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I was studying force couple and my teacher said that in this the torque about any point is same. That led me to wonder if in any case if net external force is zero then is torque about any point in universe constant?? If yes how??logically or mathematically. And if no,why it is true for force couple??

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Here is my derivation of this result. I hope you find it helpful:

Say we have n different forces $F_1, F_2, F_3... F_n$, applied at n different points. Now we pick two centers $P$ and $Q$, and express the radial vectors (1) from point $P$ to each of the n points (where forces are applied) as $r_1, r_2, ... r_n$ (2) from point $Q$ to each of the n points (where forces are applied) as $R_1, R_2, ... R_n$.

Then total torque around P is: $\tau_{p}$ = $\sum_{i=1}^{n} r_i \times F_i$ where $\times$ denotes cross product.

The total torque around Q is: $\tau_{q}$ =$\sum_{i=1}^{n} R_i \times F_i$

What we want to show is that $\tau_{p} = \tau_q $ given the constraints that: $$\sum_{i=1}^{n}F_i = 0 $$(net force is zero) and $$r_i - r_j = R_i - R_j $$for all i,j (the n points are fixed. So the relative separations do not change)

So basically you write out the summations explicitly: $$\tau_p = \sum_{i=1}^{n} r_i \times F_i = (r_1 - r_2) \times F_1 + (r_2 - r_3) \times (F_1+F_2) + (r_3 - r_4) \times (F_1+F_2+F_3) + ... + (r_{n-1}-r_n) \times (F_1+...+F_{n-1}) + r_n \times (F_1+F_2+...+F_n) $$

Similarly, $$\tau_q = \sum_{i=1}^{n} R_i \times F_i = (R_1 - R_2) \times F_1 + (R_2 - R_3) \times (F_1+F_2) + (R_3 - R_4) \times (F_1+F_2+F_3) + ... + (R_{n-1}-R_n) \times (F_1+...+F_{n-1}) + R_n \times (F_1+F_2+...+F_n) $$

By plugging in the second constraint, you see immediately that all the terms of the two summations are equal except the last one.

So we just need to confirm that:$R_n \times (F_1+F_2+...+F_n) = r_n \times (F_1+F_2+...+F_n)$

This is trivial since the first constraint says that the net force is zero! Therefore the last terms are just zero.

Therefore we conclude that $\tau_p = \tau_q$ for any p, q in space.

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  • $\begingroup$ Sorry,but I am unable to understand how you got the second constraint relation can you please explain it ??? $\endgroup$
    – Freelancer
    Oct 5, 2015 at 12:03
  • $\begingroup$ you are changing your center, so each $r_i$ is different from each $R_i$. However, the differences do not change, because the points where force is applied remain fixed $\endgroup$ Oct 5, 2015 at 12:26
  • $\begingroup$ @ZhengyanShi How did you get the idea to express the summation of torques in that manner? What is the train of thought involved? $\endgroup$
    – Draculin
    Mar 12, 2022 at 7:57

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