To determine the magnitude of the projection of force $F = 600 \, \text{N}$ along the $u$ axis, we can proceed as follows:

Step 1: Determine the Position of Point A

From the diagram, we see that point $A$ is located relative to the origin $O$ with the following distances:

• 4 meters along the $y$-axis
• 4 meters along the $z$-axis
• 2 meters along the $x$-axis

The coordinates of point $A$ are therefore $A(2, 4, 4)$.

Step 2: Determine the Direction of Force $F$

Since force $F = 600 \, \text{N}$ is applied along the line connecting $O$ and $A$, the direction of $F$ can be represented by the vector from $O$ to $A$, which is: $\vec{OA} = \langle 2, 4, 4 \rangle$

Step 3: Find the Unit Vector of $F$

To find the unit vector in the direction of $\vec{OA}$, we calculate its magnitude: $|\vec{OA}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$ Thus, the unit vector $\hat{F}$ in the direction of $\vec{OA}$ is: $\hat{F} = \frac{\vec{OA}}{|\vec{OA}|} = \frac{\langle 2, 4, 4 \rangle}{6} = \langle \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \rangle$

Step 4: Determine the $u$-axis Direction

The $u$-axis is shown to be at a $30^\circ$ angle in the $xy$-plane. Therefore, its direction vector can be represented as: $\hat{u} = \langle \cos 30^\circ, \sin 30^\circ, 0 \rangle = \langle \frac{\sqrt{3}}{2}, \frac{1}{2}, 0 \rangle$

Step 5: Project $F$ onto the $u$-axis

The projection of $F$ onto the $u$-axis is given by: $F_u = |\vec{F}| \cdot (\hat{F} \cdot \hat{u})$ where $\hat{F} \cdot \hat{u}$ is the dot product of $\hat{F}$ and $\hat{u}$.

Calculating the dot product: $\hat{F} \cdot \hat{u} = \left( \frac{1}{3} \cdot \frac{\sqrt{3}}{2} \right) + \left( \frac{2}{3} \cdot \frac{1}{2} \right) + \left( \frac{2}{3} \cdot 0 \right)$ $= \frac{\sqrt{3}}{6} + \frac{1}{3} = \frac{\sqrt{3} + 2}{6}$

Now, substitute this into the equation for $F_u$: $F_u = 600 \cdot \frac{\sqrt{3} + 2}{6} = 100 \cdot (\sqrt{3} + 2)$

Using the approximate value of $\sqrt{3} \approx 1.732$: $F_u \approx 100 \cdot (1.732 + 2) = 100 \cdot 3.732 = 373.2 \, \text{N}$

Answer

The magnitude of the projection of force $F$ along the $u$-axis is approximately: $F_u \approx 373.2 \, \text{N}$

Would you like further details on any part of this solution?

Here are some related questions you might consider:

1. How would the projection change if the angle of the $u$-axis in the $xy$-plane were different?
2. What if the force $F$ had a different magnitude?
3. How does this method apply if the axis is not in the $xy$-plane?
4. Can we find the projection of $F$ on any arbitrary axis using this method?
5. How can the dot product help in understanding projections in vector mathematics?

Tip: When working with projections, always ensure you use unit vectors for the direction of projection to simplify calculations.