F = √[(3cos90° - 5cos170°)² + (3sin90° - 5sin170°)²]

F = √[24.24615776 + 4.54439691], arctan(2.131759112/4.924038765)

F ≈ 5.37N, 23.41°

]]>∑F = √[(∑x)² + (∑y)²]

Specifically the resultant force is;

F = √[(∑x)² + (∑y)²], arctan(∑y/∑x)

I actually think that this method is easier because you can simply break all the forces into two component columns and add them up. Once you do that you can just plug them in to the above equation.

For the original equations posted this is what would happen.

F1 = 3N, 90° and F2 = 5N, 170°

F = √[(3cos90° + 5cos170°)² + (3sin90° + 5sin170°)²]

F = √[24.24615776 + 14.96328757], arctan(3.868240888/-4.924038765)

F ≈ 6.26N, -38.15°

]]>However, this only works if there are only two forces involved. Ricky's way will work for any number of forces.

]]>The vectors x direction is cos(θ). The y direction is sin(θ).

<x, y> = <cos(90°), sin(90°)>

However, this gives us a magntidue of 1. We don't want 1, we want 3:

3 * <cos(90°), sin(90°)> = <3*cos(90°), 3*sin(90°)>

And that gives us it in vector form.

Now do the same for the other vector, and you'll have to vectors:

v1 = <x1, y1> and v2 = <x2, y2>

v1 + v2 = <x1+x2, y1+y2>

v1 - v2 = <x1-x2, y1-y2>

F1= 3N at 90°

F2= 5N at 170°

]]>f2=170°

You gave a direction, but not a magnitude. You need both.

]]>Please help me!!!

ive just started doing vectors and i cant get around to doing this question:

Forces of f1=3N at 90° and f2=170° acting at a point.

Find F1 + F2 and F1 - F2

By calculation

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