find the resultant of each pair of forces
-forces of 7N east and and 12N north
(the answer to this was 13.9 N, N30°E... but i dont know how to get this answer)
2007-10-20 16:49:32 · 3 answers · asked by internet nickname 1 in Science & Mathematics ➔ Physics
to pariah z, why did you use tan THETA? (why couldn't it just be the tan of 7/12)
2007-10-20 17:41:40 · update #1
nvm, i know why you used theta now..
2007-10-20 18:00:18 · update #2
c^2= a^2 + b^2= 12^2 +7^2= 144 + 49
c= sqrt.193= 13.892N
tan[theta]= O/A
[theta]= tan^-1 (7/12)
[theta]= 30.26
13.9 N, N30°E
Drawing a right-angled triangle will help
2007-10-20 17:30:44 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The magnitude is calculated by => square root of ( force one squard + force two squared)
Now plot all three force on a graph. (Obviously the resultant force will be between the initial forces at some angle)
If you draw a line from the end of the resultant force to either the X or Y axis you will see that you made a right triangle in which you know the length of all sides. Use trig to figure out the angle.
cos(angle) = 7/13.9 = .5
Take the inverse cosine of .5 (in degree mode!) to get 30 degrees.
2007-10-21 00:43:51 · answer #2 · answered by ? 3 · 0⤊ 0⤋
arctan(1/12) = 30.256° E of N
R = 7/sin(30.256°) = 13.892 N
2007-10-21 00:22:55 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