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Vector Components, Magnitude, and Direction

Date: 07/23/98 at 06:44:19
From: Kristine Tan
Subject: Physics (a vector problem)

Vector M of magnitude 4.75m is at 58.0 degrees counter-clockwise from 
the positive x-axis. It is added to vector N, and the resultant is a 
vector of magnitude 4.75m, at 39 degrees counterclockwise from the 
positive x-axis. Find: (a) the components of N, and (b) the magnitude 
and direction of N.

I drew a graphical illustration of the problem. But I really can't 
solve it because I don't know how. 

Please help. Thank you.

Date: 07/23/98 at 11:57:54
From: Doctor Rick
Subject: Re: Physics (a vector problem)

Hi, Kristine,

I will get you started on solving this kind of problem. There are two 
tools you need to do this: (1) converting between magnitude/direction 
and components of a vector and (2) adding vectors. The first requires 
some trigonometry, so I hope you've had some.

(1) You are given the magnitude and direction of vectors M and P (the 
sum of M and N). Before you can add them, you must find their 
components. Remember this diagram:

       My+-------------* M
         |            /|
         |           / | 
         |          /  |
         |         /   |
         |        /    |
         |      L/     |
         |      /      |
   sin(a)|-----+       |
         |    /|       |
         |  1/ |       |
         |  /  |       |
         | /)a |       |
        O    cos(a)    Mx

A vector of length 1 has components (cos(a), sin(a)). By similar 
triangles, a vector M of length L has components Mx = L*cos(a), 
My = L*sin(a). Do this with both vectors M and P to get their 
components (Mx, My) and (Px, Py).

(2) You know that M + N = P. To add vectors, add their components:

   Mx + Nx = Px
   My + Ny = Py

You know Mx, My, Px, and Py, so you should be able to figure out Nx 
and Ny. These are the components of vector N.

(3) You were also asked for the magnitude and direction of vector N. 
To do this, you have to reverse step 1. Here's how, using the figure 
(remember, you'll be doing this for vector N, not vector M).

   Magnitude(M) = Mx^2 + My^2 (the Pythagorean Theorem where ^2 means     

   tangent(a) = sin(a)/cos(a) = My/Mx (by similar triangles again)

So Direction(M) = a = inverse tangent of a = arctan(a)

Those are the tools you'll need. See if you can do the job now. Write 
back if you're still confused after you've tried it.

- Doctor Rick, The Math Forum
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Associated Topics:
High School Physics/Chemistry

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