Consider thin uniform rod $AB$ of mass $m$ and length $L$ just translating with some acceleration $a$ due to two anti parallel forces $F_1$ and $F_2$ perpendicular to the rod. Force $F_1$ acts at the end $A$ where as $F_2$ acts at distance $y$ from end $A$.

Because body is just translating, Net torque on the body about any point on the rod be zero.

About Center of Mass:

$F_1.L/2 = F_2.(L/2-y)$ gives me a ratio of $F1/F2$

If I proceed with these values of F1 and F2 then net torque about the end B is not zero.

About end B of the rod:

$F_1.L = F_2.(L-y)$ gives me another value for $F1/F2$

If I proceed with these values of F1 and F2 then net torque about the center of mass is not zero.

Where am I going wrong?

EDIT: I have generalized the above situation from the following problem:

enter image description here

Solution to the above problem says "Since the rod moves translationally only, Hence Torque about $B$ is zero. Hence $N = 0$ and hence $x=2$"



I would like to tell that there was a flaw in my last answer.

Because body is just translating, Net torque on the body about any point on the rod be zero.

My previous answer told that this was wrong. This is indeed correct.

As the rod moves in a straight line, the acceleration of every point on the will be $2m/s^2$. If you want to balance the torque about $B$, you must add the pseudo force on the COM as point $B$ is accelerating with $2m/s^2$ too!

If you do that, you will get the same value for $\frac{F_1}{F_2}$.

Here in this photograph I have taken the general case which proves that torque about any point is zero which indeed gives us the same value of $\frac{F_1}{F_2}$

enter image description here

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Feb 17 '17 at 17:17
  • $\begingroup$ Please do not use pictures to post formulae, but type them in MathJax instead. $\endgroup$ – ACuriousMind Feb 17 '17 at 17:17
  • $\begingroup$ @ACuriousMind , I will do that later as due to scarcity of time, I can’t do that now. $\endgroup$ – Aaryan Dewan Feb 17 '17 at 17:19

Update as a result of more information being given in the question

The rod is subjected to a net force at its centre of mass of $F_2-F_1$ to the right and an anticlockwise couple of $(F_1-F_2)\dfrac L2 + F_2y$ as shown in the diagram below.

enter image description here

The magnitude (and direction) of the couple is independent of the point about which you measure the couple so if the bar is undergoing only translation the magnitude of the couple must be zero which gives $F_1=3\,\rm N$.
From that you can work out the torque about $B$ and hence find a value for $x$.

  • $\begingroup$ If its given that "rod is just translating" I'd still have the problem I mentioned in question. Kindly check the edit. $\endgroup$ – claws Feb 16 '17 at 11:10
  • $\begingroup$ @claws As a result of the extra information that you have given I have updated my answer. $\endgroup$ – Farcher Feb 16 '17 at 11:48
  • $\begingroup$ Method 1: If we take Torque about C.M to be zero then F1/F2 = 3/5. Implying F1 is less than F2. Then if we calculate net torque about end B then we'd get value of N. Method 2: If we take Torque about end B to be zero then F1/F2 gives another ratio which doesn't make net torque about C.M to be zero. Hence I generalized this dilemma in the problem and asked this question. $\endgroup$ – claws Feb 16 '17 at 12:02
  • $\begingroup$ @claws The question doesn't say torque about the COM is 0 though. You know the mass and acceleration of the COM and also 1 of the two forces in that direction. You should be able to solve for force without making that assumption. If you take net torque as 0 at B you also get the wrong answer for the acceleration of the COM, so there seems to be a problem with the question. $\endgroup$ – JMac Feb 16 '17 at 12:04
  • $\begingroup$ Where did you determine it was subjected to a couple and how did you find the magnitude? Is a couple moment not pure rotation without translation? This situation we aren't really free to assume either. $\endgroup$ – JMac Feb 16 '17 at 12:32

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