To find the resultant of the three concurrent coplanar forces, we can resolve each force into its horizontal (x) and vertical (y) components, then add them up to find the total components in the x and y directions. Finally, we'll use these components to find the magnitude and direction of the resultant force.

Force 1: $60 \, \text{N}$ acting $20^\circ$ east of north

• This force has components:
  • Horizontal (x-component): $F_{1x} = 60 \sin(20^\circ)$
  • Vertical (y-component): $F_{1y} = 60 \cos(20^\circ)$

Force 2: $30 \, \text{N}$ east

• This force is purely horizontal:
  • Horizontal (x-component): $F_{2x} = 30$
  • Vertical (y-component): $F_{2y} = 0$

Force 3: $150 \, \text{N}$ acting $40^\circ$ east of south

• This force is acting south and east, so its components are:
  • Horizontal (x-component): $F_{3x} = 150 \sin(40^\circ)$
  • Vertical (y-component): $F_{3y} = -150 \cos(40^\circ)$ (negative because it is southward)

Step 1: Resolve each force into components

1. Force 1:

   • $F_{1x} = 60 \sin(20^\circ) \approx 60 \times 0.3420 = 20.52 \, \text{N}$
   • $F_{1y} = 60 \cos(20^\circ) \approx 60 \times 0.9397 = 56.38 \, \text{N}$

2. Force 2:

   • $F_{2x} = 30 \, \text{N}$
   • $F_{2y} = 0 \, \text{N}$

3. Force 3:

   • $F_{3x} = 150 \sin(40^\circ) \approx 150 \times 0.6428 = 96.42 \, \text{N}$
   • $F_{3y} = -150 \cos(40^\circ) \approx -150 \times 0.7660 = -114.90 \, \text{N}$

Step 2: Add the components

Now, we sum the components in the $x$-direction and $y$-direction:

• Total horizontal component (x): $F_{\text{Rx}} = F_{1x} + F_{2x} + F_{3x} = 20.52 + 30 + 96.42 = 146.94 \, \text{N}$

• Total vertical component (y): $F_{\text{Ry}} = F_{1y} + F_{2y} + F_{3y} = 56.38 + 0 - 114.90 = -58.52 \, \text{N}$

Step 3: Calculate the magnitude of the resultant force

The magnitude $R$ of the resultant force is given by the Pythagorean theorem: $R = \sqrt{F_{\text{Rx}}^2 + F_{\text{Ry}}^2}$ $R = \sqrt{146.94^2 + (-58.52)^2} \approx \sqrt{21575.55 + 3428.43} = \sqrt{25004} \approx 158.12 \, \text{N}$

Step 4: Calculate the direction of the resultant force

The angle $\theta$ of the resultant force relative to the east (positive x-axis) is given by: $\theta = \tan^{-1}\left( \frac{F_{\text{Ry}}}{F_{\text{Rx}}} \right)$ $\theta = \tan^{-1}\left( \frac{-58.52}{146.94} \right) \approx \tan^{-1}(-0.398) \approx -21.8^\circ$ This means the angle is $21.8^\circ$ south of east.

Final Result:

The resultant force has:

• Magnitude: $158.12 \, \text{N}$
• Direction: $21.8^\circ$ south of east

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Do you want more details or have any questions?

Here are some related questions you might find useful:

1. How would you calculate the resultant of three forces acting at different angles (non-coplanar)?
2. What would happen if two forces were exactly opposite in direction?
3. How do you resolve forces acting in three dimensions?
4. What is the significance of the angle in determining the direction of a force?
5. Can the law of sines and cosines be used to find the resultant of non-concurrent forces?

Tip: When working with forces at angles, always remember to break them down into horizontal and vertical components before summing them up!