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Find the normal force by each support

  1. Jul 1, 2005 #1
    Basically, you have a rod kept horizontally on three supports. It's in equilibrium. The positions of the supports and the mass and length of the rod are given. Find the normal force by each support.

    The problem looks particularly simple. But notice that you have a shortage of an equation. You have three things to find - the three normal forces. And you have only two equations - one for translational equilibrium of the rod and one for the rotational equilibrium. So what's the problem?
    I asked this at several places and they blatantly said that the system is not solvable. But to me, non-solvability means non-uniqueness. I am not able to understand how is the system non-unique. Am I not providing all the required information? It's also surprising that the same set of information works properly with a two support system. What's happening in the transition from two to three?
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  3. Jul 1, 2005 #2


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    Take away the middle support and solve the problem. Now replace the middle support and make it adjustable so that you can gradually increase its force on the bar. This will change the force on the other supports, but equilibrium will be maintained. At some point, one of the other supports will have no force acting and you are back to a two support problem.

    This is similar to the familiar problem of a 4-legged stool or table. It only takes 3 legs to maintain equilibrium. That fourth leg may do nothing, but since the center of mass is very close to a line between two opposite legs, the stool or table will "wobble".

    In this problem, if the middle support is near the center of the rod it can support nearly all of the weight and the rod can tilt onto one or the other of the outside supports. Any combination of two supports on opposite sides of the center of mass is sufficient to achieve equilibrium. The third is not needed, but if you adjust it to apply some force it changes the force on the other two. The situation is not unique. Two of the fources are a function of the third. You must specify one of the forces to find the other two.
  4. Jul 1, 2005 #3


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    The net Torque about any point must be 0. That actually gives you an infinite number of equations (most of which are equivalent). I'll assume that the rod has mass M, take a support at the left end of the pole to be at x= 0, the middle support at x= x1 and the third support at the right end of the pole at x= x2. Call the forces at those three supports, in that order, F1, F2, F3.
    The total upward force on the rod is F1+ F2+ F3 and that must equal to the weight of the rod, Mg. F1+ F2+ F3= Mg is one equation.
    The torque about the left most support is the sum of the product of the forces times the distances from that support: (0)F1+ x1F2+ x2F3= x1F2+ x2F3 and that must be equal to the torque caused by the weight of the pole, Mg(x2/2). x1F2+ x2F3= Mg(x2)/2 is the second equation.
    The torque about the middle support is -(x1)F1+ (0)F2+ (x2-x1)F3. (Notice the negative sign on the first one: that force will "push" clockwise around the middle support.) That must be equal to the torque due to the weight: Mg((x2/2- x1). The third equation is (0)F2+ (x2-x1)F3= x1F1+ Mg(x2/2- x1.

    I will leave it to you to show that the equation showing there is no torque around the third support is equivalent to the first two torque equations.
  5. Jul 3, 2005 #4
    Hey hey! Two things give me a hint that somehow my problem is being considered a very simple problem - 1. that it has been moved from the 'Classical Physics' section to the Home Work section and 2. the reply by HallsofIvy.

    First of all, when I found HallsofIvy claiming the existence of a third equation by equating the torques about a third point, I was thinking, "No, what's this? Obviously the third eqn. is not going to be independent." And then suddenly he wrote, 'I leave it to you to prove..." What the heck?

    This is a serious issue! I know that mathematics shows that the system is not solvable. But mathematics is not everything! Mathematics is supposed to give rational results. This result doesn't appeal to the rational senses. I can't 'see' why can't we find the normal forces uniquely.

    Like, if someone asks me, "A particle is moving with a velocity 10 m/s. Find it's velocity after 10 sec.", I will 'know' that the question is not solvable. And the reason wouldn't be something like "I have one variable but no equations".

    So that's what I am trying to ask. Can anyone enlighten me about the 'physical reason' of the system in the original problem not being solvable?
  6. Jul 3, 2005 #5


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    In fact, the reason would be that you have one variable but no equation.
    The variable here is velocity and the fact that you don't have an equation to describe the velocity of the particle as a function of time is the cause of the inability to answer.
  7. Jul 4, 2005 #6


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    You haven't responded at all to my post, so perhaps you missed the point. You have more supports than are necessary to achieve equilibrium. Two of them can be adjused in an infinite variety of ways to create the necessary resultant to achieve equilibrium. Whenever you create an over-supported condition, you leave the realm of unique solutions.

    I don't know why you think the result does not appeal to the rational senses. Pehaps it is the senses that need some fine tuning. The example I gave of extending the argument to a three legged stool is common to everyones experience, and does not require complicated equations appropriate to that situation to be "rational". You don't need a fourth leg to achieve equilibrium, so you cannot know how much force the fourth leg will apply. If you know the force from one leg, then you can find the force needed by the other three. People who had no knowlege of mathematics have use three legged stools for hundreds, and probably thousands of years because they realized from experience that only three legs are required for stability and are often preferred to prevent the instability that results from providing more supports than needed. Rigid four legged stools, tables, etc., are commonly used because of the advantage of extending the area of stability that can be achieved, but only in applications where there is reasonable hope of a flat support for the four legs. Many four-point supported objects like automobiles have flexible suspensions to permit all four supports to contribute, but nobody thinks that there should be a unique solution to the division of support among the four points.

