Let's solve this problem using the component method for vector addition.

Given

• Magnitude of $\mathbf{A} = 5.25 \, \text{m}$, making an angle $20^\circ$ below the negative x-axis.
• Magnitude of $\mathbf{B} = 5.42 \, \text{m}$, making an angle $60^\circ$ above the positive x-axis.
• Magnitude of $\mathbf{C} = 3.26 \, \text{m}$, directed along the negative y-axis.

We will compute the resultant vector’s:

1. Magnitude $|\mathbf{R}|$
2. Directional angle $\theta$ (angle with respect to the positive or negative x-axis, less than $90^\circ$).

Step 1: Resolve vectors into components

1. $\mathbf{A}$

   • $A_x = -5.25 \cos(20^\circ)$ (negative x-component since it points left)
   • $A_y = -5.25 \sin(20^\circ)$ (negative y-component)

2. $\mathbf{B}$

   • $B_x = 5.42 \cos(60^\circ)$
   • $B_y = 5.42 \sin(60^\circ)$

3. $\mathbf{C}$

   • $C_x = 0$
   • $C_y = -3.26$

Step 2: Compute each component

Use trigonometric values:

• $\cos(20^\circ) \approx 0.9397$
• $\sin(20^\circ) \approx 0.3420$
• $\cos(60^\circ) = 0.5$
• $\sin(60^\circ) \approx 0.8660$

Components of each vector:

• $A_x = -5.25 \times 0.9397 \approx -4.93$

• $A_y = -5.25 \times 0.3420 \approx -1.79$

• $B_x = 5.42 \times 0.5 \approx 2.71$

• $B_y = 5.42 \times 0.8660 \approx 4.69$

• $C_x = 0$

• $C_y = -3.26$

Step 3: Sum of components

• $R_x = A_x + B_x + C_x$ $R_x = -4.93 + 2.71 + 0 \approx -2.22$

• $R_y = A_y + B_y + C_y$ $R_y = -1.79 + 4.69 - 3.26 \approx -0.36$

Step 4: Resultant magnitude

$|\mathbf{R}| = \sqrt{R_x^2 + R_y^2}$ $|\mathbf{R}| = \sqrt{(-2.22)^2 + (-0.36)^2} \approx \sqrt{4.93 + 0.13} \approx \sqrt{5.06} \approx 2.25 \, \text{m}$

Step 5: Directional angle

$\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$ $\theta = \tan^{-1}\left(\frac{-0.36}{-2.22}\right) \approx \tan^{-1}(0.162) \approx 9.2^\circ$ Since both $R_x$ and $R_y$ are negative, the vector lies in the third quadrant relative to the negative x-axis. The final angle relative to the positive x-axis is $180^\circ - 9.2^\circ \approx 170.8^\circ$.

Final Answer

1. Magnitude of the resultant vector: $2.25 \, \text{m}$
2. Directional angle: approximately $170.8^\circ$ above the positive x-axis.

Would you like a further breakdown or explanation of any step?

Related Questions:

1. How do we find vector components from polar coordinates?
2. Why do trigonometric signs depend on the quadrant?
3. What is the significance of the direction of a vector?
4. How do negative components affect the result in vector addition?
5. How can we apply this method to three-dimensional vectors?

Tip: Always check signs when resolving vectors; direction affects the final result.