# Resultant Forces in a vertical rod

This is a pretty basic question. So in the below figure is the representation of prismatic column. The forces are represented as P1 and P2. let x be an arbitrary cross section between A and B. The stress in the arbitrary (not considering self weight) is given as P1/A Where 'A' is the area of cross section of the arbitrary cross section between A,B. Similarly for an arbitrary cross section between B and C stress is given as (P1+P2)/A.

Now my question is the P2 will also induce some force on the cross section in AB region. So the stress should have been (P1+P2)/A for the cross section of X also. (Imagine this cross section as a trampoline. where a metal ball pushes it down from above and another metal ball is tied below, also pulls it down). So the net stress induced should be same (P1+P2)/A and also the net strain should be uniform in AB and BC cross section?

Is your setup required to be in static equilibrium (i.e., there is no movement of the rod at all) ? If yes, there's a very simple answer to why P2 won't affect the section x-x.

Let's take a section cut slightly above the load application point of P2:

Ask yourself - what's the force needed to keep this rod section in place ? The answer is P1:

So P2 doesn't affect P1 in this instance because of the requirement of static equilibrium.

• That what I was missing! Thanks a lot. Commented Oct 5, 2022 at 16:11

You are thinking about this in a wrong way. Just imagine you are standing on the ground and then you climb up to a car parked next to you. Should your legs feel more stress when standing on a car? You would be basically AB standing on the BC and the weight of BC should not influence your stress (unless BC is really massive like the Earth in which case you would of course be pulled by the gravity).

• You are considering there are two objects an upper and a lower one....but that's not the case. I'm talking about a single object with no gravity to pull it down. You can also refer to my example in the NMech answer's comment. Commented Oct 4, 2022 at 18:22
• @SankaraNarayan The principle is very similar. Section of the rod is not fixed in space, so the force below cannot pull on it. If every section between A and B was fixed in space, the forces P1 and P2 would in fact produce no stress there. You could try to place a stack on the table: one hand (BC), a book (P2), second hand (AB), another book (P1). Now if you relax the hands, surely you will feel more compression in the bottom hand. Commented Oct 4, 2022 at 19:55

Have a look at the image below.

By your reasoning, the person on top should feel the same stress as the person below (Which obviously is not true).

Consider the rope in the image above (its similar to your problem, if you cut the rope somewhere Between CD, with the difference that the forces are pulling instead of pushing).

(Assuming the rope has a constant cross-section, and no flaws), which part of the rope is in danger of breaking more easily? AB, BC or CD?

Do the following mind experiment:

• Does the rope to the left of point A "feel" the force that the first man is putting, or the second man for that matter? (Answer is no)

• does the rope between A and B "feel" the force of the second man ? (Answer is no)

• does the rope between A and B "feel" the force of the first man? (Answer is yes)

If you can first explain why you have those answers you should be able to understand that action and reaction does not work in the way you originally thought.

• ok. I will use your picture as an reference. Imagine there is no gravity from the ground. they are just standing on the ground and are at rest. Top and Bottom Persons are but one person whose top to hip is like the top person and the hip to toe is the bottom person Top to hip region is X and hip to bottom region is Y. The person will feel no force(no gravity). Suddenly a spacecraft from above the person pushes the person down with force P1. X and Y will experience stress P1/A. Suddenly another space craft appears and is tied in the hip region and pulls the hip region down. Commented Oct 4, 2022 at 18:17
• So won't the upper body (top to Hip) feel the compression of the spacecraft from above and the tension of spacecraft from below? So the upper body should have the force of both the spacecraft when we calculate the stress induced. Since both applies a form of stress. Commented Oct 4, 2022 at 18:18
• Also the force the spacecraft tied to the hip region exerting is P2. If you can draw a Free body diagram of the cross section in X and Y. In both the forces P1 and P2 are in the same direction (downwards). So in both regions the force must be P1+P2. and stresses must be (P1+P2)/A. I know I'm wrong and I'm missing something. but I'm trying to find out what I'm missing while reasoning this question. Commented Oct 4, 2022 at 18:31
• I didn't ask about a body which can only have tension. let's say that instead of rope there is a steel rod, people are strong enough produce a force which can deform it. Now the rope can only undergo tension whereas the steel rod can undergo compression too.. Now imagine you are the planar region between A and B Facing B. Now A would try to pull you from your back and but there are some planes in front of you which does not want you to go back hence tensile stress is developed. But B is trying to push you back but planes behind you does not want you to come backwards (compressive stress). Commented Oct 4, 2022 at 19:02
• So there are both tensile and compressive stress and A and B when you take an example of a single steel rod which can undergo both compression and tension whereas rope can undergo only tension. Commented Oct 4, 2022 at 19:04