What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 4.5 × 10
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Answer:
b > 66.41 kg/s
Explanation:
The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f
F + f = ma
-kx - bv = ma
-kx -bdx/dt = md²x/dt².
Re-arranging the equation, we have
So, md²x/dt² + bdx/dt + kx = 0
Dividing through by m, we have
d²x/dt² + (b/m)dx/dt + (k/m)x = 0
This is a second-order differential equation. The characteristic equation is thus,
D² + (b/m)D + (k/m) = 0
Using the quadratic formula, we find D.
For an overdamped system,
Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.
Substituting k = mg/x into the inequality for b, we have
b > 2√{(mg/x₀)m}
b > 2√{(m²g/x₀)}
b > 2m√{g/x₀)}
b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}
b > 1.5 kg√{19.6/s²)}
b > 1.5 kg × 4.427/s
b > 66.41 kg/s