Force in spring due to falling object

[ChristopherJ]

Hi,

I'm designing a crane that is supposed to withstand forces due to a falling object attached to the crane with a non-elastic cable. I have some troubles to calculate what force that will be due to the falling object.

I figured that one way is to simplify this system with a spring with spring constant k, extension of the spring x, non-elastic cable of length h, object with mass m, gravitational constant g.

Then, the spring is hanging vertical, attached in the upper end to a rigid mount and the cable in the free hanging end, and in the free end of the cable the object is attached. If the object is falling from the point where the cable is attached to the spring it will fall the distance h. Then the following expressions may be used.

1. Potential energy due to a height, E=mgh
2. Potential energy in the spring due to an extension, E=0.5kx^2
3. Hooke´s law due to an extension, F=kx

1 and 2 gives mgh=0.5kx^2 which gives x=sqrt(2mgh/k). This combined with 3 gives F=kx=sqrt(2mgkh).

Am I right in my reasoning and the last expression? For which cases is this then true?

I would be most grateful for response.

Regards
Christopher

 

[mfb]

The formula is good if you can neglect the additional height difference during the deceleration process. You can neglect it if F>>mg.

 

[ChristopherJ]

Okey, so you mean that h in mgh is the length of the cable plus x, i.e. h ist the distance from the point of release to where it comes to a stop? If I can't neglect that, how should I do the calculations then? Since x depends on h, which depends in x?

 

[mfb]

If I can't neglect that, how should I do the calculations then? Since x depends on h, which depends in x?

You will get a quadratic equation, which can be solved. The "new h" is just the old h (constant) plus x.

 

[ChristopherJ]

You will get a quadratic equation, which can be solved. The "new h" is just the old h (constant) plus x.

I'm not sure I'm following. If the new h is the old h plus x, then I can perform the calculations twice and the second time just add x to h? Or how would the equation look like?