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Vector Addition =(

Started by egrooker Jan 21, 2003 at 7:10 PM 1 reply 500+ views

Original Post

Jan 21, 2003 at 7:10 PM

Addition of two vectors Vectors A and B have magnitudes of 1.50 m and 2.30 m, respectively, and orientations as shown in the figure below. Using graphical methods, determine the magnitude of A+B.

x

|

|    /Avector (1.50m)
|   /
|  /
|_/_Theta=45d_______________________y
| \ Theta=30d
|  \ 
|   \Bvector (2.30m)
|          
|

the answer i get is 1.5..with a few other numbers i''m pretty sure that''s right. or is it?? help by the way i didn''t use "graphical methods"...i assume thats just guessing?

Californium

Jan 21, 2003 at 7:49 PM


A + B = C

|C| = sqrt((cos(45)*1.5 + cos(30)*2.3)^2 + (sin(45)*1.5 + sine(30)*2.3)^2);

|C| = 1.51596616099 

*This is all elementary ofcourse.

[edited by - Californium on January 21, 2003 9:06:22 PM]

grhodes_at_work

Mod

197

grhodes_at_work

Jan 21, 2003 at 8:16 PM

Your answer is wrong. Californium''s approach is correct, but the answer also wrong. Californium used the wrong angle for the bottom vector (35 vs. 30) and evaluated cos and sin using functions that treat the parameters as radians, but the sent values are in degrees. This is a critical factor in calculating the correct result numerically using a computer or calculator. The parameters have to be converted to radians before the cos() and sin() functions can be used in a computer, though most scientific calculators have different modes and can take degrees or radians. The correct answer is a bit more than 3.

Unfortunately, this reeks of homework question. This is obvious in the way the question is stated, and the specific instruction to use graphical methods (a method used to teach trig).

You''ve only recently registered with gamedev.net, so you may not have read the forum policy on homework. You can review the policy here:

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Sorry to have to do this, but I am closing the thread. In the future, please do not post homework problems here.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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