How to find the direction of a 2-dimensional vector without a prefix?

[-Dragoon-]

First of all, I'd like to thank all the kind and wonderful people here who have helped me so far in concepts I've been struggling with. I greatly appreciate it. I'd also like to wish every a happy and wonderful new year!

Homework Statement

Find the displacement of each of the situations by adding the vectors:
a) 12.5 km [E], 7.4 km [W]
b) 8.0 cm [N], 6.0 cm [W]
c) 2.0 m [E 20° S], 4.0 m
d) 100 km , 80km [ W 30° S], 20 km [N].

Homework Equations

C^2 = a^2 + b^2
C^2 = a^2 + b^2 - 2abCosC
SinA/a = SinB/b = SinC/c

The Attempt at a Solution

Alright so today I started learning the very basics of adding vector quantities. I find parts a and b relatively easy. I know that solving part a is just changing the 7.4 km [W] vector into -7.4 km [E]. For part b, I would draw the triangle and apply pythagorean theorem and finding the direction would be opp/adj = tanΘ.

Now it is part c where I start to really struggle. I've only been introduced to one method of solving for both situations in part c and d where specific directions have been given, and I don't believe it to be the most effective. First I would find displacement by applying the equation: c^2 = a^2 + b^2 - 2(a)(b)Cos(C), and I know that C would be the difference of the direction and 90°. Using the unit circle, since it is going clock-wise in the southern direction of east, that is -20°. So C would then be 70 degrees, correct? After finding the displacement, I would find the direction by applying the sine equation:SinA/a = SinB/b = SinC/c. But, for the last part I have no conventional method I can use that I believe would be effective. Any help? Is there an easier way or more effective method of solving part C, if possible? I don't have a teacher, as I do this course through correspondence, so it is quite difficult making sense of one or two examples provided in my course book.

Thanks for all the help, everyone and have a happy new year!

 

[{~}]

You need to break the first vector in c into components. We use trigonometry to do this.

 

[-Dragoon-]

You need to break the first vector in c into components. We use trigonometry to do this.

And how would I go about doing this? Frankly, I haven't learned anything of breaking vectors into components in order to simplify them.

 

[The legend]

And how would I go about doing this? Frankly, I haven't learned anything of breaking vectors into components in order to simplify them.

Basically, if you have a vector, of say magnitude 10 units at an angle of 30⁰ from the x axis, the component of that vector on x-axis will be 10*cos30. And the component of the same vector on y-axis will be 10*sin30.

Vector components are the vectors on x and y axis, which added, give the original vector.




also these will help
http://www.physicsclassroom.com/class/vectors/u3l1d.cfm#component
http://www.physicsclassroom.com/class/vectors/u3l1e.cfm

 

[-Dragoon-]

Basically, if you have a vector, of say magnitude 10 units at an angle of 30⁰ from the x axis, the component of that vector on x-axis will be 10*cos30. And the component of the same vector on y-axis will be 10*sin30.

Vector components are the vectors on x and y axis, which added, give the original vector.




also these will help
http://www.physicsclassroom.com/class/vectors/u3l1d.cfm#component
http://www.physicsclassroom.com/class/vectors/u3l1e.cfm

Ah, so we use the magnitude of 10 as the hypotneuse when breaking it down into components and to find the adjacent angle I would use 10*cos30 and for the opposite 10*sin30, correct?

So then, after I were to find these sides how would I go about solving the displacement with the given information of these sides?

 

[The legend]

Ah, so we use the magnitude of 10 as the hypotneuse when breaking it down into components and to find the adjacent angle I would use 10*cos30 and for the opposite 10*sin30, correct?

yes, it could be considered that way.

So then, after I were to find these sides how would I go about solving the displacement with the given information of these sides?

so, now that the "inclined" vector is taken care of, you can just forget about it, keeping it's components. Do the same for all given vectors, and then it's just plain addition(or subtraction, in some cases) of vectors in the same direction, which is pretty easy.

Add x and y components separately, and then you get two final components, which added give you your answer.

 

[-Dragoon-]

yes, it could be considered that way.

so, now that the "inclined" vector is taken care of, you can just forget about it, keeping it's components. Do the same for all given vectors, and then it's just plain addition(or subtraction, in some cases) of vectors in the same direction, which is pretty easy.

Add x and y components separately, and then you get two final components, which added give you your answer.

Alright, thanks so much for all the help. I just have one problem, I am not to sure on how to find the direction of the displacement using this method. I can find the magnitude and the degree without much of a problem, but I have no idea on how to find the direction. Any help?

Example, I broke part c down into components and found that the 2m south of east traveled was 1.9 m east and 0.7 m down. Now I added the 0.7 to the 4, then found the displacement using pythagorean theorem which was 5m, and the degree using tangent which was 20°. But how would I determine the direction?

 

[The legend]

Example, I broke part c down into components and found that the 2m south of east traveled was 1.9 m east and 0.7 m down. Now I added the 0.7 to the 4, then found the displacement using pythagorean theorem which was 5m, and the degree using tangent which was 20°. But how would I determine the direction?

the direction finding is pretty simple. What is the direction of the component 4.7m? And the component of 1.9m is east, as you've written.

So, the combination of both these directions gives you the direction of the final vector.