Mathematical solution for single leg hanging basket

[James Hayes]

TL;DR

Picture a basket, connected to a hanging point (ML), from two fixings (F). One of these ropes connecting these points snaps, and the basket shifts and drops on one side. Resulting in the rope snagging onto a pivot point (P). We know the mass, the length of a couple dimensions. I want to know a method of calculating the angle theta, shown in the last diagram. Point F and G and P are all fixed in place and do not move. Please see PDF for more information.

 

[BvU]

Hello @James Hayes , !

Is this homework or do you consider designing a balloon with basket ?

the basket shifts and drops on one side. Resulting in the rope snagging onto a pivot point (P)

And if nothing dampens the swinging ?
(I find the "Point F and G and P are all fixed in place" pretty questionable for G !)

Never mind, I suppose you silently assume the swinging terminates, at which time G should be directly underneath ML. Right ?

PFG is a triangle that keeps its shape, but rotates. How far is undeteermined since you give no sideways distances. Additionally, $\theta$ can not be calculated as long as ML to P is unknown

 

[James Hayes]

Hi

F, G and P are fixed. But surely the center of gravity does not need to be directly underneath the lifting point ML. The black line from ML to P to F is a rope, it does not 'fix' to P, it simply is wrapped around P. Surely, the tension in the rope pulling on fixing point 'F', around pivot point p will cause the center of gravity to shift out from being directly underneath, as this creates a force in X from point F.

thank you

 

[berkeman]

But surely the center of gravity does not need to be directly underneath the lifting point ML.

If it is not, there will be a net torque due to gravity that will cause movement, no?

 

[Dr.D]

The system will hang such as to minimize the gravitational potential energy. This requires the C M to be directly below the support, as @berkeman man has observed.

These are really small dimensions!

 

[berkeman]

These are really small dimensions!

Oh jeeze, I missed that. Especially small considering the mass...

 

[Dr.D]

Oh jeeze, I missed that. Especially small considering the mass...

View attachment 276314

It is just very high density material!

 

[Lnewqban]

Hi

F, G and P are fixed. But surely the center of gravity does not need to be directly underneath the lifting point ML.

Welcome, James!

On the contrary, points ML and G must be on the same vertical line.
That vertical line should intersect the line joining P and F somewhere.
That point of intersection depends on dimensions that are not provided by the problem.
For example, x and y distances between G and F or ML and F.

 

[The Fez]

It seems to me the rope from ML to P must be vertical in the steady state, otherwise there would be a sideways force on the body and it would swing. Therefore, Theta is zero, and ML, P and G are all on the same line vertically. The angle ML-P-F is determined by geometry.