Solve Spring Mechanics Problem with Hooke's Law - Vanchin

[vanchin]

hi all,

The problem is to find out how much lower will the end of the spring be when it reaches its new equilibrium position? One end of the spring is fixed to the ceiling and the other is attached to a 1 kg mass and the spring constant is 325 N/m.

I read this problem from text. In order to find the new equilibrium position, we must use Hooke's law, right? But, I find that the initial length is missing.

Given quantities:

F = m * a = 1 kg * 9.81 m/s2
K = 325 N/m

Unknown quantities:
l = ? (to be calculated)
lo (initial equilibrium) = ? (not given in the problem)

How can i go about to find l ( new equilibrium)? Any one please help.

regards,

vanchin

 

[poolwin2001]

You need to find only change in length : (initial equilibrium - new equilibrium).

 

[vanchin]

The initial equilibrium of the spring is not specified in the problem, then how to could the new equilibrium position be computed?

help.

vanchin

 

[mukundpa]

Think for the two eq. cases

F1 = K DL1
F2 = K(DL1 + DL2) D is for delta

can you solve for DL2

 

[vanchin]

Hi Mukundpa,

Think for the two eq. cases

F1 = K DL1
F2 = K(DL1 + DL2) D is for delta

can you solve for DL2

Yes, can solve DL2 if DL1 is given, but in the problem specification the L1 is not given. Sorry to trouble you. The problem is as follows:

A spring with a spring constant of 325 N/m hangs vertically from a ceiling. If a 1 kg mass ia attached to the end of the spring, how much lower will the end of the spring be when it reaches its new equilibrium position?

 

[Allday]

think about what the x in Hooke's law stands for. As I remember it is the displacement from equilibrium. If I tell you that some thing is x meters away from its equilibrium then what is the distance from the equilibrium point to x? In other words what is the equilibrium position of a spring with no mass hanging from it (no force).

 

[mukundpa]

If the problem says that in new equilibrium position means(sounds) previously the spring was in equilibrium with some force and an additional load of 1 kg is added due to which the equilibrium position is changed the new tension in the spring is F2 thus F2 - F1 = 1*9.8 N

subtracting the two equations we get F2 -F1 = KDL2

gives DL2 =(F2 -F1)/K

and if initially there was no load then the position of the lower point of the spring is L0
and the tension is zero
in second case the tension is 9.8N and the length is Lo + DL then

9.8 = K( DL)

I think that the confusion is only because you are trying to get the new length of the spring which is not required. the question says "how much lower will be the end of the spring" means extension.

 

[vanchin]

Hi all,

Based on the guidlines, I understand that the lower end of the spring be when it reaches its new equilibrium position is 0.0301m.

using F = k (x), where x = l - lo

=> 9.81 = 325 * x

=> x = 9.81 / 325

=> x = 0.0301m

Am I right?

 

[Allday]

looks good to me