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Static Equilibrium for an object in 3-dimension

  1. Dec 19, 2012 #1
    I have studied a few online sources about static equilibrium in a mechanical system. My overall understanding is that for an object to be in static equilibrium, two following conditions have to be fulfilled:

    1) Vector sum of all external forces that act on body must be zero.
    2) Vector sum of all external torques that act on the body, measured about any possible point must be zero.

    I fully understand condition (1), but the statement of condition (2) is not clear for me. Let's say that an object A is supported by two other objects, B and C, where each contact point, (A,B) and (A,C) could be a possible rotation axis for A. So, I can write 9 equations for unknown reaction forces:

    set 1) F_{total, x} = 0, F_{total, y} = 0, F_{total, z} = 0
    set 2) M_{total, (A,B), x} = 0, M_{total, (A,B), y} = 0, M_{total, (A,B), z} = 0
    set 3) M_{total, (A,C), x} = 0, M_{total, (A,C), y} = 0, M_{total, (A,C), z} = 0

    But we know that there exist only 6 independent equations. My question is that, if I use (set 1) and (set 2) to solve for unknown forces (I assume that the system has a solution), then can I deduce that the object A will also have zero torque about (A,C)? In other words, does a solution for (set 1) and (set 2) imply that that solution is also a solution for (set 1) and (set 3)?

    I have found a case in which, all three objects are cuboids and the contact point (A,B) is a line segment (an edge of B on a face of A) but the contact point (A,C) is a single point (a vertex of C on a face of A). In this case, the solution of (set 1) and (set 2) is not a solution for (set 1) and (set 3).
  2. jcsd
  3. Dec 19, 2012 #2
    Hello Rasoul, I'm not sure if I read your example correctly but it is statically indeterminate in the general case.
  4. Dec 19, 2012 #3
    Hi, you are right! The system is staticaly indeterminate. I use nonlinear optimization to find a solution for the system. Do you think that I should use all 9 equations in my optimization to find a consistent solution? Alternatively, is that correct if I always calculate rotation about "center of mass" of an object and if a solution found, deduce that the object is in static equilibrium (about any possible axis of rotation)? Or do I have to calculate rotation about a "possible" axis of rotation?
    Last edited: Dec 19, 2012
  5. Dec 19, 2012 #4
    Statically indeterminate means exactly that.

    There are an infinite number of solutions that satisfy the equations of static equilibrium alone.

    You have to supply other equations to solve the system.

    Are the bodies rigid?

    Then the only equations you have are those of geometric compatibility.

    You can always do an energy/stability calculation to show that some configurations are less likely than others, but they are never mechanically impossible. For instance finding a pencil balanced on its point is less likeley than finding it lying down on the table.
  6. Dec 19, 2012 #5
    Bodies are rigid and I assume that their geometrical shapes are cuboid. As an example, please see the attachement image file (the magnitude of forces are scaled for visualization purposes). I have done calculation about the edge (Box<21>, Box<11>) as the axis of rotation and the optimization solver found a solution that satisfies the static equilibrium condition, but it is intutive that the box<21> cannot be static. On the other hand, if I run optimization solver with torque calculation about contact point between the Box<21> and Box<17> (which is a single point), there is no solution for.

    What do you mean by "geometric compatibility" ?

    Attached Files:

  7. Dec 19, 2012 #6
    Geometric compatibility refers to the geometric constraints set by the system or problem in hand.

    This may be a zero differential or displacement or rotation (or other known value) that the system can take up.

    Four example in a four bar linkage if you set the position of three points of the quad, the lengths of the links describe a locus of the possible positions of the fourth point.
  8. Dec 20, 2012 #7
    I see this greatly deviated from your original question which was understanding this:

    2) Vector sum of all external torques that act on the body, measured about any possible point must be zero.

    For an object to be in static equilibrium, one of the requirements is that the object must not have a net acceleration on it. This is why (1) is so easy to see because ƩF = 0 since the a in (ƩF = ma) is 0.

    But another condition that is hardly explicitly stated in Statics textbooks is that the object must not have a net angular acceleration on it. Remember from general physics that the Torque on an object is ƩrXF where r is the distance from a force to some point O. The symbol x is to denote the cross product between the vectors r and F. It is shown that ƩrxF = I[itex]\alpha[/itex] where I is the objects moment of inertia and [itex]\alpha[/itex] is the objects angular acceleration. Because any point O can be chosen, ƩrxF is dependent on the choice of O and you will get different angular acceleration values with different choices of the point O. In Engineering, we call the Torque, Moment. Same thing, different names. Actually, most engineers refer to Moment when talking about bending and Torque when talking about twisting, but they are the same thing. Since the angular acceleration must equal 0 for the object, and the angular acceleration depends on the choice of the vector r, that must mean that ƩrxF must equal 0 for every r and thus every point O.

    If the next question is...."Well isn't ƩrxF already equal to 0 because ƩF = 0?" Not necessarily.
    Read about "couples" to learn why this is true.

    Also, a common mistake is to believe that ƩrxF = ƩrxƩF ...this is not true mathematically for all cases.
  9. Dec 20, 2012 #8
    Not done.

    So thats what the "about every possible point" means. Creating cuts will not grant you more independent equations which is what Studiot is saying. For statics, you have 6 independent equations and can solve for six independent unknowns. If the number of independent equations exceeds the number of equations, you can not solve by the equations of Statics alone. They can be solved, but you need to generate additional equations regarding how the structure deforms. This is what Studiot eluded to when he spoke of geometric compatibility.
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