I am new to physics and just started with rotational mechanics. I wanted to find the net torque on an object by different forces $\vec F_1, \vec F_2, \dots, \vec F_n$.

My Thinking : $$\vec F_T = \vec F_1 + \vec F_2 \dots$$ $$\vec\Gamma_T = \vec F_T \times \vec r$$

My first thought was to add forces right away vectorially, and then find the torque of the net force. But the textbook says to find torque of each force individually and then add them in vector form.

My question is why the idea of net force doesn't work. Is it just that we relate it with real life where net force is zero but still there is torque, or there is some mathematics behind it?

  • $\begingroup$ Open a book of classical mechanics, before try to guess and re-invent theories from scratch $\endgroup$
    – basics
    Oct 26, 2022 at 21:10
  • $\begingroup$ @basics the book I am referring is Concepts of Physics by HC Verma . There the examples mentioned is of turning of fan . Then the theory of torque comes again but in the form of angular momentum only and I think I have just asked a question not reinvented any stuff $\endgroup$ Oct 26, 2022 at 22:03
  • $\begingroup$ PS You can use \times in math for a $\times$ symbol instead of $x$ or $X$. $\endgroup$ Oct 26, 2022 at 23:46

2 Answers 2


The net torque for $n$ forces about a point is

$$\vec\tau_{net}=\vec r_{1} \times \vec F_{1}+\vec r_{2} \times \vec F_{2}+...\vec r_{n} \times \vec F_{n}\tag{1}$$

where $\vec r_{1},\vec r_{2},...\vec r_{n}$ are the position vectors from the point about which the net torque is to be computed to the lines of actions of forces $\vec F_{1}, \vec F_{2},...\vec F_{n}.$

Only if $\vec r_{1}=\vec r_{2}=...\vec r_{n}=\vec r$ are you able to say

$$\vec\tau_{net}=\vec r \times \vec F_{net}$$

That will only be the case if all the forces intersect at a single point and the position vector is directed to that point. Otherwise, you need to determine the torque contribution of each force individually per equation (1).

Hope this helps.


Simply because, $$F_1r_1+F_2r_2 \ne (F_1+F_2)(r_1+r_2) \tag 1$$, or even if you'll take an average position vector instead,

$$F_1r_1+F_2r_2 \ne (F_1+F_2)\left(\frac {r_1+r_2}{2}\right) \tag 2,$$

then still inequality is satisfied.

So in short, net torque is not the same as net force acting on some net position. There's only one exception,- when forces acts on the same point, then mathematically it's correct to say, that :

$$ F_1r+F_2r = (F_1+F_2)r \tag 3$$

So only in this case your logic is correct, but usually net torque is calculated from bunch of forces acting on different points from axis, so (3) is invalidated then.


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