The length of the spring 2 due to arrangement of the equal masses is determined as 0.238 m.
Extension of the spring 2
The extension of the spring 2 is determined from the net force acting on spring 2.
F(net) = Kx
Net force on spring 2 = sum of forces acting downwards - sum of forces acting upwards.
Net force on spring 2 = (force on 2 + force on 3) - (force on 1)
F(net) = (m₂ + m₃)g - (m₁)g
F(net) = (6.67 + 6.67)9.8 - (6.67)9.8
F(net) = 130.732 - 65.366
F(net) = 65.366 N
x = F/k
x = (65.366)/(8130)
x = 0.00804 m
The length of spring 2 = x + 0.23 m
= 0.00804 m + 0.23 m = 0.238 m
Thus, the length of the spring 2 due to arrangement of the equal masses is determined as 0.238 m.
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