    In your problem, two support points are sufficient to achieve equilibrium. Since you don't need a third support to achieve equilibrium, so there is no way to use equilibrium to yield a unique solution. Without knowing one force, there is no way of determining the others.
  8. Jul 4, 2005 #7


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    Like you have been told, the system is not solved by static equilibrium equations, and it's therefore called statically undetermined, this type of systems can be solved with compactibility equations (displacements in the body) where these displacements can be related to the forces acting on the body (stress-displacement relations) which can be linear-elastic situations (Hooke's Law) or nonlinear sistuations (Ramsberg-Osgood equations). If you want to know more about statically undetermined systems, take a Mechanics of Materials course, or for a intro look for Mechanic of Materials by James M. Gere.
  9. Jul 4, 2005 #8


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    I was thinking about an important factor, you'll probably be interested in statically undetermined systems is that the mechanics properties of the material matters, such as its modulus of elasticity, modulus of axial rigidity, isotropic, homogenous, among others are an important factor, in order to determine the internal forces and/or the reactions, unlike statically determined systems which reactions and internal forces can be calculated without knowing the properties of the material.

    In the analysis and design of structures, statically determined systems does not necessarily guarantee geometrical stability for real life applications, for example an armor with 2 mobile supports. It's also good to point out that statically undetermined structures because of their redundancy or what OlderDan was trying to explain "extra supports" can be more stable even if a structural member of the system fails.

    Most of the treatment of statically undetermined systems can be seen in a structural analysis course and mechanic of material course.

    If you find yourself interested in this topic, consult the books:

    Mechanic of Materials by James M. Gere (A good introduction of the fundamentals, and talks about statically undetermined systems)

    Structural Analysis by Aslam Kassimali (basicly your undergrad structural analysis book, a good treatment on statically determined and undetermined systems)

    Analysis of statically undetermined structures by Sterling Kinney.
  10. Jul 5, 2005 #9
    Thanks for the reply all of you. I agree with the points you all have given. Thanks for suggesting these books. May be things will be clearer once I go through them.
    It's true that the rational senses need fine tuning. And I am looking forward for that 'fine tuning'. What is not comprehensible is this:-
    You have three rigid supports of same height kept on a perfectly horizontal surface, distance between two adjacent supports being exactly 1 metre. You are holding a rigid rod of length 2.5 metres horizontally 1 metre above the three supports. Now you bring the rod down slowly and finally place it on the supports such that the first support touches the end of the rod.
    Now, I think, the information that how a system is brought into a particular state should suffice to completely determine the properties of the system in that state.
    Here I think I have described everything done to the system to bring it to the state of equilibrium. This information should be sufficient to determine every parameter of the system in that state. So according to this logic, shouldn't we be able to tell the three normal forces?
  11. Jul 5, 2005 #10


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    No, you can't, because in the beginning before the rod was on the supports, the reactions (normal forces) on rod by the supports is 0, but when you start placing the rod about the 3 supports, there's a distribution of the load, which will generate reactions by the supports, so again i repeat, the solution for statically undetermined systems are by studying the displacements (its conditions), generated by the stress due to the load on the rod.

    In order, to see the distribution of the load, try putting the rod at different angles, and use two supports to calculate the reactions, you'll see different values for the reactions, but they will still add up to guarantee static equilibrium.
  12. Jul 15, 2005 #11
    1 translational and THREE rotational(infinite actually)equilibrium(abt the 3 supports) can be generated. The solution using 3 of the equations will fit the fourth...i took a general case of keeping 2 of the supports at the end and 1 in the middle(coinciding with CoM) and solved it fully.
  13. Jul 15, 2005 #12


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    Actually it depends on the supports, if they are like hinges, you won't be able to solve it, because of the unknown forces in x, if they are like mobile supports or frictionless surfaces only having one y-component each of the reactions.

    The setup you provide of a beam with 3 supports (frictionles surfaces, probably), even coinciding the middle support with the centroid is a hyper statics of degree 1 system, not solved by statics.
  14. Jul 15, 2005 #13
    if u cant solve it, its not an unique system. But the geometry says for itself, it is unique! maybe im not getting the correct picture in mind. cud sum1 over here be sooo graceful and put up a drawing of the rod + supports???
  15. Jul 15, 2005 #14


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    Here is a diagram, and a calculation based on that diagram showing that the torque equations are not independent. Forces A, B, and C act upwards and their sum must be the weight of the rod. a, b, c, and d are the distances between the five points locating the ends of the rod and the supports. The two torque equations are calculated about the ends of the rod, but you can use any points you like to generate as many equations as you want. Add your torque equations together and see what happens.

    [tex] A + B + C = W [/tex]

    [tex] aA + \left( {a + b} \right)B + \left( {a + b + c} \right)C = W\frac{L}{2} [/tex]

    [tex] \left( {b + c + d} \right)A + \left( {c + d} \right)B + dC = W\frac{L}{2} [/tex]

    Add the last two equations

    [tex] \left( {a + b + c + d} \right)A \ \ +\ \ \left( {a + b + c + d} \right)B \ \ +\ \ \left( {a + b + c + d} \right)C = WL [/tex]

    [tex] \left( {a + b + c + d} \right)\left( {A + B + C} \right) = LW [/tex]

    Divide by the first equation

    [tex] a + b + c + d = L [/tex]

    This is just an identity. There are only two independent equations. The only additional restriction is that none of the three forces can be negative, assuming the rod is merely resting on the supports and not attached.

    Attached Files:

  16. Jul 18, 2005 #15
    ill try solving it taking torque eqns @ the supports later but if u say it cant be solved, why cant it!? its a unique system isnt it?
  17. Jul 18, 2005 #16


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    No, it is not unique. Consider the diagram, and suppose you require the forces A and B to be equal. Then the solution would be unique. Given that solution, suppose you now gradually reduce B while increasing A to maintain equilibrium until B reaches zero. Then gradually increase B while decreasing A to maintain equilibrium until A reaches zero. All you can do for the given set of positions is come up with the relationship between A and B that must be satisfied. The range of possible values of A and B will be bounded (assuming both must be upward), but the values are not unique.
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