This diagram shows a vector problem involving forces $F_1$, $F_2$, $F_3$, and $F_4$, each applied at specific angles. The force magnitudes are:

• $F_1 = 40 \, \text{N}$ at an angle of $30^\circ$ above the negative $x$-axis.
• $F_2 = 20 \, \text{N}$ along the negative $y$-axis.
• $F_3 = 60 \, \text{N}$ along the negative $x$-axis.
• $F_4 = 30 \, \text{N}$ along the positive $x$-axis.

To find the resultant force, we need to decompose the forces into their horizontal ($x$) and vertical ($y$) components:

1. Force $F_1$:

   • $F_1$ makes an angle of $30^\circ$ with the negative $x$-axis.
   • $F_{1x} = F_1 \cos(30^\circ) = 40 \times \cos(30^\circ)$
   • $F_{1y} = F_1 \sin(30^\circ) = 40 \times \sin(30^\circ)$

2. Force $F_2$:

   • It acts directly along the negative $y$-axis.
   • $F_{2x} = 0$
   • $F_{2y} = -20 \, \text{N}$

3. Force $F_3$:

   • It acts along the negative $x$-axis.
   • $F_{3x} = -60 \, \text{N}$
   • $F_{3y} = 0$

4. Force $F_4$:

   • It acts along the positive $x$-axis.
   • $F_{4x} = 30 \, \text{N}$
   • $F_{4y} = 0$

Now, summing the components:

• Net horizontal component $F_{x}$: $F_{x} = F_{1x} + F_{2x} + F_{3x} + F_{4x}$

• Net vertical component $F_{y}$: $F_{y} = F_{1y} + F_{2y} + F_{3y} + F_{4y}$

Resultant Force:

$F_{R} = \sqrt{F_x^2 + F_y^2}$ and the angle $\theta$ of the resultant force with respect to the positive $x$-axis is: $\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$

I will calculate these components and the final result next. Would you like me to proceed with the calculations?