Vectors and Projectiles Legacy Problem #16 Guided Solution

The best strategy for solving a complex problem such as this one involves using a table in order to tabulate the x and the y components of each individual vector, and then adding up all of the components in order to determine the resultants, horizontal and vertical components, then drawing the resultant from these components, determining its magnitude and direction. The table is demanded in order to do the problem in that manner. So you notice there's a table here, and in the table we've listed the three vectors that you're adding. We've also provided rows and columns in order to calculate the x and y components. Now to calculate the x and y components, you can make use of the shortcut that the x component of vector A is simply going to be magnitude of A times the cosine that that vector makes with east counterclockwise, and that's given here as 113 degrees. You can find the y component, Ay, is simply A times the sine of theta, where theta is the angle counterclockwise from east. So for the first row, I'm going to take 42 miles and 113 degrees. For the horizontal column, I'm going to go 42 times the cosine of 113. Comes out to be a negative value, meaning that it was displaced west, as we would expect. And then for the y component, I'm going to go 42 miles times the sine of 113. That comes out positive for north. And for the second vector, that's quite easy. It's just negative 61 miles, or 61 miles to the west, and for the y, it's zero. For the third vector, again, I'll have to use my calculator and use the idea that vector C, 23 miles at 253 degrees counterclockwise from east, can have an x component that's calculated as 23 times the cosine of 253 degrees, and the y component is 23 times the sine of 253 degrees. Now what I've done is filled out the top three rows of my table, and to get the bottom row, the all-important resultants row, what I'm going to do is add all of the parts. I'm going to keep in mind that some of these parts are negative and some are positive. So I'm going to sum up all the x components, and all of these x components happen to be west or negative values. So I add them up, and if you do this right, you will at least do it the way I did, you'll get 84.13 miles. Then I'm going to do the same thing for the y components, I'm going to add them up, there's only two to be added, one of them is north, and the other south, and when I add them up, I get positive 16.666, a nutty number, and that 16.666 positive means north. So what I have determined is that the resultant, whatever it is, is a vector that is going 84.13 miles towards the west, and 16.66 miles towards the north. So now I'm going to draw a diagram of this resultant from its components. So I draw a west vector, kind of a long one, 89.13 miles, and then where that one ends, I draw the vertical component northward, or upward, 16.66 miles, I'm just going to estimate this, and those are the two parts of the resultant. So I now draw the resultant from the tail of the first vector, the west vector, up to the arrowhead of the last vector, the northerly vector. Now I want to find the magnitude and direction of that vector. And that vector, that resultant, happens to be the hypotenuse of a triangle with these two sides, so to find its magnitude, I use the Pythagorean theorem, the 84.13 miles squared plus 16.666 miles squared, and square root of the whole thing. And then to find the direction of that vector, I go back to the tail of that diagonal vector, where it's making some sort of angle with west. And I find the angle that it makes with west. And to find the angle that it makes with west theta, I'm going to have to say theta equal the inverse tangent of the side opposite to the side adjacent, where 16.666 is opposite of that angle, and 85.76 is adjacent. So theta equal inverse tangent 16.66 divided by 85.76, and I find it to be 11 degrees. So I can say the direction of that vector is 11 degrees north of west, or if I want to express it counterclockwise from east, that's about 180 degrees, almost that amount around, just a little short of it by 11 degrees. That makes it 169 degrees CCW from east.