Finding Resultant Force and Equilibrant: Components Math Problem

[menal]

Homework Statement

two forces of 40N and 50N act at an angle of 60 degrees. determine the resultant force and the equilbrant

this is what i have so faar:
resultant=F1 + F2
let F1 be along the x axis
add x and y components:

F1x=40N
F1y= none since its along the x-axis.

F2x=F2cos60
F2y=F2sin60

add the forces in the y direction:=F2sin60
add the forces in the x direction:=40N+F2cos60

what would i do after this?

 

[elabed haidar]

F resultant = radical{(F2x)^2 + (F2y)^2}
that is the magnitude
as for the direstion find tan alpha where alpha is the angle between the x-axis and the resultant which is the division of F2y over F2x

 

[HallsofIvy]

Homework Statement

two forces of 40N and 50N act at an angle of 60 degrees. determine the resultant force and the equilbrant

this is what i have so faar:
resultant=F1 + F2
let F1 be along the x axis
add x and y components:

F1x=40N
F1y= none since its along the x-axis.

F2x=F2cos60
F2y=F2sin60

add the forces in the y direction:=F2sin60
add the forces in the x direction:=40N+F2cos60

what would i do after this?

You are given that F2= 50 N. Why have you not put that in?
The x component is, as you said 40+ 50 cos(60) N and the y component is 50 sin(60).

Once you have found the components, Fx and Fy, its magnitude is
$$\sqrt{F_x^2+ F_y^2}$$
(NOT $\sqrt{F_ {2X}^2+ F_{2y}^2}$ as elabed haidar has it. I think the fact that you did not put the value of F2 into the equation made him think F2 was the resultant)
and the angle the resultant makes with the F1 is
$$arctan\left(\frac{F_y}{F_x}\right)$$.

Personally, I wouldn't have used "components" at all. If you attach the two vectors "end to end" you have a triangle with two sides having length 40 and 50 and angle between them 180- 60= 120 degrees. You can use the cosine law to find the length of the opposite side (the magnitude of the resultant force) and then the sine law to find the angle the resultant force makes with whichever force you used as the first side of the triangle.

 

[menal]

alright, so i did what you said:
I used the cosine law: (40)^2+(50)^2-2(40)(50)cos120 and gt sqrt(6100)= 78.10N
then i used the sine law: (a(sinB))/b --> (40)(sin120)/78.10 and got 26.3 degrees.
is that it? i know i found the resultant force, and now i found the equilibrant too?