To find the resultant of the four forces acting on the body shown in the image, we need to break down each force into its $x$- and $y$-components, then sum these components separately to determine the overall resultant force in vector form.

Let's analyze each force based on its magnitude and direction:

1. 4000 lb Force: Positioned at a $30^\circ$ angle to the horizontal axis (angled upward to the left).
2. 5000 lb Force: Positioned at a $30^\circ$ angle below the horizontal axis (angled downward to the left).
3. 3000 lb Force: Positioned at a $30^\circ$ angle below the horizontal axis (angled downward to the right).
4. 2000 lb and 1000 lb Forces: These are vertical forces along the $y$-axis; 2000 lb is upward, and 1000 lb is downward.

Step 1: Calculate $x$- and $y$-components for each force

1. 4000 lb Force

• $F_{4000,x} = 4000 \cos(30^\circ) = 4000 \times 0.866 = 3464 \, \text{lb}$ (to the left, so $x$-component is negative)
• $F_{4000,y} = 4000 \sin(30^\circ) = 4000 \times 0.5 = 2000 \, \text{lb}$ (upward)

2. 5000 lb Force

• $F_{5000,x} = 5000 \cos(30^\circ) = 5000 \times 0.866 = 4330 \, \text{lb}$ (to the left, so $x$-component is negative)
• $F_{5000,y} = -5000 \sin(30^\circ) = -5000 \times 0.5 = -2500 \, \text{lb}$ (downward)

3. 3000 lb Force

• $F_{3000,x} = 3000 \cos(30^\circ) = 3000 \times 0.866 = 2598 \, \text{lb}$ (to the right, so $x$-component is positive)
• $F_{3000,y} = -3000 \sin(30^\circ) = -3000 \times 0.5 = -1500 \, \text{lb}$ (downward)

4. Vertical Forces (2000 lb upward and 1000 lb downward)

• $F_{2000,y} = +2000 \, \text{lb}$ (upward)
• $F_{1000,y} = -1000 \, \text{lb}$ (downward)

Step 2: Sum the $x$- and $y$-components

Total $x$-component:

$F_x = -3464 - 4330 + 2598 = -7196 \, \text{lb}$

Total $y$-component:

$F_y = 2000 - 2500 - 1500 + 2000 - 1000 = -1000 \, \text{lb}$

Step 3: Calculate the Resultant Force

The resultant force $R$ can be found using: $R = \sqrt{F_x^2 + F_y^2}$ Substituting the values: $R = \sqrt{(-7196)^2 + (-1000)^2} = \sqrt{51714016 + 1000000} = \sqrt{52714016} \approx 7262 \, \text{lb}$

Step 4: Determine the Angle of Resultant Force

The angle $\theta$ of the resultant force with respect to the $x$-axis is given by: $\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$ Substituting the values: $\theta = \tan^{-1}\left(\frac{-1000}{-7196}\right) = \tan^{-1}(0.1389) \approx 7.9^\circ \, \text{(above the negative x-axis)}$

Final Answer

• Magnitude of the Resultant Force: $R \approx 7262 \, \text{lb}$
• Direction: $7.9^\circ$ above the negative $x$-axis

Would you like a more detailed breakdown or have any questions?

Here are five related questions for further exploration:

1. How would the resultant change if one of the forces was increased or decreased?
2. What would happen if the angle of one of the forces was modified?
3. How can we use this method to determine equilibrium conditions?
4. How does changing the reference angle affect the calculation?
5. What if a force was acting at an arbitrary angle—how would that be calculated?

Tip: Always ensure to separate force components into perpendicular axes (usually $x$ and $y$) to simplify the analysis of forces in vector mechanics.