https://youtu.be/y7jq-zMzKhs
PHYS 1101: Lecture Three, Part Five
Okay, let me show you a real example here and we’re going to do some mathematics to get some real numbers. Let’s say we’re given a vector r. We’re told some angle information. This vector is at this 35-degree angle pointing a little bit to the right and down, and it has a magnitude of 80 meters. So the length of this hypotenuse represents 80 meters. We’ve got to figure out what the scalar components are, rx and ry for this vector r.
Three main steps you have to follow. Step one, at least follow the steps I showed you above. Just sketch those vectors, rx and ry. What would they be? Well, here are my axis, so I’m going to go to one end of this vector, sketch a line parallel to one axis, I’ll go with the y. At the other end I’m going to sketch a line parallel to the other axis, and then I’m going to draw in my two component vectors. The x vector is going to match the rightward trend of r, and the y component vector matches the downward trend. This is the vector rx and here’s the vector ry.
Once I’ve got those sketched in, I’m going to go off to the side here and do a real quick sketch of that picture. I’m going to add my 35 degrees and I’m going to note that the hypotenuse length is 80 meters to do my trig. I need to know what the length of this side is. Real quick I’m going to just call that ry and rx. But what I’m doing here is the trig only to get the magnitudes. I’ll shorten that to mags.
So let’s do that trigonometry. You’ll get quick at this. I encourage you to practice. In order to get ry, that’s the adjacent side to this angle, so I’ve done the algebra enough times now that I know ry is then equal to the length of the hypotenuse times the cosine of 35 degrees, meaning ry is going to be equal to 80 meters times the cosine of 35 degrees. That’s equal to about 65.5 meters.
But I’m just going to ask, are we done? The answer is no. What I’ve found here is just the magnitude of the y component. For this to represent the scalar component, ry has to have the proper sign. Let me go back to the picture here real quick. For the y component I see that this vector points in the negative y direction, so I technically have to make this scalar component ry equal to -65.5 meters. Okay, there’s my y component.
To get the x component now, that’s the side opposite to the angle, I want to use the sine for this. So rx is going to be the hypotenuse times the sine of 35 degrees. So I have to multiply 80 meters times the sine of 35, in which case this number is 45.9 meters. Am I done? Well, we give the sign some thought. I go back up here to the real vector component rx, it points to the right. That’s in the positive x-axis direction, so the scalar component rx has to be positive. It has to point to the right and I’m going to indicate that with a positive sign. I’m explicitly writing it here as positive. You don’t need to do it, but I just wanted to emphasize that point.
So that would be it. We would be done. These would be the answers, the two scalar components to that vector.
This next lecture quiz question has you put a couple of these concepts together. So I’ll let you pause here and read that and take your time. It’s asking what are the scalar components of this displacement vector A? And they emphasize it for you here. This is A. That ends the discussion on the scalar component description we’re going to use in the class. Don’t hesitate to ask me any questions. Try out the Sentra online help sessions with questions if you don’t understand that.