What is the magnitude of the third force?

Physics Forces Question - magnitude of force?

• Three forces act on a particle that moves with a constant velocity, v= 2m/s x - 7m/s y. Two of the forces acting on this particle are F1 = 4.0N x + -1.0N y + -5.0N z and F2 = 5.0N x + -5.0N y + 8.0N z. What is the magnitude of the third force? x, y and z are just variables in the question can anyone help explain this to me? thanks :D

• Answer:

  Let me clear up one point. The x, y and z bits are the three axes of a co-ordinate system. Since the particle is moving at CONSTANT velocity, THERE IS NO NET FORCE acting on it. If there was, it would be ACCELERATING! That means that the missing force must be cancelling out the other two. So we can say that, F2 - (F1 + F2) = 0 F3 = F1 + F2 I am assuming that F1 = 4x - y - 5z and F2 = 5x - 5y + 8z To add F1 and F2 you just add the corresponding x, y and z components. F3 = (4x +5x) + (-y - 5y) + (-5z + 8z) = 9x - 6y + 3z Magnitude of F3 = √(9^2 + (-6)^2 + 3^2) = √ 126 = 11.22 N to two decimal places. OK?