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December 3rd, 2009, 05:36 PM   #1
Joined: Mar 2009
From: San Bernardino, California

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Direction cosine, components problem

Hi everyone, it's been a while since I've posted here, but I am having a bit of trouble understanding some mathematical concepts being used in my vector mechanics class, particularly the subject of direction cosines. I'm very proficient with vectors but this particular concept is escaping me (alone with other mechanics based ones) due to the terrible nature of my book (Vector Statics for Engineers by Beer-Johnson). Here's my problem:


Now in my book, it says given the angle between a particular coordinate axis and the line of action of vector (measured from the positive axis) one can determine the components of the vector via the following formulas:

$\displaystyle F_x=F\cos(\theta_x)$
$\displaystyle F_y=F\cos(\theta_y)$
$\displaystyle F_z=F\cos(\theta_z)$

In this problem, I did so and got the following:

$\displaystyle F_x=200\cos(25)=$181.26
$\displaystyle F_y=200\cos(30)=$173.20
$\displaystyle F_z=200\cos(110)=$-68.40

$\displaystyle \vec {F}=<181.26, 173.20, -68.40>$

However, the answers listed by the book differ. After inspecting the solutions manual, this is what they did:

$\displaystyle F_x=200\cos(30)\cos(25)=$157
$\displaystyle F_y=200\sin(30)=$1001
$\displaystyle F_z=-200\cos(30)\sin(25)=$-73.2

This I do not understand. I understand using -cos(30) in Fz and I see where I went wrong with cos(30) instead of sin(30) on Fy, but I don't see why I need to multiply by the additional trig functions in Fx and Fz. Could someone please explain this to me?

Last edited by skipjack; February 28th, 2018 at 01:47 PM.
cmmcnamara is offline  
December 4th, 2009, 02:21 AM   #2
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Re: Direction cosine, components problem

This is an exercise in using a spherical coordinate system.

You are given a magnitude and two (or three) angles, and you have to deduce the Cartesian coordinates. The problem is: between what and what are the angles measured?

Suppose that you start with a flat sheet of paper with an x-axis and a y-axis. If the magnitude of the vector is "r", and "a" is the angle between the vector and the x-axis, you can probably tell that the coordinates are
x = r cos(a)
y = r sin(a)
Now suppose that you lift the pointed end of the vector of the paper, leaving the tail at the origin (0,0). If you lift it by an angle of "b" away from the paper, then
z = r sin(b)
but this alters the x and y values; as the vector becomes more vertical, the x and y coordinates diminish, proportionate to cos(b):
x = r cos(a) cos(b)
y = r sin(a) cos (b)
and this gives a spherical coordinate system; not the standard one, but functional.

Now, what is going on in your problem? On the basis of the solution manual, it appears that:
* 30 degrees is the angle between the vector and the XZ plane (y=0), with positive angles for y>0;
* If the vector is flattened onto the XZ plane (i.e. the y component is deleted), 25 degrees is the angle created, starting from the x axis and moving towards the z axis.
So the angles given clearly aren't direction cosines.
aswoods is offline  
December 4th, 2009, 03:58 PM   #3
Joined: Mar 2009
From: San Bernardino, California

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Re: Direction cosine, components problem

Ok....that makes a lot of sense now. I have done spherical coordinates before and I was considering their use, but I'm not sure if that's the type of notation my professor will give credit for. You are correct in the angle measurements, I have misunderstood the idea behind a direction cosine.
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