Item 5 Three identical 500 kg masses are hung by three ident

Item 5 Three identical 5.00 kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 5.10 kN/m and was 19.0 cm long before any masses were attached to it. How long is each spring when hanging as shown? (Hirt First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.) Figure 1) Figure of1

Solution

Weight of Top mass = m g = 5.00 * 9.81 = 49.05 N

Weight of Top + Middle mass = 2 * 5.00 * 9.81 = 98.1 N

Weight of all masses = 3 * 5.00 * 9.81 = 147.15 N

k = 5100 N/m

x = F/k

Top Spring stretched by Force = 147.15 N

x = 147.15 / 5100 = 2.89 cm

Top Spring length = 19.0 + 2.89 = 21.89 cm

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Middle Spring stretched by Force = 98.1 N

x = 98.1 / 5100 = 1.92 cm

Middle Spring length = 19.0 + 1.92 = 20.92 cm

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Bottom Spring stretched by Force = 49.05 N

x = 49.05 / 5100 = 0.96 cm

Bottom Spring length = 19 + 0.96 = 19.96 cm