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:



and the angle can be found from



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

Awesome thanks!

64.8N and 67.1

Quote:

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Originally Posted by HSPhysics1001

Awesome thanks!

64.8N and 67.1

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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?