Two force vectors F1 = (2.5 N) x - (3.0 N) y and F2 = - (6.0 N) x + (4.8 N) y are applied to a particle. What third force F3 would make the net, or resultant, force on the particle zero?

Magnitude
___ N
Direction
___° (from the +x axis)

To find the third force F3 that would make the net force on the particle zero, we need to find the vector sum of force vectors F1 and F2. The resultant force should have both magnitude and direction equal to zero.

First, let's add the x-components and y-components of the forces separately.

For the x-components:
F1x = 2.5 N
F2x = -6.0 N

Summing the x-components:
Fx = F1x + F2x
= 2.5 N + (-6.0 N)
= -3.5 N

For the y-components:
F1y = -3.0 N
F2y = 4.8 N

Summing the y-components:
Fy = F1y + F2y
= -3.0 N + 4.8 N
= 1.8 N

Now we have the resultant force components Fx and Fy.

To make the net force zero, both components should be zero.
Therefore, we set Fx = 0 and Fy = 0, and solve for the third force components.

0 = -3.5 N + F3x
F3x = 3.5 N

0 = 1.8 N + F3y
F3y = -1.8 N

So, the magnitude of the third force, F3, is:
|F3| = sqrt(F3x^2 + F3y^2)
= sqrt((3.5 N)^2 + (-1.8 N)^2)
≈ 3.96 N

To find the direction, we can use the arctan function.

θ = arctan(F3y / F3x)
= arctan((-1.8 N) / (3.5 N))
≈ -26.9°

However, we need to specify the direction from the +x axis, which means we need to add 180° to get a positive angle.

Direction = -26.9° + 180°
≈ 153.1°

Therefore, the third force vector F3 that would make the net force on the particle zero has a magnitude of approximately 3.96 N and is directed at approximately 153.1° from the +x axis.