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Answer :
The value of viscous damping in the suspension system that would cause the system to be critically damped is approximately 5,600 Ns/m.
The natural frequency of a system can be calculated using the following formula:
ωn = √(k/m)
In this case, each of the three landing gear suspension systems share the load evenly, so the weight supported by one suspension system is:
W = (4000 kg) / 3 = 1333.33 kg
The deflection of the suspension system, δ, is 0.2 m.
The spring constant k of the suspension system can be calculated using Hooke's Law:
k = F/δ
Since the weight is supported evenly by all three suspension systems, the force exerted by one suspension system is:
F = (1333.33 kg) x (9.81 m/s) = 13098.67 N
Therefore, the spring constant is:
k = 13098.67 N / 0.2 m = 65493.35 N/m
The mass of the system is the mass of the loaded plane plus the mass of the suspension system. Since the plane has a mass of 12,007 kg when empty and is loaded with 4000 kg, the total mass is:
m = 12,007 kg + 4000 kg = 16,007 kg
Now we can calculate the natural frequency of the system:
ωn = √(k/m)
= √(65493.35 N/m / 16007 kg)
= 1.064 rad/s
To find the value of viscous damping that would cause the system to be critically damped, we need to use the formula:
c = 2mωn
For critical damping, the damping coefficient must be equal to the critical damping coefficient, which is:
cc = 2√(km)
cc = 2√(k m)
= 2√(65493.35 N/m x 16007 kg)
≈ 5,600 Ns/m
Therefore, the value of viscous damping in the suspension system that would cause the system to be critically damped is approximately 5,600 Ns/m.
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