I have been thinking of Newton (insert bad words because of frustration) laws for hours and hours today, and I'm not going anywhere. I'm going insane! So, I made up this question.

Object A | Object B | Object C

Object A pushes on B by an amount of 60N

Object B pushes on A by an amount of 30N

Object C pushes on B by an amount of 10N

What is the resultant force of each object?

The main problem is: If two objects touch each other, they will exert simultaneous forces on each other. Moreover, there will be simultaneous reactions. But when I do the calculation, I have to start somewhere, and where I start makes a difference. So, how should I go solving the problem? And why is that way the correct way?

My attempt:

(The arrows indicate direction.)

Considering A and B:

A pushes on B by 60, so $F_{A\text{ on }B} = 60 \rightarrow$

B reacts on A by 60, so $F_{B\text{ on }A} = 60 \leftarrow$

B pushes on A by 30, so $F_{B\text{ on }A}$ becomes $90 \leftarrow$

A reacts on B by 30, so $F_{A\text{ on }B}$ becomes $90 \rightarrow$

Considering B and C:

C pushes on B by 10, so $F_{C\text{ on }B} = 10 \leftarrow$

B reacts on C by 10, so $F_{B\text{ on }C} = 10 \rightarrow$

Considering A and B (after considering C on B):

B pushes on A by 100, so $F_{B\text{ on }A} = 100 \leftarrow$

A reacts on B by 100, so $F_{A\text{ on }B} = 100 \rightarrow$


A pushes $100 \rightarrow$

B pushes $100 \leftarrow$ and $10 \rightarrow$

C pushes $10 \leftarrow$

Did I do anything wrong? Is that the right way to solve the problem?

  • $\begingroup$ What have you tried, and where exactly do you get stuck? Right now, we really have no idea what aspect of this sort of thing you need help with. Questions of the form "how do I solve this problem?" really aren't appropriate for this site - it's just like the do-my-homework questions our FAQ prohibits. The fact that it isn't an actual assigned homework problem is irrelevant. But if you can focus this on the concept that is giving you trouble, it could be perfectly fine, and I'll be happy to reopen it then. (Let me know via a comment here when you do that) $\endgroup$
    – David Z
    Mar 1, 2012 at 3:04
  • $\begingroup$ If A pushes on B by 60, then B must push on A 60. Newton's third law. $\endgroup$ Mar 1, 2012 at 3:10
  • $\begingroup$ I have added my attempt David. Please open the question. $\endgroup$
    – w4j3d
    Mar 1, 2012 at 3:33
  • 1
    $\begingroup$ B reacts on A and B pushes on A are the same thing. Think for a moment what you mean by "A pushes on B by 60" and "B pushes on A by 30". The statements make no sense, you can't split up force into original force and reaction force. Neither can you separately apply Newton's 3rd law to these. $\endgroup$ Mar 1, 2012 at 3:41
  • 1
    $\begingroup$ Comment (edit 2) That's not the point. The question makes no sense. Its like saying "That cat is black. That cat is brown. What's the color of the cat?" What is valid is this question: I push block A with a force 60N, and I push block B with 30N. Then it can be solved with an FBD pretty easily. $\endgroup$ Mar 1, 2012 at 5:17

1 Answer 1


See https://physics.stackexchange.com/a/21651/7433 (which you already have, it seems)

Aah. Your main confusion lies in the fact that you feel that object A can exert a force $F_A$ and object B can exert a force $F_B$, with $F_A\neq F_B$. This is not possible.

The forces will readjust to fix this. There is no "constant force-producing device" that exists. All force producers readjust to the circumstances.

I'm only considering two people, A and B of the same mass m. B is to the right of A.

First, I'll take a simple case an analyse it just to show you how these things are done:

You may skip to the vertical line if you want.

Let's take the case when A is on the ground, which has sufficient friction, and B is on a smooth platform. A pushes B with a force $F_A=ma$. Now it seems that B is not pushing back, so $F_B\stackrel{\text{seems to be}}{=} 0$. OK. For A to be able to push, we need a frictional force $f$. That's how we walk. Furthermore, A has to move with an acceleration $a$ to keep up with B. Now, the only external force on the system is $f$, so $f=2ma$. But, $f$ (seems to be) the only force on A. Since A is also going with an acceleration $a$, then $f=ma$. We have a contradiction. From this, we can see that $F_B=ma$ to the left. So, $F_B$ readjusted to the situation.

Now back to a situation similar to yours: OK, let's say that now A and B face each other with their palms touching. They try to stay at rest (sufficient friction). Your situation demands that A exerts a force $F_A$ on B and vice versa, with $F_A\neq F_B$. I cannot stress how impossible this is. The reason you can't find where to start is that the problem itself is nonsense. Try this with your hands. Join them together, as if you are praying. Push with whatever force you want. You will feel the same force on both hands regardless of how you push. The forces readjust.

For a simpler example, take two identical springs. These can be touted as "constant force-producers", since they produce a force proportional to their deformations. So, compress one spring by $x$, and the other spring by $2x$. Holding the other ends in place, touch the springs together. Now, for massless springs, they will instantaneously readjust so that the deformations are both $3x/2$ (you can calculate this if you want, I won't go into the details here. If the spring has a mass, it will readjust with a noticeable speed, as in everyday life. But if the spring has mass, the forces will still instantaneously readjust, just that there will be lots of internal readjustment as well between small sections of the spring.

So basically, like @DavidZaslavsky said in the linked answer, there is nothing like "action then reaction"> It's better to say that they come in pairs.


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