To determine the horizontal and vertical components of the tension force when Hector pulls the chain with 67.0 N of force at an angle of 30.0° above the horizontal, we can use basic trigonometry.

Let:

• $F = 67.0 \, \text{N}$ be the total force.
• $\theta = 30.0^\circ$ be the angle above the horizontal.

Horizontal Component:

The horizontal component $F_x$ of the force is given by: $F_x = F \cdot \cos(\theta)$ Substitute the values: $F_x = 67.0 \cdot \cos(30.0^\circ)$ $F_x = 67.0 \cdot 0.866 = 58.022 \, \text{N}$

Vertical Component:

The vertical component $F_y$ of the force is given by: $F_y = F \cdot \sin(\theta)$ Substitute the values: $F_y = 67.0 \cdot \sin(30.0^\circ)$ $F_y = 67.0 \cdot 0.5 = 33.5 \, \text{N}$

Final Answer:

• Horizontal component $F_x = 58.0 \, \text{N}$ (rounded to 1 decimal place).
• Vertical component $F_y = 33.5 \, \text{N}$.

Would you like more details, or have any further questions?

Here are 5 questions to expand your understanding:

1. What is the significance of the angle in determining force components?
2. How would the components change if the angle were increased to 45°?
3. Why is cosine used for the horizontal component and sine for the vertical?
4. How does the direction of the force affect the object being pulled?
5. What would happen if Hector pulled with the same force but in a downward direction?

Tip: Remember, cosine is used for horizontal components and sine for vertical components when resolving forces at an angle.