Calculating Force on a Vertical Wire Rope with 20kg Weight

Click For Summary
SUMMARY

The discussion centers on calculating the maximum force exerted on a vertical wire rope when a 20kg weight is dropped from a height of 1 meter. The user initially employed formulas related to potential energy and elastic potential energy but struggled to find the correct approach. The correct method involves using energy conservation principles, equating gravitational potential energy to elastic potential energy to determine the stretch in the rope and subsequently the force exerted. The elastic stretch limit of the rope is given as 47.5 kN/mm², which is crucial for ensuring the rope's integrity under load.

PREREQUISITES
  • Understanding of gravitational potential energy calculations
  • Familiarity with elastic potential energy equations
  • Knowledge of Newton's laws of motion
  • Basic principles of energy conservation
NEXT STEPS
  • Study energy conservation principles in mechanical systems
  • Learn about elastic potential energy calculations in materials
  • Explore the implications of material limits, such as the elastic stretch limit
  • Investigate the dynamics of tension in ropes and cables under load
USEFUL FOR

Engineers, physics students, and professionals involved in mechanical design or materials science who need to understand the forces acting on cables and ropes under load conditions.

DJT1967
Messages
3
Reaction score
0
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?
 

Attachments

  • WEIGHT.JPG
    WEIGHT.JPG
    21.7 KB · Views: 423
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
46
Views
7K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K