Steven W. answered • 08/24/16
Tutor
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Physics Ph.D., college instructor (calc- and algebra-based)
Hi Garry:
If you successfully got the first part of the problem, you very likely broke each vector down into components and added the components, then built the resultant vector's magnitude out of the components using the Pythagorean theorem.
If that is not the case, then just let me know how you got the magnitude.
You now have two components for the resultant vector: an x component and a y component. I got the x component of the resultant to be negative and the y component to be positive, which puts the vector in the second quadrant.
So, if you draw the x-component as an arrow, starting at the origin, with length about 4, pointing in the -x direction, and then draw the y-component as an arrow, pointing from the end of the x component upward, with length about 2, and then draw the resultant from the origin to the end of the y-component, you might be able to see that these three lines form a right triangle, and the resultant (hypotenuse) makes an acute angle with the -x axis. For this angle, the x component is the adjacent side of the right triangle (because it comes right up to the angle), and the y component is the opposite side, since it does not adjoin this angle.
Therefore, this angle has a tangent given by:
tan(θ) = Ry/Rx
where Ry and Rx are the magnitudes of the y (opposite) and x (adjacent) components of the resultant. Note that I stripped the signs off the components, because they are not germane to measuring this angle, because it is referenced to the -x axis , at the moment, instead of the +x axis. Then, to solve for θ, you have to "undo" tangent by operating on both sides of the previous equation with the inverse tangent function, tan-1. That gives:
tan-1(tan(θ)) = tan-1(Ry/Rx) --> θ = tan-1(Ry/Rx) = tan-1(1.941/3.934) = 26o
Now, this is the angle of the resultant measured with respect to the -x axis, clockwise. But it looks like your instructor is into giving answers for angles measured with respect to the +x, clockwise. So, to get from the +x axis to our resultant by swinging around counterclokwise, we have to traverse the 180o from the +x axis to the -x axis, clockwise, and then go an additional 26o from the -x axis to the resultant vector.
By this reckoning, then, the angle of the resultant is 180o + 26o = 206o
I hope this helps more, but feel free to come back again if you have further questions about this or similar topics!
Steven W.
tutor
Visual learning is definitely tough in a case where we cannot draw diagrams, but let me try to describe it. I agree that your a and y components came out correct. So, if you can draw a standard x-y set of axes, start at the origin and draw your x component as an arrow about -4 in length, pointing to the left. Then, from the end of that, draw your y component as an arrow a little under 2 units in length, pointing up. The resultant vector is then drawn as an arrow pointing from the origin to the tip of the arrow on the y component. This resultant vector points into the second quadrant.
You can tell the resultant should point into the second quadrant, because it has a negative x component, and a positive y component.
Since it points into the second quadrant, it took me a moment to figure out why the angle is 206o. As I mentioned, angles in physics are conventionally measured from the =x axis counterclockwise. So second quadrant angles should be between 90o and 180o. But then I realized, you were mentioning angles measured from the +x axis clockwise. This is the reverse of standard convention, but perhaps you have an unconventional instructor! :D
So, they way I described drawing it, you can see a right triangle, where the right angle is between the x component (drawn along the -x axis) and the y component (drawn upward from the end of the x component). One acute angle of this triangle is formed between the resultant vector and the -x axis (or, if you prefer, between the resultant and the x component, which lies along the -x axis). With respect to that angle, the x component is the adjacent side of the triangle, because that side comes right up to that angle. On the other hand, the y component does not touch that angle, so it is the opposite side.
For a given acute angle in a right triangle, the tangent of that angle is equal to the ratio of the length of the side opposite it over the side adjacent to it.
For the angle I described above, between the resultant vector and the -x axis, the tangent of that that angle is given by the formula in the previous answer, which leads to:
θ = tan-1(Ry/Rx) = tan-1(1.941/3.959) = 26.1o
Note that I stripped the signs off both components and used just the magnitude of the vectors in this case, because -- referencing the angle to the -x axis and making the triangle as I described above, the fact that the x component was negative was no longer germane -- because I am now referencing this angle to the -x axis, rather than the conventional +x axis.
So, from this, we see the resultant vector makes an angle of about 26o above the -x axis. Now, conventionally, I would say that, if we wanted to measure this angle in the standard +x axis, counterclockwise way, that angle would be between the +x axis and the resultant vector, measured counterclockwise. This would make the angle the supplement of the 26o I calculated a little bit ago, or 180-26 = 154o.
However, your instructor appears to want an answer measured from the +x axis clockwise. In that case, you would have to go 180o clockwise to get from the +x to the -x axis, an then go an additional 26o to get to the resultant vector. So the resultant is at 180o+26o = 206o, as your answer list.
I hope this helps some more. If you want to look at it with more drawing capability and live interaction, we can always have a short online session where we can share a drawing board.
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08/24/16
Garry L.
08/24/16