1
$\begingroup$

I read somewhere that torque about every point on an axis is same. But I am really confused about how this can be. Please help me and give an satisfactory answer

$\endgroup$
  • 3
    $\begingroup$ You're talking of torque about every point on the axis, or every point about an axis? $\endgroup$ – Wrichik Basu Oct 22 '17 at 4:01
2
$\begingroup$

I believe you are referring to the following: if the nett resultant of a system forces on a body is zero, then we can say that the moment of that system is independent of the point about which the moment is calculated. In symbols, suppose we have a system of forces $\vec{F}_i$ acting at positions $\vec{r}_i$, relative to our co-ordinate origin. The total moment of this system is $\vec{\tau}=\sum\limits_i \vec{r}_i\times \vec{F}_i$.

Now suppose we shift our co-ordinate origin, so that $\vec{r}_i \mapsto \vec{r}_i+\vec{r}$, for some global displacement $\vec{r}$. Then:

$$\vec{\tau}\mapsto \sum\limits_i(\vec{r}+\vec{r}_i)\times \vec{F}_i = \vec{\tau}+\vec{r}\times \sum\limits_i\vec{F}_i$$

since $\times$ distributes over $+$. But if the resultant is zero, i.e. $\sum\limits_i\vec{F}_i=0$, then the last term on the right vanishes, and we see that $\vec{\tau}$ is unaffected by our shift in origin.

$\endgroup$
1
$\begingroup$

This can be proven using vector algebra.

Let a straight line in space be your axis; a straight line in space can be defined by a point through which it passes and a vector you scale up and down in order to find every point in a specific direction; let these two object be $A_0$ and $\vec{v_0}$ respectively. Your line is the locus: $$ \mathcal{L} = \{ P \in \mathbb{R}^3 | P = A_0 + t \cdot \vec{v_0}, t \in \mathbb{R} \}$$ Let $\vec{F}$ be the vector of a force acting on the point $P_0$. Now, let $\vec{r}$ be a vector from any point on the axis (say $A_0$) to the point $P_0$. The torque about the axis is defined as: $$ \tau = \frac{(\vec{r} \times \vec{F}) \cdot \vec{v_0}}{||\vec{v_0}||} $$ Note that this is not a vector quantity. In fact, this is only the length of the torque vector; the vector itself has the same direction of the straight line. We can show this quantity does not depend on which point we choose from the straight line (meaning $\vec{r}$ does not need to stem from $A_0$), but the torque of a force on the axis has a value of zero. Imagine the point $P_0$, the point of application of the force, is on the axis. It follows that for any point we take as the "tail" of $\vec{r}$, this vector will be parallel to $\vec{v_0}$.Then the vectorial product of $\vec{r}$ with the force $\vec{F}$ is going to result in a third vector perpendicular to $\vec{r}$ and $\vec{v_0}$. But then, from the definition of scalar product, the final result is zero.

Note that we haven't taken any specified point on the axis, and the same result holds for all points. Similarly, you can show that any point on the axis will give you the same value for the torque.

$\endgroup$
-1
$\begingroup$

Keeping strictly to your question, the torque of every point on an axis is constant, and is 0.

Torque $\vec{\tau} = \vec{F} × \vec{r}$, where $\vec{F}$ is the applied linear force and $\vec{r}$ is the perpendicular distance of the point of application from the axis of rotation.

For any point on the axis, $\vec{r}$ is 0, so the torque is 0.

$\endgroup$

protected by Qmechanic Oct 22 '17 at 12:45

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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