Forces with magnitudes F1 = 42N and F2 = 35N act at a common point and make an angle of 77°12'. How big is their resultant?


F =  60.335 N


F1=42 N F2=35 N u=77+12/60=3865=77.2  x0=F1=42 y0=0 x=x0+F2 cos(u rad)=x0+F2 cos(u π180 )=42+35 cos(77.2 3.1415926180 )=49.7542 y=y0+F2 sin(u rad)=y0+F2 sin(u π180 )=0+35 sin(77.2 3.1415926180 )=34.13023 F=x2+y2=49.75422+34.1302260.335360.335 NF_{1}=42 \ \text{N} \ \\ F_{2}=35 \ \text{N} \ \\ u=77+12/60=\dfrac{ 386 }{ 5 }=77.2 \ ^\circ \ \\ x_{0}=F_{1}=42 \ \\ y_{0}=0 \ \\ x=x_{0} + F_{2} \cdot \ \cos( u ^\circ \rightarrow\ \text{rad})=x_{0} + F_{2} \cdot \ \cos( u ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=42 + 35 \cdot \ \cos( 77.2 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=49.7542 \ \\ y=y_{0} + F_{2} \cdot \ \sin( u ^\circ \rightarrow\ \text{rad})=y_{0} + F_{2} \cdot \ \sin( u ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=0 + 35 \cdot \ \sin( 77.2 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=34.13023 \ \\ F=\sqrt{ x^2+y^2 }=\sqrt{ 49.7542^2+34.1302^2 } \doteq 60.3353 \doteq 60.335 \ \text{N}

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Two vectors given by its magnitudes and by included angle can be added by our vector sum calculator.
Pythagorean theorem is the base for the right triangle calculator.
See also our trigonometric triangle calculator.

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