Solving for the Third Force in a Simple Physics Problem with Constant Velocity

[Rasine]

Three forces act on a particle that moves with a constant velocity v = (2 m/s) x - (7 m/s) y. Two of the forces acting on this particle are F1 = (-9 N) x + (10 N) y + (2 N) z and F2 = (6 N) x + (1 N) y + (9 N) z. What is the magnitude of the third force?

so what i was trying to do is take the f1z component and the f2z compoment and finnding the magnitude by doing squroot(9^2+2^2)...but that is not how you do it...

can someone please point me in the right direction?

 

[Doc Al]

First find the components of the third force, considering that the net force on the particle must be zero.

 

[Rasine]

so if the net force does not change the velocity then f1 and f2 should yeild 0 right?

 

[cristo]

No. You need to find the components of the third force; F3. Since the acceleration is zero in each direction, you know that the sum of the components of the forces in each direction must be zero; i.e. f1x+f2x+f3x=0... etc.

 

[Rasine]

so how would i go about doing this... f1 and f2 should be equal right? um...so the magnitude of f3 should make be the variable that makes f1 and f2 =0 right?

 

[cristo]

Don't think about the magnitude yet. f1 and f2 are not equal; you are given the values of them. Set up three equations, f1x+f2x+f3x=0, and the other similar two. Then you know the values for f1x and f2x, and so obtain -9+6+f3x=0. This will give you the x component of f3. Use similar equations for the y and z components.