(Firstly note that I'm not asking for solution to this problem (I've added the solution here), so don't mark this as a "Homework question". I'm asking because I've several confusions regarding the solution to this problem.)

A problem in SS Krotov's Aptitude in Physics book reads:

A horizontal weightless rod of lenght $3\ell$ is suspended on two vertical strings. Two loads of mass $m_1$ and $m_2$ are in equilibrim at equl distances from each other and from the ends of the strings. Determine the tension $T$ of the left string at the instant when the right string snaps.

With this, a figure is given:

enter image description here

And this is the solution from the end of the booklet:

At the moment of snapping of the right string, the rod is acted upon by the tension $T$ of the left string and the forces $N_1$ and $N_2$ of normal pressure of the laods of the masses $m_1$ and $m_2$. Since the rod is weightless (its mass is zero), the equations of it's translatory motion is $-T + N_1 - N_2 = 0$ and rotatory motion is $N_1\ell = 2N_2\ell$. Now from these and a bit of algebra, we get that $T = N_2$. At the moment of snapping of the right string, the acceleartions of the loads of mass $m_1$ and $m_2$ will be vertical (point $O$ is stationary, and the rod is inextensible) and connected through the relation $a_1 = 2a_2$. The equations of motion for the loads at this instant: $m_1g-N'_1 = m_1a_1$, $m_2g + N'_2 = m_2a_2$, where $N'_1$ and $N'_2$ are the normal reactions of the rod on the loads of mass $m_1$ and $m_2$. Since $N'_1 = N_1$ and $N'_2 = N_2$ a bit of algebra and newton's second law gives $T = N_2 = \frac{m_1m_2}{m_1+4m_2}g$

Now the two portions which I don't understand:

  1. Why the weird directions of the normal force ? One normal force ($N_2$) is pointing upwards, like I seen in almost everybook, but why $N_1$ is pointing downwards in the same direction of gravity ?

  2. Why the net torque or net force on the massless rod is zero ? I know arguements based on putting $m = 0$ or $I = 0$ in $F = ma$ and $\tau = I \alpha$ respectively, but then say suppose I have a wall, and from the wall a mass less rod is hanging down. I connect a heavy weight to the bottom of the rod. Now when I push the heavy weight sideways, the massless rod rotates from initially being at rest. Then why would it have an angular accleration if there's no torque acting on it ?

  • $\begingroup$ Why the downvote? $\endgroup$
    – katana_0
    Commented May 6, 2018 at 15:26

1 Answer 1


Lets start with the latter question 2. Taking F = ma with m = 0 will mean F = 0 as you have realised, but it does not mean a = 0. In fact a can have any value. 0 = 0 x 10. 0 = 0 x 10000. Fortunately massless objects only exist in physics problems where they do whatever is necessary to make the rest of the problem work out.

  1. Pick any direction you want for the normal forces (well either up or down) they can take negative values. The diagram is just drawn retrospectively from what the equations N1 = 2N2 and -T + N1 - N2 = 0 imply would give positive forces.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.