Answer:
3223 pounds
Step-by-step explanation:
If the lifting force [tex] F [/tex] varies jointly as the area [tex] A [/tex] of the wing's surface and the square of the plane's velocity [tex] v [/tex], we can express this relationship with the equation:
[tex] \Large\boxed{\boxed{F = k \times A \times v^2 }}[/tex]
where
- [tex] k [/tex] is the constant of proportionality.
Given that when [tex] A = 220 [/tex] square feet and [tex] v = 150 [/tex] miles per hour, [tex] F = 3,700 [/tex] pounds, we can use these values to find [tex] k [/tex]:
[tex] 3700 = k \times 220 \times 150^2 [/tex]
[tex] 3700 = k \times 220 \times 22500 [/tex]
[tex] 3700 = 4950000k [/tex]
Now, solve for [tex] k [/tex]:
[tex] k = \dfrac{3700}{4950000} [/tex]
[tex] k = \dfrac{37}{49500} [/tex]
Now that we have the value of [tex] k [/tex], we can use it to find the lifting force [tex] F [/tex] when [tex] A = 220 [/tex] square feet and [tex] v = 140 [/tex] miles per hour:
[tex] F = \dfrac{37}{49500} \times 220 \times 140^2 [/tex]
[tex] F = \dfrac{37}{49500} \times 220 \times 19600 [/tex]
[tex] F = \dfrac{37}{49500} \times 4312000 [/tex]
[tex] F = 3223.111111 [/tex]
[tex] F = 3223 \textsf{ (in nearest integer)}[/tex]
Therefore, when the speed is 140 miles per hour, the lifting force is approximately 3223 pounds.
