Force in spring due to falling object

In summary, the conversation is about designing a crane that can withstand the forces of a falling object attached to it with a non-elastic cable. The individual is seeking help with calculating the force of the falling object and presents a simplified system involving a spring with a spring constant, an extension of the spring, a non-elastic cable, an object with mass, and gravity. The formula for potential energy due to height is discussed, as well as potential energy in the spring and Hooke's law. The individual asks if their reasoning and the final expression are correct and for what cases it is true. Another person suggests that the formula is only accurate if the additional height difference during the deceleration process can be neglected, and if not, a
  • #1
ChristopherJ
5
0
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
 
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  • #2
The formula is good if you can neglect the additional height difference during the deceleration process. You can neglect it if F>>mg.
 
  • #3
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?
 
Last edited:
  • #4
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.
 
  • #5
mfb said:
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?
 

1. What is the force in spring due to a falling object?

The force in spring due to a falling object is the amount of force exerted on the spring when an object falls onto it. This force is caused by the weight of the object and is directly proportional to the distance the object falls and the stiffness of the spring.

2. How is the force in spring calculated?

The force in spring can be calculated using the formula F = kx, where F is the force in newtons, k is the spring constant in newtons per meter, and x is the distance the object falls in meters. This formula is based on Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed.

3. Does the mass of the falling object affect the force in spring?

Yes, the mass of the falling object does affect the force in spring. The more massive the object is, the greater the force it exerts on the spring when it falls onto it. This is because the weight of an object is directly proportional to its mass, and weight is what causes the force in spring.

4. How does the stiffness of the spring affect the force in spring?

The stiffness of the spring is directly proportional to the force in spring. This means that the stiffer the spring, the greater the force it will exert when an object falls onto it. This is because a stiffer spring requires more force to stretch or compress it, resulting in a greater force in spring.

5. Is there a limit to the force in spring due to a falling object?

Yes, there is a limit to the force in spring. This limit is known as the spring's elastic limit. If the force in spring exceeds the elastic limit, the spring will no longer return to its original shape and will be permanently deformed. It is important to consider the elastic limit when calculating the force in spring to avoid damaging the spring.

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