To find the magnitude and direction of the resultant force, we need to decompose the given forces into their respective $x$- and $y$-components, then use vector addition to combine them. Here's how we proceed:

Step 1: Decompose the forces into components

• For the 3 lb force at 51°:

  • $F_{x1} = 3 \cdot \cos(51^\circ)$
  • $F_{y1} = 3 \cdot \sin(51^\circ)$

• For the 6 lb force at 32° below the x-axis:

  • $F_{x2} = 6 \cdot \cos(32^\circ)$
  • $F_{y2} = -6 \cdot \sin(32^\circ)$ (since the angle is below the $x$-axis, the $y$-component is negative)

Step 2: Calculate the total $x$- and $y$-components

Add the components for each direction:

• Total $F_x = F_{x1} + F_{x2}$
• Total $F_y = F_{y1} + F_{y2}$

Step 3: Find the magnitude of the resultant force

Using the Pythagorean theorem, the magnitude $F_R$ of the resultant force is: $F_R = \sqrt{(F_x)^2 + (F_y)^2}$

Step 4: Find the direction of the resultant force

The angle $\theta$ that the resultant force makes with the positive $x$-axis is given by: $\theta = \tan^{-1} \left( \frac{F_y}{F_x} \right)$

Step 5: Perform the calculations

1. $F_{x1} = 3 \cdot \cos(51^\circ) = 3 \cdot 0.6293 = 1.8879 \, \text{lb}$

2. $F_{y1} = 3 \cdot \sin(51^\circ) = 3 \cdot 0.7771 = 2.3313 \, \text{lb}$

3. $F_{x2} = 6 \cdot \cos(32^\circ) = 6 \cdot 0.8480 = 5.088 \, \text{lb}$

4. $F_{y2} = -6 \cdot \sin(32^\circ) = -6 \cdot 0.5299 = -3.1794 \, \text{lb}$

Now, summing the components:

• $F_x = 1.8879 + 5.088 = 6.9759 \, \text{lb}$
• $F_y = 2.3313 + (-3.1794) = -0.8481 \, \text{lb}$

Step 6: Calculate the magnitude

$F_R = \sqrt{(6.9759)^2 + (-0.8481)^2} = \sqrt{48.6631 + 0.7193} = \sqrt{49.3824} = 7.03 \, \text{lb}$

Step 7: Calculate the angle

$\theta = \tan^{-1} \left( \frac{-0.8481}{6.9759} \right) = \tan^{-1}(-0.1216) = -6.95^\circ$ Since the angle is negative, it means the resultant is slightly below the $x$-axis.

Final Answer:

• Magnitude: $7.03 \, \text{lb}$
• Angle: $-6.95^\circ$ (with respect to the positive $x$-axis)

Let me know if you'd like further details or have questions!

Related Questions:

1. How do you calculate the components of a force at an angle?
2. What does a negative angle indicate in vector problems?
3. Why do we use the Pythagorean theorem to find the resultant force magnitude?
4. Can you explain when to use sine versus cosine for component calculations?
5. How would the solution change if the angles were measured from the negative x-axis?

Tip:

When dealing with angles below the axis, remember that the sine of the angle gives a negative component for the $y$-direction.