What are the x- and y-components of the velocity vector shown in ...

Question

What are the x- and y-components of the velocity vector shown in FIGURE EX3.20?

Accepted Answer

Hi, everyone in this practice problem, we are asked to determine the components of the force app at Newton pointing to the north along the X and the Y access of our tilted coordinate system down below. So first, what we want to do is to basically start with actually adding, identifying the different angles in our system. So first, we have our alpha here of 20 degrees, which is the guilt of our coordinate system. And then next, we will have our here which is going to be just our beta. And lastly, we will have this side angle here which I will indicate that with a gamma just like so okay. And now we know that our F here will have a projection. So the projection will actually be in the actual coordinate system access itself. So first, we will have the F Y in the Y access of our tilted coordinate system. And this is going to be our F Y or the Y component of our force. And then similarly, we will have an F X which is going to be the X component of our force along the X axis of our tilted coordinate system like so and this is going to be our F X and we're being asked, what is this FX and FY value in our system? So to do this, we obviously have to do the projection. And in this case, we want to first calculate gamma which is the angle between f the direction due north and the positive Y axis. So in this case, we know that the X and the Y axis on the tilt, a coordinate system will form a 90 degree angle. So alpha plus beta will equal to 90 degrees. So in this case, we know what the alpha is, which is 20 degrees. So you can calculate what the beta is, which is just 90 degrees minus alpha, which is 90 degrees minus 20 degrees, which it goes to 70 degrees. Next, we also know that the, the force here pointing to the north and this arbitrary horizontal line will also make a 90° angle. So essentially beta plus gamma is going to be goes through 90 degrees because there will be a right angle here and there will also be a right angle here or essentially between the X and Y of the delta coordinate system just like. So, so gamma is going to equal to 90 degrees minus beta, which is beta, we have just found that to be 70 degrees. So this is going to be 90 degrees minus 70 which will equals to 20 degrees or essentially alpha will equal to gamma. Okay. So after that, we can actually start by actually doing our projection. So first we know that our F X here F X pointing to the left, so F X is negative just by convention. So F X during the projection will equals to F remember the negative sign here multiplied by because F X is here, it's going to be multiple by see the sign because the Y, the gamma is on the opposite side. So this is going to be sign of gamma which will then corresponds to minus 25 Newton Multiplied by sine of 20°. And that will give us an FX value of -8.6 Newton. Next we have the F Y which is pointing up. Therefore, F Y is going to be positive. So F Y will equals to F without the negative sign Because the gamma and the projection is on the same side, then this is going to be co signed gamma which will equal to mutant Multiplied by co sign of 20°FY Will then equals two B 23.5 Ethan. And that will be the answer to this problem, an X value of minus 8.6 and an F Y value of 23.5 which will corresponds to option B the one that we have here. So option B will be the answer to this practice problem and that will be all for this particular video. If you guys have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that will be all for this video. Thank you.

What are the x- and y-components of the velocity vector shown in FIGURE EX3.20?

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