What is the net force (or single equivalent force) of the following three forces

F1- 50N East, F2-30 N NE and F3 60N W40degreesN.
(Answer as a size and direction)

When adding forces you start by breaking each force into orthogonal consituent parts - i.e a force component acting north + force component acting east. Then you add all the north comnponents of the three forces to get a total north componenet, and do the same for the east components of all three forces. Finally combine these totals into a single force with magnitude and direction. Post back with your attempt and we'll be happy to check it for you.

How do I deal with F2 and F3? Im not sure what to do because one is going NE and the other is going a 40 degree angle.

My attempt:

F1-50N East

F2-
I decided to solve by finding the x and y components and because its at going NE used 45 degrees as the angle.

Y-comp. 30Sin(45)=21.2N
X-comp. 30Cos(45)=21.2N

F3-
Y-comp 60Sin(40)=38.6
X-comp 60Cos(40=46N-Which is negative because its going West

Anding them together
X-Comp- 50N+21.2N+-46N=25.2N to the East
Y-Comp-21.2N+38.6N=59.8N North

Looking good so far. Last step is to combine the horizontal and vertical components of force. The magnitude of the resutant force vector comes from Pythagorus:

|F| = \sqrt{F_x^2 + F_y^2}

and the angle can be found from

\theta = Arctan(Fy/Fx)

Be careful when doing the arctangent calculation to be sure which quadrant \theta should be in.

Awesome thanks!

64.8N and 67.1

Awesome thanks!

64.8N and 67.1

Be careful to define how the angle you got of 67.1 degrees is measured. Is it a bearing from north? Or degrees north of east?