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 add the given forces, F1 and F2, and then find the negative of their sum. The magnitude and direction of the resultant force will give us the values we need for F3.
1. Add the two given forces, F1 and F2:
F1 = (2.5 N) x - (3.0 N) y
F2 = - (6.0 N) x + (4.8 N) y
F1 + F2 = (2.5 N - 6.0 N) x + (-3.0 N + 4.8 N) y
= -3.5 N x + 1.8 N y
2. Find the negative of the sum of the forces:
- (F1 + F2) = -(-3.5 N x + 1.8 N y)
= 3.5 N x - 1.8 N y
The magnitude of the third force, F3, is the magnitude of the negative of the sum of the forces, which is given by the Pythagorean theorem:
Magnitude of F3 = √[ (3.5 N)^2 + (1.8 N)^2 ]
Therefore, the magnitude of F3 is approximately:
Magnitude of F3 ≈ √[ 12.25 N^2 + 3.24 N^2 ]
≈ √[15.49 N^2 ]
≈ 3.94 N
To determine the direction of F3, we can use the inverse tangent (arctan) function with the components of the force:
Direction of F3 = arctan(1.8 N / 3.5 N)
Therefore, the direction of F3 is approximately:
Direction of F3 ≈ arctan(0.514)
≈ 28.2° (from the +x axis)
Hence, the third force F3 that would make the net force on the particle zero has a magnitude of approximately 3.94 N and is directed at an angle of approximately 28.2° from the +x axis.