(II) For the vectors shown in Fig. 3–41, determine
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Question

(II) For the vectors shown in Fig. 3–41, determine

(b) 2 A (→ above A) ― 3 B (→ above B) + 2 C (→ above C) , and

Accepted Answer

Welcome back everyone in this problem, we want to find P minus two Q utilizing the vectors P and Q shown in the diagram for our answer choices. A says that the resultants magnitude is 66 along the positive Y axis. B says it's 66 along the positive X axis. C says it's 18 along the negative X axis. And D says it's 18 along the negative Y axis. Now, what are we trying to figure out here? Well, we're trying to find P minus two Q. So let's call our resultant vector R P minus two Q. Now if we're going to find the magnitude of the resultant, first of all, we need to figure out its individual components. So let's see if we can find the components of P, find the components of Q and then put them into our resultant to eventually solve for P minus two Q. OK. Now, if we're going to do that, that means we need to split P and Q into X and Y components. So what do we know about the components of a vector? Well, recall that for a vector A at an angle of theta degrees to the horizontal. OK. Then we know that the X component of our vector is going to be a cos theta where A is the length or magnitude of the vector. Likewise, we know that the Y component is going to be equal to a sine theta. So what this is telling us is that once we have the length and anger the vector makes with the horizontal, then we should be able to find its X and Y components. If you notice here, we have that information both for vectors P and Q. So we can use that to figure out their X and Y components and then use it to figure out X and Y components of the resultant. So I'm going to make a table here with the X and Y components for both of our vectors. OK? And we're finding the components for P and the components for queue. Now, for vector P, we know that its length is 38 units and it's at 61 degrees to the horizontal. So its X component is going to be 31 K 61 and its wire component is going to be 31 s sorry, 38. Where did I get 31 from? It's 38 plus 61 and 38 S 61. And we know that both of those values are positive because P is going eastward and upward, both in positive X and Y directions. Next, let's take vector Q notice that vector Q is going in the positive Y direction but in the negative X direction. So the negative, the X component will be negative while the Y component will be positive. Since Q is 29 units long at that 35 degree angle to the X axis or to the horizontal axis, then Q is going to be negative 35 cost 29 sorry, negative 29 cost 35. So you have to make sure you get your angles and sides right? And its Y component is going to be positive 29 sine 35. OK? Now let's hold on to these values for a bit because remember our resultant is not the sum of P and Q but it's rather P minus two Q, that's our resultant. So since we have pxpyqx and Qy, then for the Victor R, this implies that R which is equal to P minus two Q is going to be equal to PX minus two multiplied by QX. OK? I should probably put that in red. Let me do that. Oh no, let me leave it in, in black because these are two different. I don't want to confuse the P and Q here. So that's P minus two QX and it's going to be py minus two Qy since we know what they are, we can substitute because no, that means we can get our resultant to be 38 cost 61 minus two multiplied by negative 29 cost 35. OK? And the Y component to be 38 sin 61 minus two multiplied by positive 29 sine 35. OK. So now let's substitute to see if we get our vector R. Now if you put these expressions into your calculator, then you'll get the X component of R to be equal to 65.94 units. And then you're going to get the Y component of R to be negative 0.02 units. OK? So Rx is 65.94 and Ry is negative 0.02. So now we found the X and Y components of our vector, we will be able to find our resultant, the magnitude of our vector or the magnitude of our resultant R because no, what we're saying if we were to picture it, what we're really saying here is that we have our vector R that is 65.94 units in the positive X direction and negative 0.02. That means it's in the negative Y direction. So our vector probably looks something like this. OK? Where if I were to draw it something like that, so we have our sides 65.94 negative 0.02. So now if we're going to find the resultant or the magnitude of our vector R, then that means the magnitude of I is going to be equal to the square root of the square of the X component, 65.94 squared plus the square of the Y component negative 0.02 squared. Now, if we work those out, OK, if we, again, if we put this into our calculator and evaluate this expression, we should get the magnitude of R to be equal to 65.9 units which writing it like our answers to the nearest unit would be 66 units. OK. So that's the magnitude of our vector R. All we need to do now is to figure out which axis it's along if it's along the positive or negative X axis or along the positive or negative Y axis. Well, we kind of have an idea of how it looks here from our sketch, but let's verify it by finding the direction of R. In other words, let's try to find the angle that R makes with our horizontal axis, the anger theta. OK. And we know that if it's, if it's very close to zero, then it's going to be along the X axis. If it's close to along the positive X axis, close to 90 it will be along the positive Y axis. As a matter of fact, let's just do a sketch of that. So if it's here, remember that for our axis from 0 91 83 60. So if our vector is close to any of these, for example, here it's close to zero positive X, close to 90 positive Y close to 180 negative X close to 270. Sorry, that would be negative. Y you understand. So this can give us an idea of which one it is close to. So we can find the angle. OK? Because no, the anger that I is set to is going to be equal to the inverse tangent of the magnitude of interest tangent of the magnitude of rxrry divided by Rx, which is going to be equal to the inverse tangent of negative 0.02 divided by 65.94. Now, when we put this expression into our calculator, we get our angle to be 0.02 degrees which is approximately zero degrees. So since the angle it makes with the horizontal is approximately zero degrees, that means it's laying along the positive X axis. So its magnitude is 66 it's along the positive X axis. And if we look back at our answer choices, the correct answer is choice B thanks a lot for watching everyone. I hope this video helped.

Struggling with Physics?

(II) For the vectors shown in Fig. 3–41, determine

(b) 2 A (→ above A) ― 3 B (→ above B) + 2 C (→ above C) , and

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