A particle of mass m is attached to a fixed point A by a model spring of natural

Question

A particle of mass m is attached to a fixed point A by a model spring of natural length l0 and stiffness mg/l0, and to another fixed point B vertically below A by another model spring, of natural length l0 and stiffness 3mg/l0, where g is the acceleration due to gravity. The distance between point A and B is 3l0. The particle moves along the vertical line between A and B as shown.

If m = 1.89 kg and l0 = 0.767 m, calculate the equilibrium position in metres, relative to A. Give your answer to 3 decimal places, and take g = 9.81 ms-2.

A particle of mass m is attached to a fixed point A by a model spring of natural length l0 and stiffness mg/l0, and to another fixed point B vertically below A by another model spring, of natural length l0 and stiffness 3mg/l0, where g is the acceleration due to gravity. The distance between point A and B is 3l0. The particle moves along the vertical line between A and B as shown. If m = 1.89 kg and l0 = 0.767 m, calculate the equilibrium position in metres, relative to A. Give your answer to 3 decimal places, and take g = 9.81 ms^-2.

Explanation / Answer

distance AB = 0.767 * 3 = 2.301 meters

Ka = 1.89 * 9.81 / 0.767 = 24.1732724902216428 = 24.17327

Kb = 3 * 24.17327 = 72.51981

Let the equilibrium position in metres, x from A
Ta-tension in spring A; Tb in spring B

Ta = Tb + mg

Ka*(x - 0.767) = Kb*(2.301 - x - 0.767) + 1.89 * 9.81

24.17327(x - 0.767) = 72.51981(1.534 - x) + 18.5409

24.17327x - 18.54089809 = 111.24538854 - 72.51981x + 18.5409

24.17327x + 72.51981x = 111.24538854 + 18.5409 - 18.54089809

96.69308x = 111.24539045

x = 1.1505 meters ---> ANSWER

So, the equilibrium position is at 1.1505 meters from A

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