Spring Extension and Force Distribution in Ideal Mass-Spring Systems

AI Thread Summary
In an ideal mass-spring system with two masses connected by a spring, the extension of the spring depends on the forces acting on it. When both masses are equal, the spring extends by a length 'x' such that the force exerted by the spring on each mass is kx, where k is the spring constant. If the masses are unequal, the system may not reach equilibrium, as the forces would differ, leading to potential collapse. The tension in the spring remains constant at both ends, regardless of the external forces applied, meaning it cannot sustain different forces on either side. Ultimately, the equilibrium condition requires equal forces acting on both ends of the spring for it to function correctly.
sachin123
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Please consider the following case:

Spring connected at both ends by some masses.Everything is ideal.
Find the extension of the spring.
[URL]http://img101.imageshack.us/i/unledpi.jpg/[/URL]
http://img101.imageshack.us/i/unledpi.jpg/

My big question is ...if the spring extends by a length say 'x',then will it pull both sides by kx where k is its spring constant or,will it pull each side by a different force?
(like kx/2)
Can you also reason why?

Lets take 1st case.For equilibrium kx=m1g=m2g.This happens only if both the masses are equal.
Otherwise is the system bound to collapse?
 
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sachin123 said:
My big question is ...if the spring extends by a length say 'x',then will it pull both sides by kx where k is its spring constant or,will it pull each side by a different force?
The stretch of the spring just depends on the tension it is placed under. Just like a massless string under tension, each end is pulled with the same force.
 


Okay,then consider the 1 st situation I put up with both the masses same say 'm'.
Now,let extension of spring be 'x'.
As you said,the spring pulls both the blocks with the same force kx and this force keeps each block in equilibrium.That makes kx=mg so x=mg/k.
But,considering tensions,the tension in both the strings have to be mg,so as to pull the spring in opposite directions by forces mg each.
So spring is under a net force of 2mg.
So the extension must be
2mg/k.
Now where did I go wrong?
 


sachin123 said:
But,considering tensions,the tension in both the strings have to be mg,so as to pull the spring in opposite directions by forces mg each.
So spring is under a net force of 2mg.
So the extension must be
2mg/k.
Now where did I go wrong?
Where you are going wrong is in thinking that if a string (say) is pulled by a force F at each end, that the tension is 2F not F. Not so. The tension in a string or a spring is the force that it exerts at each end.

Imagine this. Attach a spring to the ceiling. Hang a mass m from it. What's the tension in the spring? With what force does the ceiling pull up on the spring?
 


Okay,I got the case when same force acts at both ends.
How about this?I take a spring in my hand vertically,and pull the upper part by 5N amd the lower part with the other hand by 10 N.
As you said:
"The tension in a string or a spring is the force that it exerts at each end."
now no matter by what length the spring expands,it will pull both the ends with the same force.But my hand pulls them with different forces.Now how can it ever stay in equilibrium?
(I find the same problem when I do series spring combination problems)
 


sachin123 said:
How about this?I take a spring in my hand vertically,and pull the upper part by 5N amd the lower part with the other hand by 10 N.
Sorry, but it's physically impossible to pull a massless spring with different forces on each end.
 


Okay then.Thank you
 
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