An aircraft flies with its wings tilted as shown to fly in a horizontal circle of radius R. The aircraft has mass 4000 kg and moves with a constant speed of 250 m/s. With the aircr... An aircraft flies with its wings tilted as shown to fly in a horizontal circle of radius R. The aircraft has mass 4000 kg and moves with a constant speed of 250 m/s. With the aircraft flying in this way, the two forces acting on the aircraft in the vertical plane are the force L acting at an angle 35° to the vertical and the weight W. (a) State the direction of the net force on the aircraft. (b) What is the vertical component of force L? (c) Calculate the force L. (d) Determine the aircraft's centripetal acceleration. (e) Calculate the radius R of the circular motion. Read more

Steps to Solve

1. Direction of the Net Force

Since the aircraft is moving in a horizontal circle, the net force must be directed towards the center of the circle. This net force is the centripetal force, which causes the circular motion. Therefore, the direction of the net force is horizontal and towards the center of the circle.

2. Vertical Component of Force L

The vertical component of the lift force $L$ must balance the weight $W$ of the aircraft since there is no vertical acceleration. The weight $W$ is given by $W = mg$, where $m$ is the mass and $g$ is the acceleration due to gravity ($g \approx 9.8 , \text{m/s}^2$). The vertical component of $L$ is $L\cos(35^\circ)$. Thus, $L\cos(35^\circ) = W$.

3. Calculate Weight W

Let's calculate the weight of the aircraft $W = mg = 4000 , \text{kg} \times 9.8 , \text{m/s}^2 = 39200 , \text{N}$

4. Vertical component of Lift Force L

Since $L\cos(35^\circ) = W$, the vertical component of the Lift force is equal to $W$ that we have found in the previous step: Vertical component of $L$ = $39200 , \text{N}$.

5. Calculate Force L

Using the equation $L\cos(35^\circ) = W$, we can solve for $L$: $L = \frac{W}{\cos(35^\circ)} = \frac{39200 , \text{N}}{\cos(35^\circ)}$

Calculating $L$: $L = \frac{39200}{0.819} \approx 47863.25 , \text{N}$

6. Aircraft's Centripetal Acceleration

The horizontal component of the lift force $L$ provides the centripetal force $F_c$. The horizontal component is $L\sin(35^\circ)$. $F_c = L\sin(35^\circ) = ma_c$, where $a_c$ is the centripetal acceleration.

We calculate $L\sin(35^\circ)$: $L\sin(35^\circ) = 47863.25 , \text{N} \times \sin(35^\circ) = 47863.25 , \text{N} \times 0.574 \approx 27466.9 , \text{N}$

Now we can calculate the centripetal acceleration $a_c = \frac{F_c}{m}$: $a_c = \frac{27466.9 , \text{N}}{4000 , \text{kg}} \approx 6.87 , \text{m/s}^2$

7. Calculate the Radius R

We know that centripetal acceleration is also given by $a_c = \frac{v^2}{R}$, where $v$ is the speed and $R$ is the radius. We can solve for $R$: $R = \frac{v^2}{a_c} = \frac{(250 , \text{m/s})^2}{6.87 , \text{m/s}^2}$

Calculate $R$: $R = \frac{62500}{6.87} \approx 9097.53 , \text{m}$