Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible.

Answer: FR=2.330kN

Explanation:

Write down x and y components.

Fx = FSin30°

Fy = FCos30°

Choose the forces acting up and right as positive.

∑ (FR) = ∑ (Fx)

(FR) x = 5-Fsin30° = 5-0.5F

(FR) y = Fcos30°-4 = 0.8660-F

Use Pythagoras theorem

F2R = √F2-11.93F+41

Differentiate both sides

2FRdFR/dF = 2F - 11.93

Set dFR/dF to 0

2F = 11.93

F = 5.964kN

Substitute value back into FR

FR = √F2 (F square) - 11.93F + 41

FR=√ (5.964) (5.964) - 11.93 (5.964) + 41

FR = 2.330kN

The minimum force is 2.330kN