# Homework Help: Force equivalency in a physical simulation

1. Nov 18, 2017

### remettub

1. The problem statement, all variables and given/known data

Question 1: given a rigid tetrahedron, with vertices A, B, C and D anchored to their initial locations with an arbitrary but uniform spring constant, will Cases 1, 2, and 3 below result in the same behavior of the system?

- Case 1: a force of constant magnitude (red vector) applied to point A, and whose orientation relative to the tetrahedron is constant (if the tetrahedron moves, the force vector moves with it, as if there is a theoretical rocket engine fixed to the tetrahedron at point A).

- Case 2: the individual components of the force vector from Case 1, in the direction of points B, C, and D respectively (brown vectors), applied to point A in the same manner as before.

- Case 3: the same components from Case 2, but now applied respectively to points B, C, and D (green vectors) in the same manner as before.

If "yes" to question 1:

Question 2: if the rigid object is also anchored with same spring constant at additional points (the object is no longer necessarily a tetrahedron), will all cases still result in the same behavior? The additional points have no rocket-like forces applied to them.

Question 3: if the object is not perfectly rigid, will all cases still result in a similar behavior?

2. Relevant equations

A set of force vectors applied to a given point will have the same effect as a single force vector which is the sum of the vectors in the set applied to the same point.

A force has the same effect on a system regardless of of where it is positioned along its line of action.

These are principles that apply to static systems, however my assumption is that since the forces always preserve their orientation relative to the tetrahedron, they will be equivalent at any given instant in time and therefore the static principles will apply.

3. The attempt at a solution

I'm attempting to use this scenario to verify the accuracy of a physical simulation. My assumption is that they should all result in the same behavior, and (since the simulation has a small damping factor) the tetrahedron should stabilize to the same location in all cases. However when I run Case 3, the tetrahedron ends up in a very different location, so I'm posting this question here to check my assumptions.

Please let me know if you believe this post is more appropriate on a different forum. It is not homework, but I believe it may fall under the category of "homework-like."

2. Nov 18, 2017

### CWatters

If I have understood correctly. Case 1 and 2 are equivalent because a vector is equal to the vector sum of its components.

I would not expect case 3 to behave the same for the simple reason that if you move the point of application of a force you typically end up applying a different net torque. For example in case 1 the net torque about point A is zero. In case 3 it might not be zero as you have forces acting at B, C and D.

I'm not a mathematician but someone else might be able to give you a proof.

3. Nov 18, 2017

### CWatters

Actually now I look at it again I think I'm wrong because the torque about A is also zero in case 3.

4. Nov 19, 2017

### PhanthomJay

Good catch. In general, the behavior of the system will not be the same when the forces are applied at different joints. However, when the components of the force are applied at different joints but colinear with the original components, reactions and movement and rotations will be the same.

5. Nov 19, 2017

### scottdave

I guess it depends on the direction of the reaction forces of the springs. Since no numbers are given, then I think an extreme case in which the "rocket" is pointing in some tangent direction to all the edges would cause the maximum amount of torque. It is not clear how the springs are suppose to react. I guess the simulation is setup so that springs can only act in one direction?
The problem statement did not specify if it is a regular tetrahedron, so in an extreme case, I would imagine a situation where it is formed like a spike. Applying the rocket at different locations may be easier to visualize different behaviors, then.

6. Nov 20, 2017

### remettub

The springs provide a reaction force proportional to the distance from the current position and the initial position, and in the direction of the initial position. So they can act in any direction.

I guess the discrepancy I'm seeing must be from an error in my code. Thanks for all your replies!