How torque about every point on Axis is same? 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 
 A: 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.
A: 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.
A: Suppose you have an axis and you select one point on the axis from where the torque is calculated. Then the torque by definition is:
$$\vec{\tau}_1 = \vec{r}_1 \times \vec{F} $$
Now choose a second point on the axis. The vectors pointing from the different origins to the point where the force is applied are related by:
$$\vec{r}_2 = \vec{r}_1 + \Delta \vec{r}$$
The torque from this point second point is:
$$\vec{\tau}_2 = \left(\vec{r}_1 + \Delta \vec{r} \right) \times \vec{F} $$
and exploiting the property of cross product:
$$\left(\vec{a} + \vec{b} \right) \times \vec{c} = \vec{a} \times \vec{c} + \vec{b} \times \vec{c} $$
$$\vec{\tau}_2 = \vec{\tau}_1 + \Delta\vec{r} \times \vec{F}$$
The last term on the right hand side of the above equation drops to zero since the vector $\Delta \vec{r}$ is parallel to the axis. Therefore:
$$\vec{\tau}_1 = \vec{\tau}_2$$
A: Torque is defined by the minimum distance (perpendicular distance) of a point to the axis. If the point is on the axis, the perpendicular distance is zero, and torque is zero.
Consider first a force $F$ on a point A, and the torque it produces about the origin O

The magnitude of the torque equals to $|\tau| = d\, |F|$ where $d$ is the minimum distance to the line.
Now consider sliding the force over to points B (or C). The magnitude of the torque remains the same because you have not changed the line of action of the force.
A forced line of action is defined by the direction of the force vector, and the point the force goes through.
By sliding the force vector along its line of action, you are not changing the nature of the problem. The corollary to this by defining the torque $\tau$ about the origin you essentially define what the minimum distance $d$ is for a force $F$.
In terms of vectors and math, torque is
$$ \tau = \vec{r} \times \vec{F} $$
where $\vec{r}$ is the position of point A relative to the origin.
But $\vec{r}$ can be any point along the line of action, and it won't change the result due to the action of the vector cross product $\times$.
Consider another point a distance $h$ from A, but along the direction vector $\hat{e}$ that is parallel to $\vec{F}$ and take the torque
$$ \require{cancel} \tau =( \vec{r} + h \hat{e} ) \times \vec{F} = \vec{r} \times \vec{F} + h ( \cancel{ \vec{e} \times \vec{F} }) = \vec{r} \times \vec{F}  $$
And not to answer your question, if the reference point (the origin) by which torques are measured is on the line of action of the force, then the minimum distance $d$ is zero and torque $|\vec{\tau}|$ is zero also.
Mathematically point A would be some distance $\vec{r} = h \hat{e}$ from the origin, and as you saw above $\vec{e} \times \vec{F}=-0$ since both vectors are parallel.
You can work the location of the line of action of a force by using the following calculation. In any case, the expression below will yield the point on the line closest to the origin (point B)
$$ \vec{r}_B = \frac{ \vec{F} \times \vec{\tau} }{ \| \vec{F} \|^2 } $$
This is like the reverse of $\vec{\tau} = \vec{r} \times \vec{F}$. It is trivial to show that for $\vec{r}_B=0$, torque $\vec{\tau}=0$ must also be zero. Or in words, for the line of action to pass through the origin, the net torque must be zero.
There is an alternative though when the net torque is parallel to the force. When $\vec{\tau} = h \vec{F}$, with $h$ again being a distance measurement called the pitch, again $\vec{r}_B = 0$.
A: 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.
