Let $\vec{B}$ = (5.0 m, 30 degrees counterclockwise from vertically up). Find the x- and y-components of $\vec{B}$ in each of the two coordinate systems shown in FIGURE EX3.21.

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Hi everyone. In this particular practice problem, we are asked to determine the components of the position factor along the X and the Y access where we have first a normal coordinate system And second a tilted coordinate system where it is tilted 15° from the horizontal axis counterclockwise. And we will have a car which will have a position factor of 3m with a direction of 45° clockwise from the North. Okay. So first, what we want to do is to just identify the actual position factor that we have in this case. So what we have is a position factor of three m and the direction of 45 degrees clockwise from the north. So we know that the north is going to be the one upwards. So 45 degrees clockwise from that, we want to indicate that in both coordinate access and the position factor R is actually going to be represented just like this in the normal, in the normal coordinate system. With this angle here Being data of 45°. And also this angle here will then also be data Of 45° as well because we have a right angle here of 90°. And that will be it for the normal coordinate system. And then next, we want to draw the same thing on our tilted coordinate system. In this case, we have this arbitrary access to help us kind of draw our system. But I'm going to draw our arbitrary y access our vertical access to make it easier for us. And we cannot want to draw exactly the same thing here. So it will be 45° from the um from the North. So essentially this angle right here is actually going to be theta equals 45°. And this angle right here is also going to be theta equals 45 degrees. However, what we want to know is what is this angle right here, right? This is the angle that we are curious about and I'm going to call that the fire here. So Phi in this case is then going to Ashley just be um the data minus the delta angle here. So five equals data minus 15 degrees where data is 45 degrees minus 15 degrees and our five is going to be 30 degrees just like. So Okay. Now that we figure out our system that we have, we're gonna start with the easy 1 1st, which is the normal coordinate system. So in the normal coordinate system here, we can obviously project our, our so that it will have a Y and an X component right, like just like this and the R access, pointing to the right. So our access positive and the Ry is pointing outwards. So are Y is positive and the components Rx and Ry in the normal coordinate system can then be found by. So during the projection, we have data of 45 degrees are X will equals two R multiplied by co sign of 45 degrees or of data which will equals to three m, which is the magnitude of the position factor multiplied by coast side of 45 degrees. That will be 2.1 m. So this is the normal um coordinates are X and then we have the R Y which we can find by multiplying the R with the sine of the angle which is 4 to 5 degrees. So the RS three m Sign of and I will correspond to also these two are the same because sine and cosine of 45° are the same. This is going to be 2.1 m and these two values are going to be positive. Okay. So this is going to be looking at the normal coordinate. However, looking at the tilt that coordinate, we pretty much want to do a similar thing. But the only thing that we want to know this here is the fact that our angle is going to be a little bit different. Okay. So doing our doing our projection, our projection is going to be a little bit different. Then our projection is actually going to be on the tilted access itself where we will have this to be our Ry here. And our, our act is going to be just like this, this is going to go on our X. Okay. So our X can be found by multiplying our, as we can tell, we have a triangle weight of five value of 30 degrees. So this is going to be our multiplied by co sign of five, Which will be three m multiplied by a co sine of 30°, which will correspond to a value Of 2.6 m. As we can see here as well. Our ex is going to the rights or act as positive. And then our way is going to be our multiplied by sine of Phi which will be three m multiplied by sine of 30 degrees, which will correspond to an R Y value of 1.5 m. And Ry is pointing upwards. So are Y as positive And that will essentially be the answer to our problem in this particular practice problem. So in the normal coordinate Rx is going to be plus 2.1, Ry is also going to be plus 2.1. And in the tilt according to our X is going to be plus 2.6 and RY is going to be plus 1.5, which will actually correspond to option C that we have in this problem. So option C is going to be the answer to this particular problem and that will be all for this video. If you guys have any sort of confusion, please make sure to check out our other similar lesson videos on similar topics and that will be all for this video. Thank you.

Let $\vec{B}$ = (5.0 m, 30 degrees counterclockwise from vertically up). Find the x- and y-components of $\vec{B}$ in each of the two coordinate systems shown in FIGURE EX3.21.

Verified video answer for a similar problem:

Key Concepts

Vector Components

Coordinate Systems

Trigonometric Functions

Let B→\overrightarrow{B} = (5.0 m, 30 degrees counterclockwise from vertically up). Find the x- and y-components of B→\overrightarrow{B} in each of the two coordinate systems shown in FIGURE EX3.21.

Verified video answer for a similar problem:

Key Concepts

Vector Components

Coordinate Systems

Trigonometric Functions

Let $\vec{B}$ = (5.0 m, 30 degrees counterclockwise from vertically up). Find the x- and y-components of $\vec{B}$ in each of the two coordinate systems shown in FIGURE EX3.21.