Let's break down the task to solve it step by step:

1. The magnitude and direction of the total force along the vertical:

We'll sum the vertical components of each force. First, let's break each force into its vertical and horizontal components. The general formulas are:

• Vertical component: $F_y = F \cdot \sin(\theta)$
• Horizontal component: $F_x = F \cdot \cos(\theta)$

Where $F$ is the magnitude of the force and $\theta$ is the angle it makes with the horizontal axis.

For each force:

• $F_1 = 2\,N$ at $38^\circ$ above the positive x-axis:

  • $F_{1x} = 2 \cdot \cos(38^\circ)$
  • $F_{1y} = 2 \cdot \sin(38^\circ)$

• $F_2 = 6\,N$ at $42^\circ$ below the negative x-axis:

  • $F_{2x} = 6 \cdot \cos(42^\circ)$ (to the left)
  • $F_{2y} = 6 \cdot \sin(42^\circ)$ (downwards)

• $F_3 = 7\,N$ at $25^\circ$ above the positive x-axis:

  • $F_{3x} = 7 \cdot \cos(25^\circ)$
  • $F_{3y} = 7 \cdot \sin(25^\circ)$

• $F_4 = 9\,N$ at $110^\circ$ below the positive x-axis:

  • $F_{4x} = 9 \cdot \cos(110^\circ)$
  • $F_{4y} = 9 \cdot \sin(110^\circ)$

2. Magnitude of the Resultant Force:

The resultant force components $F_x$ and $F_y$ are the sum of the horizontal and vertical components:

• $F_x = F_{1x} + F_{2x} + F_{3x} + F_{4x}$
• $F_y = F_{1y} + F_{2y} + F_{3y} + F_{4y}$

Once we have $F_x$ and $F_y$, we calculate the magnitude of the resultant force $F_R$ using:

$F_R = \sqrt{F_x^2 + F_y^2}$

3. Direction of the Resultant Force:

The direction (angle $\theta_R$) of the resultant force is given by:

$\theta_R = \tan^{-1}\left(\frac{F_y}{F_x}\right)$

4. Sketch the Resultant Force:

This requires sketching the force vector according to the calculated magnitude and direction.

I'll proceed with the detailed calculations.### Solution Summary:

1. Vertical Force Components:

   • The total vertical force is obtained by summing all vertical components:
     • $F_{1y} = 1.23 \,N$
     • $F_{2y} = -4.01 \,N$
     • $F_{3y} = 2.96 \,N$
     • $F_{4y} = 8.46 \,N$
     • Total vertical force $F_y = 8.63 \,N$

2. Magnitude of the Resultant Force:

   • Total horizontal force $F_x = 9.30 \,N$
   • Total vertical force $F_y = 8.63 \,N$
   • Magnitude of the resultant force $F_R = 12.69 \,N$

3. Direction of the Resultant Force:

   • The angle of the resultant force is $\theta_R = 42.86^\circ$ above the positive x-axis.

4. Sketch:

   • Sketch the resultant force as a vector at an angle of approximately $42.86^\circ$ with a magnitude of $12.69 \,N$.

Would you like further details or explanations for any part of this solution?

5 Related Questions:

1. How do you calculate the components of a force in any direction?
2. How does the direction of a force affect its components?
3. Can you explain the role of the tangent function in determining the angle of a vector?
4. How would the calculations change if the direction of one of the forces were reversed?
5. Why is the angle $\tan^{-1}(F_y / F_x)$ always measured from the x-axis?

Tip:

When calculating forces in different directions, always ensure your angle references are consistent (either all from the positive x-axis or another agreed reference). This avoids confusion when adding components.