# Tension in streched spring

• physicshawk
The unstretched length of the portion of the spring above the putty is 2.5 meters, and the unstretched length of the portion of the spring below the putty is 1.5 meters. The spring constants for these springs are k = 80 Newtons/meter and k = 0 Newtons/meter. The force exerted by the upper spring on the putty is 80 Newtons in a downward direction, and the force exerted by the lower spring on the putty is 0 Newtons in a downward direction. The total spring force on the putty is 80 Newtons in a downward direction.f

#### physicshawk

A point A of the ceiling is directly above a point B of the floor 2.5 meters below. A and B are joined by a light spring of natural length 2 meters; the tension in the spring is then 20 Newtons. A Lump of Putty of weight 4 Newtons is now attached to the spring at the point 1 meter above the floor. At what height above the floor will the Putty rest in equilibrium?

My working and diagram;
http://www.flickr.com/photos/90383971@N05/10050116394/in/photostream/
My Ans:
First i found natural length of the spring.
Without the Putty:
through Hooke's Law, we know
T= λx ⁄ l
where T = tension, l= natural length, λ = modulus of elasticity, x = extension
λ = Tl ⁄ x
λ = 20 * 2 / 0.5
λ = 80 Newtons

So this is the modulus of elasticity of the spring. From here the problem becomes complicated for me. Because after the Putty is attached, the spring is streched from the top and i do not know how to bring into my calculation how the spring below the Putty affects the extension. I'm lost, Could you PLEASE help.
The answer in my book is given 0.976 meters (which is the height of the putty above the floor)

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Note that in this case, the upper part of the string extend slightly, whereas the lower part of the spring will be compressed somewhat (relative to the upper part), so that you do NOT have an equally distributed elongation of the spring. Note that the upper part of the string pulles upwards, the lower part downwards. Calling the length by which the stretched lengths of the springs changes, you get the equation
k(2.5+l*-2)-k(2.5-l*-2)-mg=0 for the equilibrium position of the putty.

Note that the equilibrium tension in the spring for the case of no putty vanishes identically as a parameter (as it should)

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1 person
Why the factor of 2 in your calculation of λ ?

Why the factor of 2 in your calculation of λ ?
The 2 is the initial length 2 meters.

If the putty is attached to the stretched spring 1 meter above the floor, what fraction of the original spring length (2m) is below the putty (before the putty is attached)? What fraction of the original spring length is above the putty. What is the unstretched length of the portion of the spring above the putty? What is the unstretched length of the portion of the spring below the putty? The portion of the spring above the putty and the portion below the putty can be looked upon as two separate springs, with these unstretched lengths. What are the spring constants k = λ/l for each of these two springs? If the putty moves down by Δx, what is the force that the upper spring exerts on the putty, and in what direction. What is the force that the lower spring exerts on the putty, and in what direction. What is the total spring force on the putty, and in what direction?

1 person