Tetrahedron with 3 points fixed, and force applied to 4th

  • #1
SilasHokanson
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Homework Statement:
Given a perfectly inelastic tetrahedron in 3space with coordinates:
a = < a1, a2, a3 >
b = < b1, b2, b3 >
c = < c1, c2, c3 >
d = < 0, 0, 0 >

Given a force vector F is applied to point D
Assume points A, B and C are fixed
Assume tetrahedron is completely still (is in static & rotational equilibrium)
Calculate the normal forces on points A, B and C in order to maintain said equilibriums
Relevant Equations:
X is Cross Product
Torque_Net = 0
Force_Net = 0
Torque = Force X Distance
My approach to this problem is to recognize that the tetrahedron being still means that net torque is zero and net force is zero.
Fd is given

Fa + Fb + Fc = -Fd
Fa X a + Fb X b + Fc X c = <0,0,0>
This can be split up into a series of 6 equations, 2 for each component.
However, this is where I get stuck. 2 equations for each component is not enough to solve each system, since solving a system of 3 variables (x,y,z) necessitates 3 equations?

This is not technically a homework problem, it is a problem I posed myself, however I am convinced it must be solvable, if I imagine the situation in my head. A tetrahedron with 3 of its faces fixed, and a load applied to the fourth cannot move, since fixing 3 points is enough to determine an objects position in 3 space. Therefore there must be a definite normal force being applied on each of the 3 vertices, in terms of all variables given in this problem?

I have also attempted solving this problem generally, as in using actual numbers for some positions instead of variables, but it doesn't make the problem any less difficult.

I may be in over my head here, or I may just be missing some critical information. I'm fairly certain a third "value" that I could set to zero would be enough to solve this problem? Or perhaps I'm just setting it up incorrectly.
 

Answers and Replies

  • #2
haruspex
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The situation is statically indeterminate.
Even without the applied force, there are nonzero combinations of forces at A, B, C which produce equilibrium.
 
  • #3
SilasHokanson
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The situation is statically indeterminate.
Even without the applied force, there are nonzero combinations of forces at A, B, C which produce equilibrium.
I thought about this, but I don't understand how it is possible?

Because if I were to physically take a tetrahedron in 3D space, fix 3 points with load blocks, and apply a force to the fourth, surely it would produce definite forces on the load blocks? How could it not?

I think the reason for this is I'm setting the problem up incorrectly from the beginning
 
  • #4
haruspex
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how it is possible?
Start with the 2D version, triangle ABC. Equal and opposite forces at two vertices will balance.
For the tetrahedron, apply any two forces at A, B acting in the plane of ABC and balance with a third such force at C.
 
  • #5
SilasHokanson
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Start with the 2D version, triangle ABC. Equal and opposite forces at two vertices will balance.
For the tetrahedron, apply any two forces at A, B acting in the plane of ABC and balance with a third such force at C.
Just because the forces balance doesn't mean the forces exist. Your body isn't randomly subjected to 5000N forces in opposite directions, cancelling each other out, just because those forces could exist doesn't mean they do
 
  • #6
haruspex
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Just because the forces balance doesn't mean the forces exist. Your body isn't randomly subjected to 5000N forces in opposite directions, cancelling each other out, just because those forces could exist doesn't mean they do
You are missing the point. Your equations permit those forces, so you cannot obtain a unique solution from them.
E.g. in the 2D (triangle) case, force F applied at C, given any solution for the forces at A and B you can obtain further solutions by adding equal and opposite forces at those points.
To get a solvable set of equations you need to add constraints. For the triangle, you could insist that A, say, has no force component in the AB direction. Or that the two have the same force component (same sign) in that axis.
It is not clear to me how to adapt that to 3D.
 
  • #7
Lnewqban
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  • #8
haruspex
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Welcome @SilasHokanson !

Perhaps you could solve it by constraining the directions of the internal forces to the spatial orientations of each of the six edges, assuming those can only transfer tension or compression loads, like members of a space truss do.

Please, see space trusses in this link:
https://engineering.purdue.edu/~aprakas/CE297/CE297-Ch6.pdf
That changes nothing since even with freely jointed corners a tetrahedron is rigid.
 
  • #9
Lnewqban
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That changes nothing since even with freely jointed corners a tetrahedron is rigid.
I must have understood the problem incorrectly, then.

To me, the OP describes a tripod having an external force of arbitrary direction applied onto the apex (point D), and three fixed anchored ends of legs (points A, B and C).
Being non-coplanar with the A-B-C anchored base, that force at point D induces moments that result on forces at each of the other three points.
 
  • #10
haruspex
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I must have understood the problem incorrectly, then.

To me, the OP describes a tripod having an external force of arbitrary direction applied onto the apex (point D), and three fixed anchored ends of legs (points A, B and C).
Being non-coplanar with the A-B-C anchored base, that force at point D induces moments that result on forces at each of the other three points.
You have not misunderstood. The rods can be taken to be freely jointed.
 
  • #11
SilasHokanson
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You are missing the point. Your equations permit those forces, so you cannot obtain a unique solution from them.
E.g. in the 2D (triangle) case, force F applied at C, given any solution for the forces at A and B you can obtain further solutions by adding equal and opposite forces at those points.
To get a solvable set of equations you need to add constraints. For the triangle, you could insist that A, say, has no force component in the AB direction. Or that the two have the same force component (same sign) in that axis.
It is not clear to me how to adapt that to 3D.
Exactly what I was thinking. My equations permit these forces.
What equations am I missing to further constrain the object?

It's fairly intuitive that an experiment like this would have a definitive answer if performed IRL, meaning I am not adequately modeling the problem itself.
 
  • #12
haruspex
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It's fairly intuitive that an experiment like this would have a definitive answer
You would need to consider elastic deformation of the rods.
Consider a vertical massless rod, with elastic constant k, constrained by supports A at the top, B at the bottom.
A weight W is attached at its middle.
Let ##F_A,F_B## be the forces exerted by the supports and x be the displacement of the weight, up positive in all cases,
##F_A=-kx, F_B=-kx, W+F_A+F_B=0##
##W=2kx##
##F_A=F_B=W/2##.

The analysis for the tetrahedron will be rather more taxing.
 

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