Magnitude of resultant of three force as small as possible

[redtop2]

Homework Statement

determine the magnitude of the force F3, so that the resultant FR of the three force is as small as possible. F1= 20KN [E36.9N] F2= 12KN F3= x KN [W45N]

Homework Equations

Fx= xcosθ Fy= xSinθ FR = F1 + F2 + F3

The Attempt at a Solution

i used component method for this question and separated each vector in component.

and i got, F1x= 20KN(Cos 36.9) = 16KN F1y= 20KN (Sin 36.9) = approx. 12KN
F2x = 0KN since it has no horizontal component F2y= -12KN since the force points at south

so i thought FRx= F1x+ F2x= 0+16= 16KN and FRy= 0 KN (F1y + F2y = 12 + (-12) = 0KN)

therefore, resultant of two force has only x component which is 16KN so it points horizontally to the right along x-axis so in order to have smallest possible resultant of the vectors when F3 added would be 0KN all together.

so it has to have 16KN in the West side (- x axis) to cancel the resultant of F1+F2 but because the F3 has angular direction of [W45N] it has to have same magnitude along both horizontal and vertical since cos 45 and sin 45 has same value. it would been much easier if F3 had no direction and the answer would been 16KN [W] but i am stuck from here. please help me.

 

[voko]

You did well calculating the resultant of forces 1 and 2. Now you need to calculate the resultant of the resultant which you calculated with force 3; specifically, you need to compute its magnitude and then see what value of $x$ minimizes it.