To solve the problem of finding the components of the force $P = 10 \, \text{kN}$ in the $x$, $y$, $u$, and $v$ directions, we need to analyze the vector diagram shown in the image.

Step 1: Understanding the Force Components

Given:

• $P = 10 \, \text{kN}$
• The angle $P$ makes with the positive $u$-axis is $30^\circ$.
• The $u$-axis is inclined at $60^\circ$ to the $x$-axis.

We need to find:

1. $P_x$ (Component in the $x$-direction)
2. $P_y$ (Component in the $y$-direction)
3. $P_u$ (Component in the $u$-direction)
4. $P_v$ (Component in the $v$-direction)

Step 2: Resolving Force into $u$ and $v$ Components

Since $P$ is aligned along $u$ and makes an angle of $30^\circ$ with the $u$-axis:

• $P_u = P \cos(30^\circ) = 10 \cos(30^\circ)$
• $P_v = P \sin(30^\circ) = 10 \sin(30^\circ)$

Step 3: Converting $P_u$ and $P_v$ into $x$ and $y$ Components

Using the inclination of $u$ to $x$ and $y$ axes:

• $P_x = P_u \cos(60^\circ) + P_v \sin(60^\circ)$
• $P_y = P_u \sin(60^\circ) - P_v \cos(60^\circ)$

Step 4: Calculations

Let's compute each component:

1. $P_u$: $P_u = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66 \, \text{kN}$

2. $P_v$: $P_v = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5 \, \text{kN}$

3. $P_x$: $P_x = 8.66 \times \cos(60^\circ) + 5 \times \sin(60^\circ) = 8.66 \times \frac{1}{2} + 5 \times \frac{\sqrt{3}}{2}$ $P_x \approx 4.33 + 4.33\sqrt{3} \approx 8.66 \, \text{kN}$

4. $P_y$: $P_y = 8.66 \times \sin(60^\circ) - 5 \times \cos(60^\circ) = 8.66 \times \frac{\sqrt{3}}{2} - 5 \times \frac{1}{2}$ $P_y \approx 7.5 - 2.5 = 5 \, \text{kN}$

Summary of Components:

• $P_x \approx 8.66 \, \text{kN}$
• $P_y \approx 5 \, \text{kN}$
• $P_u \approx 8.66 \, \text{kN}$
• $P_v = 5 \, \text{kN}$

Would you like more details or have any questions about these calculations?

Here are 5 related questions:

1. How do you convert vector components from one coordinate system to another?
2. Why is it important to resolve forces into components in engineering?
3. What are some practical applications of vector resolution in mechanics?
4. How does the angle between axes affect the calculation of vector components?
5. Can vector components have negative values? If yes, under what conditions?

Tip: When resolving forces, always double-check the angles used, as they are critical in determining the correct components.