and it's Y component is going down also in negative direction. So both components will be negative since its length is 25 units and it's at 45 degrees to the horizontal or is going to be negative 25 cost 45 degrees for its X component and it Y component is going to be a negative 25 sine 45 degrees. So now we have the individual components of each vector. That means we can add them all up OK to get the result on Victor. So in this case, when we add up vector's PQ and O, we'll get the resultant vector of X and we'll also get sorry, the X component of the result and vector and we'll also get the Y component of the result and vector. So let's put these into our calculator for the X component. That's 42 costs 58 minus 31 cost, 33 minus 25 costs 45. Now, when you do that, you should get the X component to be negative 21.4199 which we notice all of our answers are to the nearest one. So that would be negative 21. That's the X component. Next for the wire component, we're going to add 42 sine 5831 sine 33 and the negative 25 sine 45. And when we do that, we should get the Y component to be equal to 34.8241 or rounded to the nearest 1, 35 units. So these would be the X and Y components, X and Y components of our result and vector. Now that we have the X and Y components, all we need to do now is to figure out the magnitude and the direction. So let's put that here and keep that in black so far. This is part two, OK? For its magnitude and direction. OK. Then we're talking about our Victor, remember if I just sketch it here. It's negative 21. So it's going 21 units to the left and 35 units up because it's positive. OK? And we're trying to figure out what a magnitude would be. And we're also trying to figure out what theater, the Anger Theater is. Well, again, recall it forms a right triangle which means that we cannot apply the Pythagorean theorem because recall that the Pythagorean theorem says the square of S OK. So buy the Pythagorean theorem, the square of S is going to be equal to the square, the sum of the squares of the other two sides. So that means S is going to be equal to the square root of that sum. So we're going to add the squares of the X and Y components and find their square root. So that means the magnitude of vector S then is going to be equal to the square root of negative 21 squared plus 35 units squared. So again, we can put this into our calculator. And when we do that uh work it to, to the nearest one, we should find that we get the magnitude of S to the 41 units. Let me write it in the same line. So the magnitude of S is 41 units. Next, we're talking about the anger that's uh or, or the anger that is formed for a resultant vector relative to the positive Y axis. So we can use our tangent function to help us because recall that for a vector for a vector, the anger theta is equal to the inverse tangent of its Y component divided by its X component. So in this case, that's going to be equal to the inverse tangent of 35 divided by negative 21. And now again, we can put this value into our calculator. And when we do that, OK, we get it to be equal to 58.4 degrees with respect to the negative X axis because not if we think about it, OK? If we respect the negative X axis, if we think about it here, that angle is formed with the negative X axis. But remember we want the anger with respect to the positive Y axis or relative to the positive Y axis. So that means relative to the positive Y axis by the th but don't tell anybody I had trouble spreading positive. OK. Let's just keep that between us. So relative to the positive, why access OK. Which that axis, if I were to put it on a diagram, that axis would have been right here since the entire thing forms 90 degrees, we can subtract our anger from 90 degrees to get its the anger relative to the positive X axis. In other words, the anger that would be right here. OK. And let's call that anger s theta. So relative to the positive Y axis, it would be equal to 90 degrees minus 58.4 degrees which is 31.6 degrees or to the nearest degree, 32 degrees. Therefore, our resultant vector is 41 units at an angle of 32 degrees relative to the positive Y axis and its X and Y components are negative 21 and 35. If we look back at our answer choices, that means the correct choice is choice. A thanks for watching everyone. I hope this video helped.

Struggling with Physics?

(II) Three vectors are shown in Fig. 3–41. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with the +𝓍 axis.

<IMAGE>

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Struggling with Physics?

(II) Three vectors are shown in Fig. 3–41. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with the +𝓍 axis.<IMAGE>

Watch next

Adding Vectors by Components practice set

(II) Three vectors are shown in Fig. 3–41. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with the +𝓍 axis.

<IMAGE>