What is the equilibrium position of a ball attached to two springs?

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The equilibrium position of a ball attached to two identical springs, each with a spring constant k and unstretched length L/2, is determined by balancing the forces exerted by the springs against the weight of the ball. The correct equilibrium position is given by the formula \(\frac{1}{2}L(1+\frac {Mg}{kL})\). To derive this, one must analyze the forces acting on the ball when it is displaced from the midpoint (L/2) and apply Newton's second law to establish the conditions for equilibrium.

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A small ball of mass M is attached t two identical springs of constant k, which are attached to the floor and the roof. The springs have unstretched lengh L/2. At what position will the ball remain at rest?

The back of the book gives the answer.
[tex]\frac{1}{2}L(1+\frac {Mg}{kL})[/tex]

But the text doesn't do a good job explaining how to arrive at that answer. I've stared at it for 20 minutes. Any thoughts?
 
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When the ball is exactly in the middle (at L/2) what forces do the springs exert on it? And at a point a distance X beneath that middle point? What does X have to be so that the net spring force balances the ball's weight?
 
Okay draw the forces...Apply Newton's second law.And then use the fact that the body is equilibrium.

The answer in the book gives the body-ceiling distance...


Daniel.
 

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