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Answer :
The bank angle of the airplane is approximately 11.4°.
To find the bank angle of the airplane, we can use the concept of centripetal force. When an airplane is making a turn, the lift force from its wings provides the centripetal force required to keep it in a circular path.
The formula for centripetal force is:
[tex]\[ F_c = \frac{{m \cdot v^2}}{{r}} \][/tex]
Where:
- [tex]\( F_c \)[/tex] is the centripetal force,
- m is the mass of the airplane,
- v is the velocity of the airplane, and
- r is the radius of the turn.
The lift force [tex](\( F_L \))[/tex] acting on the airplane can be resolved into two components: [tex]\( F_L \cos \theta \)[/tex] acting perpendicular to the horizontal component of weight, and [tex]\( F_L \sin \theta \)[/tex] providing the centripetal force.
[tex]\[ F_c = F_L \sin \theta \][/tex]
Using the lift force equation:
[tex]\[ F_L = mg \][/tex]
[tex]\[ F_c = mg \sin \theta \][/tex]
Equating this to the centripetal force formula:
[tex]\[ mg \sin \theta = \frac{{m \cdot v^2}}{{r}} \][/tex]
Solving for [tex]\( \theta \)[/tex], the bank angle:
[tex]\[ \theta = \arcsin \left( \frac{{v^2}}{{gr}} \right) \][/tex]
Given [tex]\( v = 60.0 \, \text{m/s} \)[/tex] and [tex]\( r = 185 \, \text{m} \),[/tex] and [tex]\( g = 9.81 \, \text{m/s}^2 \),[/tex] we can calculate [tex]\( \theta \)[/tex]
[tex]\[ \theta = \arcsin \left( \frac{{(60.0)^2}}{{(9.81) \cdot (185)}} \right) \][/tex]
[tex]\[ \theta \approx \arcsin(0.197) \][/tex]
[tex]\[ \theta \approx 11.4^\circ \][/tex]
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