My teacher showed us the following diagram.

He said: When a box is placed on a surface, a weight force $W = mg$ acts on the surface, to which, the surface applies Normal Reaction $N$ on the box so that the box does not falls through the surface. He then says that ALL forces occur in action-reaction pair, therefore, the box would also apply a reaction force $N_{reaction}$ on the surface.

But if all forces occur in action-reaction pair how come there are only 3 forces in his FBD, shouldn't there be an even number of forces? Did he miss out drawing a reaction force one of them?

A box on a surface

  • $\begingroup$ What does the $N$ in the upper direction mean ? And I think $N_{reaction}$ should be in the upper direction . $\endgroup$
    – Users
    Commented May 23 at 15:14
  • 2
    $\begingroup$ The Earth is pulled up by the Box by the same force as the weight of the Box. That is the pair if you so wish to draw it. It will, however, just inevitably, cause you to confuse yourself. $\endgroup$ Commented May 23 at 15:16
  • $\begingroup$ The pair to gravity is acted upon the earth, and the diagram omits this force. $\endgroup$ Commented May 24 at 14:21
  • $\begingroup$ "The Earth is pulled up by the Box by the same force as the weight of the Box" Mind is blown!! $\endgroup$ Commented May 24 at 18:54

4 Answers 4


When you do a free body diagram, you should focus only on forces and moments acting on the system of your interest. As an example, if you're studying the equilibrium of the box, you should only draw forces acting on the box, namely its weight and the reaction from the surface to the box.

Anyways, your professor is right: you should always expect pairs of forces when you look at all the bodies of the world. In this example there are two pairs (related by action/reaction principle) of forces:

  • the one due to the contact between the box and the ground, and you've already drawn both the force acting from the ground on the box and the one from the box acting on the ground;
  • the one due to gravitation: here you've only drawn the weight of the box, but this is the force of attraction exerted by the mass of the Earth (assuming we're on Earth) on the box. The other force of this pair is the force of attraction acting on the Earth due to the presence of the box

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First off, number of forces is a vague term. You can break a force like 10N into two smaller force of 5N. For example number of forces between a block and table is infinite and is distributed all over the surface. You are talking about the net force exerted by one body, like the normal force taken through the COM of the body.

Now, as forces appear in action-reaction pair, the number of ALL forces on ALL the bodies which are exerting force must be even, because action-reaction forces act on different bodies, we must consider all the bodies, thus creating two equal and opposite forces.

So, the normal force N is exerted BY table ON the block. So, the block will also exert the same magnitude of force ON the table. However, you missed a force. You considered gravity, which is a force between block and Earth. So the 4th force is the force exerted BY block ON the earth with equal magnitude(mg).

The Earth also exerts a normal force on the table and also a gravitational force. In return, the Table will also push the Earth with the same normal force and simultaneously pull the earth with the same gravitational force acting on it. Thus, the total number of forces can be said to be even on all the bodies.

A general tip for mechanics: Always be sure to select a system on which forces will act. Many students make the mistake of not clearly defining a system and then apply force of the system on itself.


As @basics already answered, the diagram is missing the equal and opposite force $W$ acting on the earth by the box. But this omission is not in error, but rather a convenience as the motion of the earth isn't important for this problem, and when drawing an FBD we draw only the forces acting on one body.

In this sense, the diagram is a bit misleading because it includes the force $N_{\rm reaction}$ acting on the earth/surface.

I wanted to elaborate a bit more on a tangent topic of the nature of contact forces. Although a single force $N$ is drawn, this isn't an accurate picture, as there is a distribution of pressure acting over the entire surface area.


Only when you sum up all the little contributions of each part of the contacting surfaces you can evaluate the contact force ($N_{\rm total}$ above). But also, there is an equipollent torque reaction that can occur if needed that would shift the contact force distribution to one side or another.

So the fact there is a reaction force and reaction torque is only an idealization done for simplicity, and the true nature of contacts is that there are infinite tiny forces acting in many different locations.

Also indicated above is the fact that when two flat surfaces meet, the contact forces tend to concentrate on the edges. This is the reason in game simulations contact forces are split on the edge of the contacting surface and applied there.


There are two forces acting on the box, which are its weight $W$ and the normal force from the surface $N$.

There is at least one force acting on the surface, which is the equal and opposite reaction to the normal force, which is shown as $N_{reaction}$ in your diagram. We do not know how many other forces are acting on the surface. If the surface is actually the ground (i.e. the surface of the Earth) then there are many, many other forces acting on the Earth, but they are not of interest to us. If the surface is a table that in turn rests on the ground then there are three forces acting on the table; $N_{reaction}$, its own weight, and another normal force from the ground.

If the box is in equilibrium then we can say that the net force acting on it must be zero, so $W$ and $N$ are equal in magnitude and opposite in direction. However, if the box is not in equilibrium (if it is in an accelerating lift, for example) then the magnitude of $N$ may be greater than $W$ (if the lift is accelerating upwards) or less than $W$ (if the list is accelerating downwards. The force $N_{reaction}$ is always equal and opposite to $N$, so its magnitude may similarly be greater than or less than $W$.

You may be wondering if forces come in equal and opposite action/reaction pairs then where is the equal and opposite force corresponding to the box's weight $W$ ? Since the force $W$ is exerted on the box by the Earth, then the box must exert an equal and opposite force on the Earth - in other words, the box attracts the Earth with the same force as the Earth attracts the box. However, the Earth is so massive that the effect of this force on the Earth is negligible and we usually ignore it. If, however, we were thinking about the motion of the Moon then we could not ignore the force that it exerts on the Earth.


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