Two forces

The two forces F1 = 580N and F2 = 630N have the angle of 59 degrees. Calculate their resultant force F.

Result

F =  1053.418 N

Solution:

F1=580 N F2=630 N A=0  B=59   x=F1 cos(A rad)+F2 cos(B rad)=F1 cos(A π180 )+F2 cos(B π180 )=580 cos(0 3.1415926180 )+630 cos(59 3.1415926180 )=904.47399 y=F1 sin(A rad)+F2 sin(B rad)=F1 sin(A π180 )+F2 sin(B π180 )=580 sin(0 3.1415926180 )+630 sin(59 3.1415926180 )=540.0154 F=x2+y2=904.4742+540.015421053.41821053.418 NF_{1}=580 \ \text{N} \ \\ F_{2}=630 \ \text{N} \ \\ A=0 \ ^\circ \ \\ B=59 \ ^\circ \ \\ \ \\ x=F_{1} \cdot \ \cos( A ^\circ \rightarrow\ \text{rad}) + F_{2} \cdot \ \cos( B ^\circ \rightarrow\ \text{rad})=F_{1} \cdot \ \cos( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ ) + F_{2} \cdot \ \cos( B ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=580 \cdot \ \cos( 0 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) + 630 \cdot \ \cos( 59 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=904.47399 \ \\ y=F_{1} \cdot \ \sin( A ^\circ \rightarrow\ \text{rad}) + F_{2} \cdot \ \sin( B ^\circ \rightarrow\ \text{rad})=F_{1} \cdot \ \sin( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ ) + F_{2} \cdot \ \sin( B ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=580 \cdot \ \sin( 0 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) + 630 \cdot \ \sin( 59 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=540.0154 \ \\ F=\sqrt{ x^2+y^2 }=\sqrt{ 904.474^2+540.0154^2 } \doteq 1053.4182 \doteq 1053.418 \ \text{N}

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