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We are given that the lifting force \( F \) varies jointly as the area \( A \) of the wing's surface and the square of the plane's velocity \( v^2 \). This implies the relationship:

\[
F = k \cdot A \cdot v^2
\]

where \( k \) is a constant of proportionality.

We know:
- \( A = 130 \, \text{square feet} \),
- \( F = 20,200 \, \text{pounds} \) when \( v = 160 \, \text{miles per hour} \).

First, we will solve for the constant \( k \) using the given values.

\[
20,200 = k \cdot 130 \cdot (160)^2
\]

Now solve for \( k \):

\[
k = \frac{20,200}{130 \cdot (160)^2}
\]

Calculate the denominator:

\[
130 \cdot (160)^2 = 130 \cdot 25,600 = 3,328,000
\]

Now,

\[
k = \frac{20,200}{3,328,000} \approx 0.00607
\]

Next, we use this value of \( k \) to find the lifting force when \( v = 200 \, \text{miles per hour} \). The equation becomes:

\[
F = 0.00607 \cdot 130 \cdot (200)^2
\]

Simplify the expression:

\[
F = 0.00607 \cdot 130 \cdot 40,000 = 0.00607 \cdot 5,200,000 = 31,564
\]

Thus, the lifting force is approximately **31,564 pounds**.

Would you like further details or explanations? Here are 5 questions to explore more:

1. What happens to the lifting force if the velocity doubles?
2. How does the area of the wing affect the lifting force?
3. Can we derive a formula for the lifting force in terms of area and velocity?
4. How would the lifting force change if the area increased by 10%?
5. Why does the lifting force depend on the square of the velocity rather than the velocity itself?

**Tip:** When solving proportionality problems, always focus on isolating the constant first. This helps apply the relation to new conditions easily.

The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the plane’s velocity, v. The lift of a wing with an area of 130 square feet is 20,200 pounds when the plane is going at 160 miles per hour. Find the lifting force if the speed is 200 miles per hour.

Solve for the lifting force on an airplane wing where force varies jointly with area and velocity squared. Given an area of 130 square feet, a lifting force of 20,200 pounds at 160 mph, find the force at 200 mph.

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