Vectors and Projectiles Legacy Problem #15 Guided Solution

A habit, and not necessarily a good one, which I've observed for quite some time in students is the tendency to use a calculator to solve a problem, and just a calculator. The failure to write things down, to diagram, to document, to use a sheet of paper. Now if you're consistently missing a problem in the same problem, you're going to have great difficulty tracing through your path on a calculator and trying to find out what you did. Now I recognize that calculators are changing through the years, and there's more than one line now, and there's many lines and probably will continue to change, but still, commit yourself to documenting, diagramming, organizing, and writing things down on a sheet of paper. That allows you to go back and trace through your problem solution to see where the error is, and in the end it may save much time. Do that here for Dora's problem, a very difficult vector addition problem that has many, many steps, and you'll notice in the solution page here that there is a table given that lists each of the four vectors, and then two more columns, one for the x component of each vector and one for the y component of each vector. The idea is that as you go through this problem and organize yourself, you're going to use this type of structure and write down for each vector the x and the y components. Now for the first vector, 68 meters south, there's no x component, you just put 0, and for the y component, you put 68 meters south, or negative 68, I personally prefer negative 68. Ultimately, you will use your calculator and you're going to talk to your calculator about direction in terms of positive and negative, so you have for the first vector, 0, east west, and negative 68 meters. Now for the next vector, you have magnitude and you have a direction, and what's very useful here is you have the direction counterclockwise from east, so use that to help you find x and y for this vector. For x, it's just 112 times the cosine of 155 meters, of 155 degrees, that gives you a negative value, meaning that that x component is west, so you get negative 101.51. Repeat the process for the y component, going 112 times the sine of 155 degrees, and you will get positive 47.33 for the y component, that positive means north. Now for the next vector, it's quite easy, 0, east west, and 34 meters south, or I like negative 34, and for the last vector, the x is going to be calculated as 182 times the cosine of 343, you're going to get a positive there listed in your table, and the positive there means east, and when you go 182 times the sine of 343 degrees, you get a negative at that point, the negative means south. Now take a deep breath, you've got a lot of the work done, now what you have to do is add up all of the parts to find the resultant's x and y components, and that's the last row of the table. So add up all the parts, keeping in mind that you have some positives and negatives here. By the time you finish that, you ought to do it yourself as practice, but this is what I got when I did it myself. Under the x column for the R row, the resultant has a horizontal component, a positive 72.54, and under the vertical column, the sums of all those y components is negative 107.88, where the negative means south. So I know my resultant, whatever its magnitude and direction is, has as its parts 72.54 meters east, and 107.88 meters south. So I make a triangle that stretches out horizontally, go ahead and do it, 72.54 meters, and then from where that ends, it goes south 107.58 meters, and I draw the resultant from the tail of that first easterly vector to the arrowhead of that second southerly vector. I'm going to try to find the magnitude and direction of that diagonal hypotenuse. I get the magnitude quite easy from Pythagorean theorem, and I get the direction using Sohcahtoa, using trigonometry. To get that direction, I go back up to that tail of that diagonal vector, and I find theta at the tail, the angle it makes with west. I go inverse tangent on the side opposite 107.88, divided by the side adjacent, 72.54, and I get my value for theta. It ends up for me being about 67 degrees, and then what I do with, sorry, it ends up for me being about 56.1 degrees. What I do with that value is I then subtract it from 360 to get myself a counterclockwise from west vector.