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

Solution Steps

First, we convert each force from polar coordinates to Cartesian coordinates using the formulas: \[ F_x = F \cos(\theta) \] \[ F_y = F \sin(\theta) \]

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

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

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

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

Next, we sum the \( x \)-components and \( y \)-components of all forces: \[ \sum F_x = F_{1x} + F_{2x} + F_{3x} + F_{4x} = -655.4 - 573.4 + 307.8 + 563.8 = -357.2 \, \text{N} \] \[ \sum F_y = F_{1y} + F_{2y} + F_{3y} + F_{4y} = 458.9 - 401.5 - 845.7 - 205.2 = -993.5 \, \text{N} \]

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

Final Answer