Resolving the electric force into x y components?

[Engr.abshir]

Homework Statement

A 64 microC charge is locate 30 cm to the left of a 16 micro C charge. What is the resultant force
on a -12 micro C charge positioned exactly 50 mm below the 16 micro C charge?

Homework Equations

coulomb law f=k*qq/(r^2),resultant force,resolve into x and y component.

The Attempt at a Solution

the first three steps are straightforward
first step
using Pythagoras rule gives s=58.3 mm and theta = 59.0 degree
the force b/w q1 and q2 using coulomb law i calculated
F13 = 2033 N.
similarly ,the force between q2 and q3, gives correctly
F 23= 691 N.
i am having trouble understanding these steps about resolving it into components.i need some revision here.

Fx = 0- F13 cos 59.0 (where does the zero come from?)
= -(2033 N) cos 59.0
which gives Fx = -1047 N
also
Fy = F23 + F13 sin 59.00(why add force 23)
= 691 N + (2033 N) sin 590
Fy = 2434 N
i can do the rest.

 

[Engr.abshir]

also it says
F13=2033 N ,59 degree north of West.
F23=691 Newton,Upward.

I don't get how these directions come and what they mean in the answers.

 

[Engr.abshir]

anyone help here.i am revising tons of questions and this is standing in the way or preventing me from proceeding .
Thank you
PF

 

[haruspex]

This is one of those threads where it all seems so straightforward that I might misunderstand where your difficulties lie, so bear with me.

Fx = 0- F13 cos 59.0 (where does the zero come from?)

The 0 represents the x component of the force between q2 and q3.

Fy = F23 + F13 sin 59.00(why add force 23)

It looks like "F12" means the force exerted on q2 by 1, etc. In the y direction, you have both F23 and a component of F13.

I don't get how these directions come and what they mean in the answers.

F13 acts along the hypotenuse of the triangle. As you calculated, that is at 59 degrees to the x axis; specifically, 59 degrees N of W.

anyone help here

Unfortunately, if you make a second post to your own thread it no longer shows as "unanswered", so tends to take longer to get an answer. Better to edit your initial post.

 

[Engr.abshir]

greatly appreciated.