Calculating Force on a Vertical Wire Rope with 20kg Weight

[DJT1967]

OK here goes...

I have a 20kg weight attached to a 10 meter long wire rope.
The rope is fixed vertically to a cross beam.
The rope has an elastic stretch limit of 47.5 kN/mm^2.
If I lift the 20kg weight up 1 meter and then let it go what would the maximum force exerted by weight be on the wire rope?

It's been a very very long time since I studied Newton so forgive me if I'm barking loudly up the wrong tree!

I started of with Potential Energy = Mass X Gravity X Height but this didn't seem to cut the mustard so I then went for:

Elastic Potential Energy = 1/2 x (k Spring Constant) x (L Spring Length)^2

But I'm not sure if this is the correct equation to use...

Impulse = (Mass x Velocity Initial) - (Mass x Velocity Final)

Which again doesn't seem right but maybe I'm getting confused...

Tension = Mass x Gravity x SQRT(2xHeight/Length Increase)

So I have a selection of formulas that I think I should be using but I don't think I have the correct data to complete the formulas.

So this is what I came up with but I don't really know if I'm going along the right lines or not.

Velocity = sqrt 2 x9.81x1 = 4.429446918 m/s
Potential Energy = 20x9.81x1 = 196.2 Joules
Momentum = 20 x 4.429446918 = 88.58893836kg m /s
Force = Change in Momentum / Time = 88.58893836 / 0.1 Second = 885.8893836

So just how far wide of the mark am I?

 

[Dale]

Use energy conservation, no need for momentum considerations here. The gravitational potential energy (h = 1m + x) at the top goes to elastic potential energy at the bottom (the rope stretches by x). Solve for x then put x into the elastic stretch formula to get the force.

 

[ogb p]

I appreciate your efforts to solve this problem and your use of various equations to approach it. However, the correct equation to use in this case would be the formula for elastic potential energy, as the wire rope has an elastic stretch limit. This formula is as follows: Elastic Potential Energy = 1/2 x (k Spring Constant) x (L Spring Length)^2. In this case, the spring constant is the elastic stretch limit of 47.5 kN/mm^2 and the spring length is the change in length of the wire rope, which is 1 meter. So the potential energy would be 1/2 x 47.5 x (1)^2 = 23.75 kJ.

To find the maximum force exerted on the wire rope, we can use the formula for tension: Tension = Mass x Gravity x SQRT(2xHeight/Length Increase). In this case, the mass is 20kg, gravity is 9.81 m/s^2, height is 1 meter, and length increase is the same as the spring length, which is 1 meter. So the maximum tension would be 20 x 9.81 x SQRT(2x1/1) = 196.2 N.

It is important to note that this tension is the maximum force exerted on the wire rope, but it may not be the actual force experienced by the rope, as it depends on factors such as the strength and durability of the rope and the angle at which the weight is released. Further experimentation or calculations may be needed to determine the actual force experienced by the rope.