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Suppose there is a chain with 5 links in it where each link has a mass of $m=0.1kg$, and the chain is being accelerated upward at $2.5 m/s^2$. I want to find the net force on each link in the chain.

I would have thought that the net force is the sum of the force from gravity (i.e., its weight) with the upward acceleration. But the book answer has 0.25N, indicating that only the upward acceleration is relevant.

After thinking about it, I realized that the downward force on each link is being offset since each link is supported by the link above it (with the top one supported by the rope accelerating it upward). Thus for each $W=mg$ force vector downward, there is a normal force going upward to offset it because of the normal force from the supporting link above it, leaving only the upward acceleration of the chain as a whole as the net force, which gives $F=(0.1kg)(2.5m/s^2)=0.25 N. $ Is my reasoning correct?

Thanks for any help.

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The reason each link will have the same net force is because:

  • Each link has the same acceleration, and;
  • Each link has the same mass.

Since the net force is $\Sigma F=ma$ and $m$ and $a$ are the same for each case, the net force on 1 link is the same as the net force on each other link.

So yes, each link will have a net force of $\Sigma F_\mathrm{link}=T_i-T_i'-mg=ma$ where $m$ is the link mass, and $T_i$ is the force pulling up that specific link (each link will have its own magnitude for $T_i$), and $T_i'$ the force pulling up the link below (Newton's third law action-reaction pair).

I would have thought that the net force is the sum of the force from gravity (i.e., its weight) with the upward acceleration.

The net force on the system is $\Sigma F_\mathrm{system} = T-Mg$ where $T$ is the upwards force pulling everything up and $M$ is the mass of the five chain links, $M=5m$.

The net force of a link is $\Sigma F_\mathrm{link}=T_i-T_i'-mg$ where $m$ is the mass of 1 chain link and $T_i$ is a force pulling up that specific chain link (different for each chain), and $T_i'$ is the force pulling up the link below.

The following FBD should clear it up:

enter image description here

You can also calculate the net force on the system, namely:

enter image description here

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  • $\begingroup$ Nice diagram, thank you for this thorough answer. May I ask what software you used to make your FBD? Thanks again. $\endgroup$
    – Hank Igoe
    Commented Jan 5, 2021 at 3:41
  • $\begingroup$ I use Microsoft Visio 2016. My version is a full license, but they offer subscription based plans as well. There are however many alternatives, a lot of them free. See here for more physics.stackexchange.com/questions/401/…. $\endgroup$
    – user256872
    Commented Jan 5, 2021 at 3:56
  • $\begingroup$ @Hank Igoe EDIT: In my initial sketch, I forgot to put the reaction forces. Fixed now. $\endgroup$
    – user256872
    Commented Jan 5, 2021 at 4:04
  • $\begingroup$ @Hank Igoe I updated my post once more. It should be much clearer (I did make a small error by forgetting to include the 3rd law action-reaction pairs, but it's fixed now, hope it was not confusing :) $\endgroup$
    – user256872
    Commented Jan 5, 2021 at 6:19

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