Resultant force

Calculate mathematically and graphically the resultant of a three forces with a common centre if:

F1 = 50 kN α1 = 30°
F2 = 40 kN α2 = 45°
F3 = 40 kN α3 = 25°

Correct result:

F =  113.1225 kN
f =  65.613 °

Solution:

$$\begin{gathered} F_1=50\text{ kN } \\ {F}_2=40\text{ kN } \\ {F}_3=40\text{ kN } \\ {a}_1=30^{\circ } \\ {a}_2=a_1+45=30+45=75^{\circ } \\ {a}_3=a_2+25=75+25=100^{\circ } \\ x=F_1\cdot \cos a_1^{\circ }+F_2\cdot \cos a_2^{\circ }+F_3\cdot \cos a_3^{\circ }=F_1\cdot \cos 30^{\circ }+F_2\cdot \cos 75^{\circ }+F_3\cdot \cos 100^{\circ }=50\cdot \cos 30^{\circ }+40\cdot \cos 75^{\circ }+40\cdot \cos 100^{\circ }=50\cdot 0.866025+40\cdot 0.258819+40\cdot (-0.173648)=46.7081 \\ y=F_1\cdot \sin a_1^{\circ }+F_2\cdot \sin a_2^{\circ }+F_3\cdot \sin a_3^{\circ }=F_1\cdot \sin 30^{\circ }+F_2\cdot \sin 75^{\circ }+F_3\cdot \sin 100^{\circ }=50\cdot \sin 30^{\circ }+40\cdot \sin 75^{\circ }+40\cdot \sin 100^{\circ }=50\cdot 0.5+40\cdot 0.965926+40\cdot 0.984808=103.02934 \\ F=\sqrt{x^2+y^2}=\sqrt{46.7081^2+103.0293^2}=113.1225\text{ kN} \end{gathered}$$

$f={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (y\mathrm{/}x)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arctan (103.0293\mathrm{/}46.7081)=65.613^{\circ }=65^{\circ }36^{\mathrm{\prime }}47\mathrm{"}$

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