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Chapter 4: Problem 24

The vertical component of the acceleration of a sailplane is zero when the air pushes up against its wings with a force of \(3.0 \mathrm{kN}\). (a) Assuming that the only forces on the sailplane are that due to gravity and that due to the air pushing against its wings, what is the gravitational force on the Earth due to the sailplane? (b) If the wing stalls and the upward force decreases to \(2.0 \mathrm{kN}\) what is the acceleration of the sailplane?

Short Answer

Expert verified
Answer: The gravitational force acting on the sailplane is 3000 N. When the upward force decreases to 2000 N (2.0 kN), the acceleration of the sailplane is approximately 3.27 m/s虏 downward.

Step by step solution

01

Identify the given information and variables

For the first part of the exercise, we are given that the upward force due to air pushing against the sailplane's wings is \(3.0\mathrm{kN}\). We will call this force \(F_{air}\). We need to find the gravitational force, which we will call \(F_g\). For the second part, when the wing stalls, the upward force decreases to \(2.0\mathrm{kN}\). We will call this force \(F'_{air}\). We need to find the acceleration of the sailplane, which we will call \(a'\).
02

Calculate the gravitational force on the sailplane

When the vertical component of the acceleration of the sailplane is zero, the upward force due to air pushing against the wings equals the gravitational force. We can write this as: \(F_{air} = F_{g}\) Now, we can substitute the given value of the air force, \(3.0\mathrm{kN} = F_{g}\) So, the gravitational force on the sailplane is \(F_{g} = 3.0\mathrm{kN}\).
03

Convert the gravitational force to Newtons

Since we are dealing with force in kN and want to find the acceleration in m/s虏, we should convert the force to Newtons. We can do this by multiplying by 1000. \(F_{g} = 3.0\mathrm{kN} \times 1000 = 3000\mathrm{N}\) So, the gravitational force is \(3000\mathrm{N}\).
04

Calculate the mass of the sailplane

We can find the mass of the sailplane using the gravitational force and the acceleration due to gravity, \(g = 9.81\mathrm{m/s^2}\). We can use the formula: \(F_{g} = m \times g\) Substituting the values, we get, \(3000\mathrm{N} = m \times 9.81\mathrm{m/s^2}\) Now, solve for the mass, \(m\): \(m = \frac{3000\mathrm{N}}{9.81\mathrm{m/s^2}}= 306.01\mathrm{kg}\) So, the mass of the sailplane is approximately \(306.01\mathrm{kg}\).
05

Calculate the net force on the sailplane when the wing stalls

When the wing stalls, the upward force decreases to \(2.0\mathrm{kN}\) or \(2000\mathrm{N}\). The net force on the sailplane can then be calculated as the difference between the gravitational force and the new air force: \(F_{net} = F_{g} - F'_{air}\) \(F_{net} = 3000\mathrm{N} - 2000\mathrm{N} = 1000\mathrm{N}\) So, the net force on the sailplane when the wing stalls is \(1000\mathrm{N}\).
06

Calculate the acceleration of the sailplane when the wing stalls

Now, we can find the acceleration of the sailplane, \(a'\), using Newton's second law of motion: \(F_{net} = m \times a'\) Substituting the values, we get, \(1000\mathrm{N} = 306.01\mathrm{kg} \times a'\) Now, solve for the acceleration, \(a'\): \(a' = \frac{1000\mathrm{N}}{306.01\mathrm{kg}} = 3.27\mathrm{m/s^2}\) So, the acceleration of the sailplane when the wing stalls is approximately \(3.27\mathrm{m/s^2}\) downward.

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