I've tried to think on how a force acting on a single object that is in contact with another makes the system of the two accelerate and wanted to know if I'm thinking correct.

Consider the situation: In vacuum, two objects, A and B are in contact with each other horizontally in a plane as shown in the image below. A force of magnitude F acts on object A.

Diagram showing object A and B and force F acting on A

My explanation on the acceleration of the system:

Force F on A causes A to also apply a force with magnitude F on B, and according to Newton's 3rd law of motion, B also applies an equal and opposite force (-F) on A causing A to not accelerate in any direction.
Here, B does accelerate towards the right for a moment due to F acting rightwards, but this acceleration doesn't last long as B loses contact with A thus there is no force acting on B and B continues to move rightwards in some constant velocity. On the other hand, as A loses contact with B, F causes it move rightwards and collide with B and such collisions repeat again and again causing the system to accelerate rightwards.
(These collisions and time when contact is lost between objects would be of very small magnitudes.)

The same thing can also be thought for A and B connected by a string and a force F acting on A, pulling it. Here the string will relax and get taut repeatedly making the system of the two objects accelerate.

Note: My proposition that A also exerts a force of magnitude F on B is based on the fact that we consider the normal force (reaction force) on an object on earth's surface to be $mg$ and that means the object is also applying $mg$ on Earth's surface while $mg$ is the force applied by Earth on the object.

Am I wrong here, why? What could be the correct explanation for how the system accelerates in such a case?

  • $\begingroup$ The force between A and B cannot be F, if it was so then A would not accelerate because the sum of forces actin on A would be zero $\endgroup$
    – user65081
    Commented Oct 8, 2021 at 19:36
  • $\begingroup$ @Wolphramjonny Yes, A would not accelerate only till B is in contact with A, but as B loses contact with A due to F, B moves while A also starts moving when B loses contact as it doesn't experience any opposite force from A. $\endgroup$
    – Silica19
    Commented Oct 9, 2021 at 4:39
  • $\begingroup$ If the masses are the same, then the contact force is equal to F/2, and B never accelerates faster than A $\endgroup$
    – user65081
    Commented Oct 9, 2021 at 5:08
  • $\begingroup$ @Wolphramjonny Can you please explain why force experienced by B is $\frac{F}{2}$? I'm really confused. $\endgroup$
    – Silica19
    Commented Oct 9, 2021 at 5:19
  • $\begingroup$ Do you know how to make a diagram of the forces and write newton's second law for each mass? $\endgroup$
    – user65081
    Commented Oct 9, 2021 at 5:23

2 Answers 2


Alright, you have described one hypothesis.

Here is another hypothesis:

  • Force $F$ pushes on ball A and some of this force is "spent" on accelerating ball A.
  • The rest of the force is propagated to ball B and is "spent" on accelerating ball B.

If my hypothesis is correct, then notice that the contact force $x$ will be smaller than $F$. Newton's 2nd law for balls A and B: $$\text{Ball A: }\qquad F-x=m_Aa\\ \text{Ball B: }\qquad x=m_Ba\quad \\\Updownarrow \\ F-x=m_A\underbrace{\frac{x}{m_B}}_a \quad\Leftrightarrow\quad F=\frac{m_A}{m_B}x+x \quad\Leftrightarrow\quad F=\left(\frac{m_A}{m_B}+1\right)x \quad\Leftrightarrow\quad x=\frac{F}{\frac{m_A}{m_B}+1}$$

In the simple case of equal balls, $m_B=m_A$, then B apparently feels exactly half the force that A feels from the left, $x=F/2$.

On the other hand, if your hypothesis is correct, then the force between the balls should be $F$.

So, which of our hypotheses is correctly describing the scenario? We could maybe measure it with a force-meter at the contact point of A and B (or for a more dramatic effect we could place a fragile object, like a glass piece or similar, between them which can endure a force of $F/2$ but not $F$.)

I do remember doing such experiments during engineering school and do remember seeing close to the expected $F/2$. I don't have a reference at hand, though, and you are very welcome to doubt my testimony. So feel free to do the experiment yourself.


Force F on A causes A to also apply a force with magnitude F on B, and according to Newton's 3rd law of motion, B also applies an equal and opposite force (-F) on A causing A to not accelerate in any direction

You are incorrect in thinking that force F on A causes A to also apply a force F on B. The equal and opposite forces that A and B exert on one another per Newton's 3rd law are internal forces of the AB system whereas the force F is an external force on the AB system. You need to consider them separately.

As long as A and B remain in contact with one another while F is applied to A, they will move together with the same acceleration $a$ according to Newton's 2nd law per the equation:


In order to determine the forces that A exerts on B, $F_{AB}$, and the equal and opposite force that B exerts on A, $F_{BA}$, you need to draw a free body diagram (FBD) on each and apply Newton's 2nd law to A and/or B individually. In this example since $F$ acts directly on A the most direct approach is to apply Newton's 2nd law to A which will determine $F_{BA}$. Then $F_{AB}$ will be equal and opposite to that. So, applying Newton's 2nd law to A we get


From which you can solve for $F_{BA}$. Then of course, per Newton's 3rd law, $F_{AB}$ is equal in magnitude and opposite in direction to $F_{BA}$.

I want to understand that if there is an constant external force (F) on an object (C) and if that object comes in contact with another object (D) in its way of motion, then at the instant when they come in contact does C apply a force F on D. It is related to my 'earth-object' question because in my asked question, the gravity is the external force on the object and it is causing the object to apply the force (action force) of the same magnitude as the gravitational force on Earth's surface and thus the normal force (reaction force) is of the same magnitude.

The answer to the first sentence is no because a collision is involved. The force C exerts on D and the equal and opposite force D exerts on C during the collision is an impact force. It is not simply the passing along of the force $F$ from C to D. Let’s use the earth-object example.

Let the object C have a mass $m$ released at a height $h$ near the surface of the earth. The force on C is therefore the force of gravity, or $F_{g} = mg$. Let the object D be the earth, which is in the way of the motion of the falling object C. Likewise, the object C is in the way of the motion of the earth which is also being pulled "up"by C, though infinitesimally.

When the object C impacts the earth it will have a kinetic energy just prior to impact of $mgh$ which is the loss of gravitational potential energy (assuming no air resistance). Let the earth bring object C to rest over a distance $d$ (the depth of the penetration of C into the earth). Per the work energy theorem the net work done on object C equals its change in kinetic energy. Or, neglecting the loss of potential energy over the distance $d$,


$$F_{imp} = mgh/d$$


$F_{imp}$ = the average impact force

With $d<h$ the average impact force on C will be greater than the force of gravity on C. The less the penetration for a given $h$, the higher the impact force. Moreover, the impact force that C exerts on the earth is equal and opposite to the impact force the earth exerts on C, per Newton’ s 3rd law. Once C comes to rest, the equal and opposite forces will simply be $mgh$.

Hope this helps.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – SuperCiocia
    Commented Oct 11, 2021 at 19:53

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