Here's a general approach to vector problems that I've found useful, using the magnitude-angle notation.
There is the original vector, 150$\angle$165°, with x-component $150 *$cos$(165°)$and y-component $150 *$sin$(165°)$
This is to be broken down into two non-orthogonal "components", F$_v$$\angle$90°, and F$_u$$\angle$195°
Each of these two last can be separately broken down into x and y components in the same way as the 150 N force, above.
Then the 150 N x-component equals the $F_v$ x-component plus the $F_u$ x-component, and the 150 N y-component equals the $F_v$ y-component plus the $F_u$ y-component. Two equations, two unknowns, all angles known , so just solve. Oh, and one equation has only one unknown!
2
user53153
On
Method 1 is correct. In method 2 you calculated orthogonal projections of the force $F_2$ onto the given axes. However, the components of a force coincide with orthogonal projections only when the axes are mutually orthogonal.
Agreed, @5PM.
Here's a general approach to vector problems that I've found useful, using the magnitude-angle notation.
There is the original vector, 150$\angle$165°, with x-component $150 *$cos$(165°)$and y-component $150 *$sin$(165°)$
This is to be broken down into two non-orthogonal "components", F$_v$$\angle$90°, and F$_u$$\angle$195°
Each of these two last can be separately broken down into x and y components in the same way as the 150 N force, above.
Then the 150 N x-component equals the $F_v$ x-component plus the $F_u$ x-component, and the 150 N y-component equals the $F_v$ y-component plus the $F_u$ y-component. Two equations, two unknowns, all angles known , so just solve. Oh, and one equation has only one unknown!