Simple closed force system -- Mechanics

[Maria]

Homework Statement

There is a system of 4 points located along a circle of radius R. Points are connected by undeformable ropes (pink on the picture). There is a force applied to each point. Scheme:
https://we.tl/MsCEViCQdB
I need to find resulting force in the system.

The Attempt at a Solution

My question is: is it correct to sum up the forces to get the resulting force (I know all angles and this will be just a sum of 4 vectors)? Like this:
https://we.tl/O7KNxVicNZ
Or there is something more complicated behind?

 

[kuruman]

There is no picture. Can you post it?

 

[Maria]

Hi,
I will be happy to do it, but I don't see how. When I press on add image it only suggests me to add an url. So I have uploaded it to one of the fileexchange servers, but apparently this is not working. FAQ also is not helpful.
Just is case the link to file exchange: left - scheme, right - forces summation
https://we.tl/O7KNxVicNZ

 

[kuruman]

If you have the image stored as a file you click UPLOAD to the right of PREVIEW... (lower right), navigate to your file and then click THUMBNAIL. That's for future reference. Anyway, I was able to view your image. The resulting force is the sum of all the forces added as vectors. To do that you need to know the angles and the magnitudes of the forces. You need to add all four x-components together to get the x-component of the resultant and all four y-components together to get the y-component of the resultant. That's it.

 

[Maria]

Thank you!
I was wondering would something change if connection between points will change? If there will be multiple connections to one point?
And if I would like to calculate a force acting on one point (lets say 1), should I just sum up 3 forces F1 of the point and F4 and F2 of the neighbour point?

I am just trying to figure out how to find in which direcion I should move each point to allow them equilibrate and so that resulting force would be 0, after some iteration (points can move only alond the cicle).

 

[kuruman]

You did not state what you know and what you are looking for. One needs to know the magnitudes and directions of the forces. Do you know what they are? Do they depend on the distance between points? Is this perhaps a force table experiment where you have strings pulling on a ring in a radial direction?

 

[Maria]

Forces are always perpendicular to the tangent of the circle at the point where they are applied. They are 1 unit and I randomly seed points to equlibrate them then. Circle there is just to show where points are allowed to move but is not a physical object.
Actually, will something change if I have a circle as a ring, except there will be addititional forces acting along the the ring tangent? I thought to add it after I solve a simplier problem :)

 

[kuruman]

Circle there is just to show where points are allowed to move but is not a physical object.

Actually, it helps to have a real wire circle and imagine the points to be beads that are free to move around. The wire allows the beads to move freely on the circle in a tangential direction and prevents them from moving in a radial direction. Regardless of how you model this, the most basic problem to understand how forces add as vectors. Imagine walking in a straight line in some direction a certain distance. Then you change direction and walk another distance, and another and another, four changes in direction until you end up where you started. Your overall displacement is zero because you end up where you started. This is another way of saying that the sum of the four individual displacement vectors is zero. Note that if you were to draw the four vectors as arrows with the end of one at the tip of the other, you will get a closed quadrilateral loop representing the path you took. The same idea applies to all vectors including forces: if you have N vectors adding up to zero, they must form a closed N-sided polygon.

So in this case you need to find a quadrilateral that can have a circle passing through its four apices. The length of the ropes between adjacent beads can be viewed as proportional to the force acting on a particular bead. So the sides of the quadrilateral must all be equal, which makes it a rhombus. The additional constraint that the points are on circle makes the rhombus a square. So a square is the only shape that satisfies all the constraints. This means that each bead experiences the same radial force. This is the obvious symmetrical solution and probably of little interest to you.

If you relax the constraint that the forces be equal, you can construct a quadrilateral of unequal sides inside a circle as long as two of diagonally opposite angles are 90^o each. Then the beads will be in translational equilibrium, i.e. the center of the circle will not accelerate. However, there will be an unbalanced torque on the four-bead system which will give the system a spin about the center. Shown below is a picture of what I mean. The opposing right angles are DAB and DCB and you may imagine that the dotted lines are light rigid rods connecting the beads. The force on each bead is along the length of the rod attached to it on one side and proportional to the length of the rod. The diagram on the left shows force arrows connected to the sides of the quadrilateral that are proportional to them. The diagram on the right has swapped the forces (A →D, D→C, C→A), keeping their magnitudes and directions fixed, so that the sum will still be zero. I leave it up to you to figure out how many different such swaps are possible.

 

[Maria]

Thank you for elaborate reply!
I understood this all from the beginning. However, I have written a script which should equilibrate 4 randomly located points on the circle, but I do not get square in the end.
I did it in 2 ways:
1. I calculate net force by simply adding up 4 forces acting perpendicular to tangent of the circle. At each step I move one by one, each point in the direction opposite to the net force (all 4 forces) and along the circle outline. I recalculate net force after each point move and condition to finish for me is when net force =0. As a result I get long rectangles.
2. At each step, one by one, for each point I calculate force from other 3 points and move current point in opposite direction and along the circle outline. Here I get rectangle, but when I use more points I never get equilateral shapes.

So, maybe I should add more forces, some acting along the tangent (shouldn't they cancel each other?)? Or maybe I should somehow take into account direction of connections of the current point to other. But this will always be only 2 points..

 

[kuruman]

I guess I am having trouble visualizing what you are trying to do. Can you post what you consider a solution to the problem using methods 1 and 2? Are all the forces supposed to be perpendicular to the tangent of the circle (radial direction) at all times and add to zero? What other constraints do you have? In the simplest situation, any four radial forces will cancel in pairs so there is an infinity of solutions. In other words, $\vec{F}_A +(-\vec{F}_A)+\vec{F}_B +(-\vec{F}_B)$ will always be zero no matter how you choose the magnitudes and directions. If you want to find the rectangle they form, you need to fix $\vec{F}_A$ and $\vec{F}_B$ and then use your methods to find $\vec{F}_C$ and $\vec{F}_D$ such that the sum is zero. Is that what you are doing? What measures have you taken to prevent your script from going around in circles trying to find a solution? Something has to be fixed when there is a multitude of possible solutions.