aham925925 said:
Homework Statement
Find the Magnitude and the direction of the resultant vector obtained by adding displacements of 5.0m W and 5.0m NW
Homework Equations
I think for these problems, pythagoras' theorem is needed (c^2=a^2+b^2) as well as the basic trigonometric formulae (sin = O/H etc)
The Attempt at a Solution
I've tried methods of representing the vectors as arrows and putting them "head-to-tail" but I'm finding it difficult to represent a vector in the north direction as well as the north-west direction. I'm not entirely sure on the proper way to start attempting this question
Draw a picture. As on a map, East is to the right, West to the left, North upward,and
South downward. So your 5.0 m W vector is draw horizontally to the left. "North-West" is half way between N and W so it is drawn (from the tip of the W vector) 45 degrees upward and again 5.0 m long. Their sum is the vector from the base of the W vector to the tip of the NW vector. Look at that and you should see a triangle- but not a right triangle. Yes, you could resolve the two vectors into " xy-components" (i.e. East and North values)- for the W vector that is trivial. For the NW vector only slightly harder. Then add the component. However, the problem specifically asks for so then you would have to convert back to that form.
Here, I think it is simplest to use the "cosine law". You know two sides of the triangle and the angle between them. If you call those sides a and b, and the angle between them [itex]theta[/itex], then the opposite side, the "magnitude" of the vector, c, is given by [itex]c^2= a^2+ b^2- 2ab cos(\theta)[/itex]. You could use the sine law to find the angle but here that is much easier. Since this is obviously an isosceles triangle (the two given sides are both 5.0 m) the two unknown angles must be the same; and the sum of those two and the 135 degree angle must be 180 degrees.
Becareful how you write the direction. If you use [itex]\theta[/itex] itself, notice that the angle is from the horizontal (West) line up to the new vector- It must be given as that angle "North of West". Or you could use its complement- the angle from the new vector to the vertical (North) line, [itex]90- \theta[/itex] and call it "West of North".
