Consider a system at rest with three bodies and let an external force applied. The bodies' masses are: $m_x$, $m_y$ and $m_z$. The bodies move together when a force is applied.

So on my mind, the same force should be applied to each of these boxes,because boxes conduct forces.But when we use newton's second law to analyze the system, it turns out they are not same.

$$\sum F_\text{system}=(m_x+m_y+m_z)a=F$$

but for individual objects:

$$\sum F_x=m_xa$$ or

$$\sum F_y=m_ya$$

Both are smaller than F, which doesn't make sense to me. Please explain how this works. System

  • $\begingroup$ Have you done a free body diagram for each box? $\endgroup$
    – Bob D
    Commented Nov 15, 2020 at 11:03
  • 1
    $\begingroup$ @Tuna Yılmaz what do you mean by boxes conduct forces ? $\endgroup$
    – Ankit
    Commented Nov 15, 2020 at 11:10
  • $\begingroup$ No,because I feel confused. $\endgroup$ Commented Nov 15, 2020 at 11:10
  • 1
    $\begingroup$ Here's another way to think about this. Imagine this was all a single object; when you apply a force, you give the body the acceleration $a = F/(m_X+m_Y+m_Z)$ (all parts have the same acceleration). Now, cut off the X box (so that you're just left with Y+Z). To supply the same acceleration, a force of smaller intensity is enough: $F_{Y+Z} = (m_Y+m_Z)a$. Now cut off the Y box. Again, to supply the same acceleration, it's $F_Z = m_Za$, smaller still. $\endgroup$ Commented Nov 15, 2020 at 19:48
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    $\begingroup$ @TunaYılmaz Feeling confused is one of the best reasons to draw free body diagrams, and drawing free body diagrams is one of the best responses to feeling confused. Even if your free body diagrams were wrong, they would help others point out where you're mistaken. $\endgroup$
    – JiK
    Commented Nov 16, 2020 at 0:16

2 Answers 2


Right so I understand your confusion because I've been there before. First of all have in mind that, in this situation, if you push the X box, all boxes will move together thus having same acceleration $a$.
If what you say was true (that all boxes should be under the same force), since each box has a different mass, their acceleration would be different (following $F=m_i·a_i$). Now imagine $m_x \lt m_y \lt m_z$. In this situation, when applying a force to X box, all the boxes would "separate" because their accelerations would be different. That would of course be nonsense.

I understand what you say by "boxes conduct forces" and it's "true", but not in the way you think. They don't "conduct" the net force they recieve. They "conduct" the force they recieve less the amount needed to accelerate themselves.
Applying 2nd Newton law clarifies it. Drawing a free body diagram for each box:

Free body diagram for each box So for X box we have $F$ and $F_{xy}$. This last one is consequence of Newton's 3rd law. As we push box X, box X will push box Y. Box Y will then push box X back. Same in Y box's diagram.
For Z box, we only have the force of Y box pushing it.

Now applying Newton's 2nd law for each box. $$ F-F_{xy}=m_x·a\\ F_{xy} - F_{yz} = m_y·a\\ F_{yz} = m_z · a $$ You can see it now: X box pushes Y box following $F_{xy}=F-m_x·a$
And Y box pushes Z box following $F_{yz} = F_{xy} - m_y·a = F - (m_x+m_y)·a$

  • $\begingroup$ The label of the rightmost box should be … $\endgroup$
    – gboffi
    Commented Nov 15, 2020 at 23:08

You need to do a free body diagram (FBD) on boxes X and Y applying Newton’s 2nd and 3rd laws with the constraint that all boxes have the same acceleration.

For example, take box X. An FBD on X will have $F$ acting to the right and $F_{yx}$ acting to the left where $F_{yx}$ is the force that box Y exerts on box X. Then per Newton’s second law net force on X will equal it’s mass times the acceleration $a$. Or


Do the same for box Y. For box Y the force exerted on it by box X will equal the force box Y exerts on X per Newton’s third law.

When done you will have 2 equations and 2 unknowns. The net force on box Z will simply equal the value of the second unknown $F_{zy}$

Hope this helps


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