Let's solve the problem step by step.

Given forces:

• $F_1 = 164 \, \text{N}$ (to the right, along the x-axis)
• $F_2 = 286 \, \text{N}$ (to the right, along the x-axis)
• $F_3 = 263 \, \text{N}$ at $60^\circ$ above the horizontal (this force has both x and y components)
• $F_4 = 161 \, \text{N}$ (to the left, along the negative x-axis)

Step 1: Resolve each force into components.

• $F_1$ is entirely in the positive x-direction, so:
  $F_{1x} = 164 \, \text{N}, \quad F_{1y} = 0$

• $F_2$ is entirely in the positive x-direction, so:
  $F_{2x} = 286 \, \text{N}, \quad F_{2y} = 0$

• $F_3$ has both x and y components:
  $F_{3x} = 263 \times \cos(60^\circ) = 263 \times 0.5 = 131.5 \, \text{N}$
  $F_{3y} = 263 \times \sin(60^\circ) = 263 \times 0.866 = 227.1 \, \text{N}$

• $F_4$ is entirely in the negative x-direction, so:
  $F_{4x} = -161 \, \text{N}, \quad F_{4y} = 0$

Step 2: Sum the components.

X-components:

$F_{x,\text{total}} = F_{1x} + F_{2x} + F_{3x} + F_{4x} = 164 + 286 + 131.5 - 161 = 420.5 \, \text{N}$

Y-components:

$F_{y,\text{total}} = F_{1y} + F_{2y} + F_{3y} + F_{4y} = 0 + 0 + 227.1 + 0 = 227.1 \, \text{N}$

Step 3: Find the magnitude of the resultant force.

Using the Pythagorean theorem: $F_{\text{resultant}} = \sqrt{(F_{x,\text{total}})^2 + (F_{y,\text{total}})^2} = \sqrt{(420.5)^2 + (227.1)^2}$ $F_{\text{resultant}} = \sqrt{176820.25 + 51516.41} = \sqrt{228336.66} = 477.81 \, \text{N}$

Step 4: Find the angle of the resultant force.

The angle $\theta$ is given by: $\theta = \tan^{-1} \left( \frac{F_{y,\text{total}}}{F_{x,\text{total}}} \right) = \tan^{-1} \left( \frac{227.1}{420.5} \right)$ $\theta = \tan^{-1}(0.54) = 28.2^\circ$

Final Answer:

The resultant angle of the forces is $28.2^\circ$ above the horizontal.

Would you like further details or have any questions?

Relative Questions:

1. How do you resolve forces into components in general?
2. What is the significance of using trigonometric functions in force resolution?
3. How does changing the angle of $F_3$ affect the resultant force?
4. Can the resultant force ever have a negative angle? Why or why not?
5. How would the result differ if one of the forces was removed?

Tip:

Always break forces into components along standard axes to simplify vector addition.