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
