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To solve this problem, we need to resolve the 600 N force along the given axes \( u \) and \( v \) using the **sine law**.

### Given:
1. **Force (F)**: 600 N
2. **Angle between \( F \) and the \( u \) axis**: 30°
3. **Angle between \( F \) and the \( v \) axis**: 30°
4. **Angle between \( u \) and \( v \) axes**: 120° (since \(180^\circ - 30^\circ - 30^\circ = 120^\circ\))

### Steps:
We apply the **sine law** for forces:
\[
\frac{F}{\sin \theta_{uv}} = \frac{F_u}{\sin \theta_{Fv}} = \frac{F_v}{\sin \theta_{Fu}}
\]
where:
- \( F \) is the resultant force (600 N),
- \( F_u \) is the component along the \( u \) axis,
- \( F_v \) is the component along the \( v \) axis,
- \( \theta_{uv} = 120^\circ \) is the angle between the \( u \) and \( v \) axes,
- \( \theta_{Fv} = 30^\circ \) is the angle between \( F \) and the \( v \) axis,
- \( \theta_{Fu} = 30^\circ \) is the angle between \( F \) and the \( u \) axis.

### Calculations:

1. **Calculate \( F_u \):**
   \[
   F_u = F \cdot \frac{\sin \theta_{Fv}}{\sin \theta_{uv}}
   \]
   \[
   F_u = 600 \cdot \frac{\sin 30^\circ}{\sin 120^\circ}
   \]

   Using \( \sin 30^\circ = 0.5 \) and \( \sin 120^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \),
   \[
   F_u = 600 \cdot \frac{0.5}{0.866} \approx 600 \cdot 0.577 = 346.41 \, \text{N}
   \]

2. **Calculate \( F_v \):**
   \[
   F_v = F \cdot \frac{\sin \theta_{Fu}}{\sin \theta_{uv}}
   \]
   \[
   F_v = 600 \cdot \frac{\sin 30^\circ}{\sin 120^\circ}
   \]

   Since this setup is symmetric, we get the same calculation:
   \[
   F_v = 600 \cdot 0.577 = 346.41 \, \text{N}
   \]

### Answer:
- \( F_u = 346.41 \, \text{N} \)
- \( F_v = 346.41 \, \text{N} \)

Would you like more details or have any questions?

### Related Questions:
1. What is the cosine law and how does it relate to resolving forces?
2. How would you approach the problem if the angles were different?
3. How does the sine law work in non-right triangles?
4. What would change if the force direction or magnitude were altered?
5. How would you decompose forces if the axes were perpendicular?

### Tip:
For non-right triangles, the sine law is an efficient method to resolve forces, especially when you know the angles between each component.

For the 600 N force shown in Figure (1), use the sine law to determine the magnitudes of its components acting along the u and v axes.

Learn how to resolve a 600 N force using the sine law along specified u and v axes with a step-by-step breakdown. This problem involves applying trigonometry, including the sine law, to calculate force components along two axes. Suitable for high school physics students.

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