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Imagine a long flexible rod with uniform mass. The rod is supported by a pivot at its center and it is at equilibrium. How do we explain why the rod sags at the far ends?

If we attempt to draw a free body diagram of the situation, the weight and normal contact force will both balance out at the pivot. Why then will there be a bending of the rod at the ends?

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3 Answers 3

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enter image description here

the equation for a flexible rod is (Euler-Bernoulli beam Theory) :

$$E\,I\,\frac{d^4 w(x)}{dx^4}=q\tag 1$$ where

  • $w(x)~$ deflection of the rod
  • $q~$ the distributed load
  • $E~$ elastic modulus
  • $I~$ is the second moment of the area of the rod cross section
  • $x~$ rod length coordinate

solve equation (1) with the proper conditions you obtain .($~x(0)~$ is the pivot position)

enter image description here

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You are supposing that the rod is not deformable. First, you should know that in a rod the weight is distributed throughout all the bar. Indeed it is also important to understand that this system isn't in equilibrium while the rod sags at the ends as the movement that the rod has downwards is due to the fact that the normal force from the pivot is less than the weight. But why does this happen? Because materials such as steel are deformable and applying shear stress will deform them. The fact that there is some weight in the ends and the fact that the rod itself can not support all the weight makes it deform and as I said before this will not break any law as the normal force will be lower than weight.

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  • $\begingroup$ But after the rod has sagged, it will be in equilibrium right? Since there is no translational or rotational acceleration $\endgroup$
    – john
    Commented Mar 10, 2022 at 9:01
  • $\begingroup$ Exactly, then normal force will be equal to weight. $\endgroup$
    – Marc Barceló
    Commented Mar 10, 2022 at 10:16
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    $\begingroup$ +1. But note that a real rod would still not be at equilibrium because a weight is being suspended, leading to continued time-dependent deformation. We can see this occurring in plastics (and even some metals) even at familiar loads, time scales, and temperatures. The effect is idealized away as negligible in introductory mechanics classes. $\endgroup$
    – Chemomechanics
    Commented Mar 10, 2022 at 17:22
  • $\begingroup$ True, the amplitude of the movement is usually negligible so I skipped It. Sorry for not mentioning It. $\endgroup$
    – Marc Barceló
    Commented Mar 11, 2022 at 8:12
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Because not only application of force matters, but also a torque. Each elementary mass section $dm$ of rod at distance r from a pivot point is affected by elementary torque of $$ d\vec \tau = \vec r \times d\vec F = \vec r \times \vec g~dm $$

Torque increases to the ends of rod, thus rod is bended like a parabola.

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    $\begingroup$ Like a quartic, rather. The deflection scales as ${\sim}L^4$. $\endgroup$
    – Chemomechanics
    Commented Mar 10, 2022 at 17:24
  • $\begingroup$ @Chemomechanics Good point. But in general it also depends on mass distribution in a rod and gravitational field profile, so one case doesn't capture all situations. $\endgroup$
    – Agnius Vasiliauskas
    Commented Mar 11, 2022 at 7:57

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