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The diagram shows a force of 250 lbs applied horizontally, and the wire needs to resist this force at an angle of 30°. To find the force in the wire, \( F \), we can resolve the forces using trigonometry.

The wire is subject to a tension force \( F \) at an angle \( \theta = 30^\circ \), and it needs to counterbalance the horizontal force of 250 lbs. The horizontal component of the tension force is given by:

\[
F_{\text{horizontal}} = F \cdot \cos(\theta)
\]

Since this component must equal the 250 lbs force:

\[
F \cdot \cos(30^\circ) = 250 \text{ lbs}
\]

Now, solving for \( F \):

\[
F = \frac{250}{\cos(30^\circ)}
\]

We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \), so:

\[
F = \frac{250}{0.866} \approx 288.68 \text{ lbs}
\]

Therefore, the wire must resist a force of approximately **288.68 lbs**.

Would you like further details on this solution or any clarifications?

Here are 5 related questions to expand this concept:

1. How would the required force change if the angle increased to 45°?
2. What happens to the wire tension if the horizontal force increases to 300 lbs?
3. How does changing the angle to 90° affect the tension in the wire?
4. What is the relationship between the angle and the force required for different horizontal forces?
5. Can this type of problem be applied to real-life examples like bridge suspensions?

**Tip:** For any system involving forces at an angle, always break the force into components (horizontal and vertical) to analyze the situation efficiently.

What force does the wire need to resist a 250 lb horizontal force at a 30° angle?

Learn how to calculate the tension in a wire needed to resist a 250 lb horizontal force acting at a 30° angle. This step-by-step guide explains force resolution using trigonometry, with relevant formulas and key concepts suitable for high school physics students.

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