Question

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The Particle is at rest find the magnitude and direction of F.

F1=6N
F2=10N

Answer 1

Could anyone just describe to me how to attack this problem?

Answer 2

You need to break down each force into its components (along x and y).

Then equate all the x-components and solve for what Fx must be to satisfy this.
Do the same for the y-components and solve for Fy.

At rest means Fnet = 0. That is the motivation for this approach.

Answer 3

F1= 6N and 30 degrees I get to ((-5.2) ; 3) (x ; y)
F2=10N and 45 degrees i get to ((-7.1) ; (-7,1))

and those together is ((-12.3) ; (-4.1)

it should be F1+F2=F?
so F= ((-12.3) ; (-4.1) and the magnitude 13N and direction 19.2 degrees?

Answer 4

F is both positive in x and y, so your vector F is correct except the values should be positive.

The magnitude is unaffected, so 13 N is correct.

I get an angle of 18.4 degrees, so i'm not sure if you've rounded off or something, but you're close enough for me.

You can do it 3 ways:
angle = arctan(4.1/12.3) = 18.4 degrees
angle = arcsin(4.1/13) = 18.4 degrees
angle = arccos(12.3/13) = 18.9 degrees

They all vary slightly due to rounding error, but are generally the same.

Answer 5

Good work! :D

Answer 6

Thanks alot! :)