Why Can't I Use Standard Cosine Calculations for Non-Orthogonal Axes Forces?

[goldfish9776]

Homework Statement

in this question , why can't i find Fu = 600cos 30= 520N ?
why can't I find Fv = 600cos 120 = -300N ?

Homework Equations

The Attempt at a Solution

Fu = 600cos 30= 520N ?Fv = 600cos 120 = -300N ?
[/B]

 

[SteamKing]

Homework Statement

in this question , why can't i find Fu = 600cos 30= 520N ?
why can't I find Fv = 600cos 120 = -300N ?

Homework Equations

The Attempt at a Solution

Fu = 600cos 30= 520N ?Fv = 600cos 120 = -300N ?[/B]

Well, for one thing, it appears that Fu and Fv are not the sides of a right triangle. It's hard to tell from the attached image.

For solving oblique triangles, the law of sines or the law of cosines must be used. For more information, see the article below:

https://en.wikipedia.org/wiki/Solution_of_triangles

 

[goldfish9776]

Well, for one thing, it appears that Fu and Fv are not the sides of a right triangle. It's hard to tell from the attached image.
Why can't I do in this way?

For solving oblique triangles, the law of sines or the law of cosines must be used. For more information, see the article below:

https://en.wikipedia.org/wiki/Solution_of_triangles

 

[goldfish9776]

Well, for one thing, it appears that Fu and Fv are not the sides of a right triangle. It's hard to tell from the attached image.

For solving oblique triangles, the law of sines or the law of cosines must be used. For more information, see the article below:

https://en.wikipedia.org/wiki/Solution_of_triangles

Why can't I do in this way??

 

[SteamKing]

Why can't I do in this way??

As has already been explained, the forces Fu and Fv are not the sides of a right triangle.

The triangle which contains Fu and Fv has angles of 120°, 30°, and 30°. Do you see any right angles there?

Haven't you learned the difference yet between an oblique triangle and a right triangle?

The solution in your text clearly shows the use of the Law of Sines.

Please study the material I linked to previously. If you have any questions about that, I will be happy to answer them.

 

[gneill]

Why can't I do in this way??

As was mentioned already, The u and v axes are not orthogonal (do not meet at a right angle). So you can't treat them as if they do.

If the using the laws of sines or cosines is throwing you off, you could approach the problem as two equations in two unknowns.

Suppose that F_u and F_v are two vectors with the given directions and you need to find F_u and F_v such that they sum to a 600 N horizontal resultant. Impose axes x-y on the image such that the x-axis coincides with the horizontal 600 N force. Then write equations summing the x and y components of F_u and F_v accordingly.