Questions: F1=800 145^circ F2=700 215^circ F3=900 290^circ F4=600 340^circ

F1=800  145^circ  F2=700  215^circ  F3=900  290^circ  F4=600  340^circ
Transcript text: $F_{1}=800 @ 145^{\circ} \\ F_{2}=700 @ 215^{\circ} \\ F_{3}=900 @ 290^{\circ} \\ F_{4}=600 @ 340^{\circ}$
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Solution

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Solution Steps

Step 1: Convert Forces to Cartesian Coordinates

First, we convert each force from polar coordinates to Cartesian coordinates using the formulas: Fx=Fcos(θ) F_x = F \cos(\theta) Fy=Fsin(θ) F_y = F \sin(\theta)

For F1=800N F_1 = 800 \, \text{N} at 145 145^\circ : F1x=800cos(145)=800cos(145)=800×(0.8192)=655.4N F_{1x} = 800 \cos(145^\circ) = 800 \cos(145^\circ) = 800 \times (-0.8192) = -655.4 \, \text{N} F1y=800sin(145)=800sin(145)=800×0.5736=458.9N F_{1y} = 800 \sin(145^\circ) = 800 \sin(145^\circ) = 800 \times 0.5736 = 458.9 \, \text{N}

For F2=700N F_2 = 700 \, \text{N} at 215 215^\circ : F2x=700cos(215)=700cos(215)=700×(0.8192)=573.4N F_{2x} = 700 \cos(215^\circ) = 700 \cos(215^\circ) = 700 \times (-0.8192) = -573.4 \, \text{N} F2y=700sin(215)=700sin(215)=700×(0.5736)=401.5N F_{2y} = 700 \sin(215^\circ) = 700 \sin(215^\circ) = 700 \times (-0.5736) = -401.5 \, \text{N}

For F3=900N F_3 = 900 \, \text{N} at 290 290^\circ : F3x=900cos(290)=900cos(290)=900×0.3420=307.8N F_{3x} = 900 \cos(290^\circ) = 900 \cos(290^\circ) = 900 \times 0.3420 = 307.8 \, \text{N} F3y=900sin(290)=900sin(290)=900×(0.9397)=845.7N F_{3y} = 900 \sin(290^\circ) = 900 \sin(290^\circ) = 900 \times (-0.9397) = -845.7 \, \text{N}

For F4=600N F_4 = 600 \, \text{N} at 340 340^\circ : F4x=600cos(340)=600cos(340)=600×0.9397=563.8N F_{4x} = 600 \cos(340^\circ) = 600 \cos(340^\circ) = 600 \times 0.9397 = 563.8 \, \text{N} F4y=600sin(340)=600sin(340)=600×(0.3420)=205.2N F_{4y} = 600 \sin(340^\circ) = 600 \sin(340^\circ) = 600 \times (-0.3420) = -205.2 \, \text{N}

Step 2: Sum the Cartesian Components

Next, we sum the x x -components and y y -components of all forces: Fx=F1x+F2x+F3x+F4x=655.4573.4+307.8+563.8=357.2N \sum F_x = F_{1x} + F_{2x} + F_{3x} + F_{4x} = -655.4 - 573.4 + 307.8 + 563.8 = -357.2 \, \text{N} Fy=F1y+F2y+F3y+F4y=458.9401.5845.7205.2=993.5N \sum F_y = F_{1y} + F_{2y} + F_{3y} + F_{4y} = 458.9 - 401.5 - 845.7 - 205.2 = -993.5 \, \text{N}

Step 3: Calculate the Magnitude of the Resultant Force

Finally, we calculate the magnitude of the resultant force using the Pythagorean theorem: Fresultant=(Fx)2+(Fy)2=(357.2)2+(993.5)2 F_{\text{resultant}} = \sqrt{(\sum F_x)^2 + (\sum F_y)^2} = \sqrt{(-357.2)^2 + (-993.5)^2} Fresultant=127,584.84+987,052.25=1,114,637.091055N F_{\text{resultant}} = \sqrt{127,584.84 + 987,052.25} = \sqrt{1,114,637.09} \approx 1055 \, \text{N}

Final Answer

Fresultant1055N \boxed{F_{\text{resultant}} \approx 1055 \, \text{N}}

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