Case 1:

Suppose there is a rod hung vertically from the ceiling. It experiences gravitational force W (its weight).It's Young's Modulus is $Y$. Now I know that the end connected to the ceiling experiences a lot more force as it has to bear the entire rod. Whereas the bottom doesn't have to bear anything. So clearly there is non uniformity in stress. In increases gradually $(\propto length)$ from the bottom. So the net strain should be $$ \delta L=WL/(2AY)$$

Case 2:

There is a rod kept on a smooth horizontal surface and is pulled by a Force $F$ on one end. The other end is free. My book says the tension experienced by a differential length at a distance $l$ from the pulled end decreases linearly. The reasoning is that the differential experiences a force experienced by the complete right hand part $l$. Treating this as a statistical phenomena only, the net elongation is $$\delta L=FL/(2AY)$$

Next the book says that if we pull by an equal force F from the other side too, the elongation would be double. so now the elongation is $$\delta L=FL/(AY)$$

My issue is, isn't Case 1 a particular case of the modified case 2 (pulled by 2 equal forces F). Here F=W. I say this because the rod is hung. It is clearly in equilibrium. The ceiling too exerts a reaction force equal to W. So the hung rod is also pulled by equal forces on 2 sides. Using the general formula explained in case 2, the elongation should be double of the one written.

I am very uncertain about Case 2. I think it is conceptually very clumsy. But it's a general formula mentioned by the book.

I don't want to learn a wrong concept so please clear this confusion. Either I am missing something or the book is wrong.

  • $\begingroup$ An assumption I'm making about Case 2 single force, in that since the surface is smooth and there isn't an opposing force, the rod is therefore accelerating. From a brief scan, the book seems correct. When the rod is hung and pulled only by its weight, a cross-section very close to the ceiling experiences a pulling force close to the weight of the whole rod (W). The cross-section close to the bottom of the rod though experiences nearly no pulling force, since any downward-pulling force (which equals upward reaction force) comes entirely from the weight of the remaining rod below. $\endgroup$
    – Melvin
    Jan 26, 2019 at 6:26
  • $\begingroup$ The stress profile along the rod is therefore linear, starting from weight of rod at one end and ending at 0 on the other. The area below the curve is therefore half that of what it would have been, if the rod was pulled equally on both ends by the same force, as in Case 2. Hence the 2 in the denominator in Case 1. $\endgroup$
    – Melvin
    Jan 26, 2019 at 6:29
  • $\begingroup$ I do understand that, but if we treat weight as just another force F, won't it become a modified case 2 situation. We can simply imagine the vertical rod is a horizontal one that is being pulled by force W on both sides (ceiling reaction and its own weight). So there should not be a 2 in the denominator. why do they differ then? $\endgroup$ Jan 26, 2019 at 6:38
  • $\begingroup$ The difference between case 2 duo and case 1, is in where the force is effectively applied. In case 2, this is clearly at the opposite end. In case 1 however, we can think of the gravitational force as applying a force W to the centre-of-mass of the rod, which for a uniform rod is at the centre and not at the end. This is equivalent to case 2 duo, if one had applied the pulling force for one side at the middle of the rod. So in the formula for case 1, you can think of the 1/2 factor as coming not from W, but from L, since the effective length is halved. That is one way of thinking about it. $\endgroup$
    – Melvin
    Jan 26, 2019 at 6:44
  • $\begingroup$ I had actually thought of that. So the weight acts on the centre of mass which is at $L/2$ distance. This also accounts for the 2 in the denominator. But is it a correct interpretation? Because here we totally neglect the internal distribution of stress which is vital to the solution. $\endgroup$ Jan 26, 2019 at 6:44

2 Answers 2


Assuming Hooke's law holds in every case and assuming rod's mass is uniformly distributed and assuming gravitational field is acting uniformly on the rod, what you said about case 1 is right.
Assuming there is no friction, When only 1 force is applied to 1 end of the rod in case 2. enter image description here

Here the rod must be accelerating to the right.
Now if we take a very thin strip of the rod there will be a tension T acting on the strip on the left end and there must be a force greater than T acting on the right end for the strip to accelerate to the right (The force i denoted in the diagram as T + dT which is again a tension).
Using this we can say, the tension must decrease as you go farther away from from the end experiencing the force, or in other words the tension must increase as you go closer to the right end of the diagram. If you want to prove $ \delta L=FL/(2AY)$ try applying Hooke's law to the strip (also I wonder why can you apply Hooke's law to the strip as forces acting on it are unequal on both sides? What I'm thinking is that maybe as $dx$ approaches 0, $dT$ also approaches 0, so forces acting on the sides of the strip will tend closer to being equal, making the use of Hooke's law more accurate.) to find the extension of the strip and then add all the extensions of all the strips up across the entire rod (integrate it). You'll end up getting this

$$\delta L = \int_0^L \frac{T \,dx}{AY} $$

But how do we solve this integral as T also changes with x. This indicates that T depends on x so lets try to find that relationship. notice $T = m'a = \rho xAa$ and $F = ma = \rho LAa$ where $\rho$ is density of the rod. Dividing the two equations.

$$\frac{T}{F} = \frac{x}{L}$$ $$T = \frac{F\,x}{L}$$

This is the relationship we were seeking. We can put this back into the integral and solve it to prove $ \delta L=FL/(2AY)$

But when we pull the rod by an equal force F on each side the rod will not accelerate. The tension on each side of a very tiny strip of the block must be same for the strip to not accelerate, and that tension must be equal to F. So using Hooke's law you can prove $ \delta L=FL/(AY)$.
Now qualitatively you can see case 1 and the case in case 2 with a force F acting on each end of the rod are different, as in case 1 the internal tension of the rod varies due to different strips of the rod having to bear different weights under it, while in the case in case 2 the internal tension is same throughout the rod.
Proof is in the picture below.

enter image description here

  • $\begingroup$ Now im wondering what happens when there is friction $\endgroup$
    – Saif
    Nov 28, 2023 at 3:53
  • $\begingroup$ I'm thinking instead of F it should be F - f, following a similar argument i said above but physics.stackexchange.com/questions/222310/… it seems different according to this $\endgroup$
    – Saif
    Dec 16, 2023 at 10:42

In case 1, gravitational force vary along the length. Just above the bottom most point there is almost negligible pull which increases linearly as we go up and becomes maximum at the hinged point. However, in the situation where we apply equal forces from both sides there is net force acting on either end. I guess that's how the result are different in the two cases.


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