The problem involves finding the magnitude and direction of the vector sum of three forces given in the diagram. The three forces are:

• $\vec{F}_1 = 300 \, \text{N}$ at an angle of $45^\circ$ above the negative $x$-axis,
• $\vec{F}_2 = 155 \, \text{N}$ at an angle of $53^\circ$ below the negative $x$-axis,
• $\vec{F}_3 = 200 \, \text{N}$ at an angle of $30^\circ$ above the positive $x$-axis.

To solve this problem graphically:

1. Resolve each force into components: For each force $F_i$, resolve it into its $x$- and $y$-components using trigonometry:

   • For $\vec{F}_1$: $F_{1x} = -300 \cos(45^\circ), \quad F_{1y} = 300 \sin(45^\circ)$
   • For $\vec{F}_2$: $F_{2x} = -155 \cos(53^\circ), \quad F_{2y} = -155 \sin(53^\circ)$
   • For $\vec{F}_3$: $F_{3x} = 200 \cos(30^\circ), \quad F_{3y} = 200 \sin(30^\circ)$

2. Sum the components: Find the total $x$-component ($F_x$) and $y$-component ($F_y$) of the resultant force by summing the components of all three forces: $F_x = F_{1x} + F_{2x} + F_{3x}$ $F_y = F_{1y} + F_{2y} + F_{3y}$

3. Calculate the magnitude and direction of the resultant force:

   • Magnitude of the resultant force $F$ is given by: $F = \sqrt{F_x^2 + F_y^2}$
   • The direction (angle $\theta$) of the resultant vector relative to the positive $x$-axis is: $\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$

Would you like me to compute the exact values for the components, magnitude, and direction?

Here are some related questions:

1. How are force vectors resolved into components in general cases?
2. What is the significance of the angles when resolving forces?
3. How can you determine if the graphical sum method yields a correct result?
4. What happens if forces are not balanced in magnitude and direction?
5. How does the principle of superposition apply to vector addition?

Tip: Always resolve vectors into their components before performing any summation; it simplifies the math considerably!