To solve this problem, we need to break down the force vector $\mathbf{F}$ into its components in terms of the unit vectors $\mathbf{i}$ and $\mathbf{j}$.

Given:

• Magnitude of the force, $|\mathbf{F}| = 20 \, \text{N}$
• The force points to the right and downward, making an angle of $30^\circ$ with the vertical axis.

Since the angle is with the vertical axis and we are asked to express the force in terms of horizontal $\mathbf{i}$ and vertical $\mathbf{j}$ components:

1. The vertical component (along $\mathbf{j}$) is given by $\mathbf{F}_j = |\mathbf{F}| \cos(30^\circ)$, but since the force is pointing downward, it will be negative.
2. The horizontal component (along $\mathbf{i}$) is given by $\mathbf{F}_i = |\mathbf{F}| \sin(30^\circ)$, and it will be positive as the force points to the right.

Steps:

• The horizontal component (rightward direction):
  $\mathbf{F}_i = 20 \sin(30^\circ) = 20 \times \frac{1}{2} = 10$

• The vertical component (downward direction):
  $\mathbf{F}_j = 20 \cos(30^\circ) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$

  Since the vertical component is downward, we make it negative:
  $\mathbf{F}_j = -10\sqrt{3}$

Final answer in terms of $\mathbf{i}$ and $\mathbf{j}$:

$\mathbf{F} = 10 \mathbf{i} - 10\sqrt{3} \mathbf{j}$

You can enter this as:
F = 10i - 10*sqrt(3)j

Would you like any further clarifications on this?

Related questions:

1. How would the components change if the angle was measured with the horizontal axis instead of the vertical?
2. How do we derive the values of sine and cosine for standard angles like $30^\circ$?
3. What are the properties of unit vectors $\mathbf{i}$ and $\mathbf{j}$?
4. How would the answer change if the force pointed upward instead of downward?
5. Can you generalize how to find the components of any vector given an arbitrary angle?

Tip:

When breaking down vectors, always check the direction of the components to correctly assign positive or negative signs based on the coordinate system.