# 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°

Result

F =  113.122 kN
f =  65.613 °

#### Solution:

$F_{1}=50 \ \text{kN} \ \\ F_{2}=40 \ \text{kN} \ \\ F_{3}=40 \ \text{kN} \ \\ a_{1}=30 \ ^\circ \ \\ a_{2}=a_{1}+45=30+45=75 \ ^\circ \ \\ a_{3}=a_{2}+25=75+25=100 \ ^\circ \ \\ x=F_{1} \cdot \ \cos( a_{1} ^\circ \rightarrow\ \text{rad})+F_{2} \cdot \ \cos( a_{2} ^\circ \rightarrow\ \text{rad})+F_{3} \cdot \ \cos( a_{3} ^\circ \rightarrow\ \text{rad})=F_{1} \cdot \ \cos( a_{1} ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )+F_{2} \cdot \ \cos( a_{2} ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )+F_{3} \cdot \ \cos( a_{3} ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=50 \cdot \ \cos( 30 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )+40 \cdot \ \cos( 75 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )+40 \cdot \ \cos( 100 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=46.7081 \ \\ y=F_{1} \cdot \ \sin( a_{1} ^\circ \rightarrow\ \text{rad})+F_{2} \cdot \ \sin( a_{2} ^\circ \rightarrow\ \text{rad})+F_{3} \cdot \ \sin( a_{3} ^\circ \rightarrow\ \text{rad})=F_{1} \cdot \ \sin( a_{1} ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )+F_{2} \cdot \ \sin( a_{2} ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )+F_{3} \cdot \ \sin( a_{3} ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=50 \cdot \ \sin( 30 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )+40 \cdot \ \sin( 75 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )+40 \cdot \ \sin( 100 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=103.02934 \ \\ \ \\ F=\sqrt{ x^2+y^2 }=\sqrt{ 46.7081^2+103.0293^2 } \doteq 113.1225 \doteq 113.122 \ \text{kN}$
$f=\dfrac{ 180^\circ }{ \pi } \cdot \arctan(y/x)=\dfrac{ 180^\circ }{ \pi } \cdot \arctan(103.0293/46.7081) \doteq 65.613 \doteq 65.613 ^\circ \doteq 65^\circ 36'47"$

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