GoldPheonix said:
Unfortunately, this is just one of those "why does something move if a net force is acted upon it?" questions. The answer is in the definitions:
\tau = F \times r
It's just the recorded observation, and therefore, it happens to "just work that way." Sorry, there's not really a proof for this, it's just what the experimental evidence shows will happen and therefore its the definition we use.
Half true. However, similarly as Newton's third law can be derived out of conservation of momentum (they are equivalent), also this torque stuff can be derived out of conservation of angular momentum (then you can wonder if conservation of angular momentum can be derived somehow, and these questions continue quite far!). In general setting the proof would be quite difficult, it can probably found in good books about mechanics, but I can outline a highly simplified proof.
synergyguru, I have couple of questions. Do you know the definition of angular momentum
<br />
\boldsymbol{L}=\sum_{k=1}^{N} \boldsymbol{x}_k\times\boldsymbol{p}_k,<br />
and do you understand that for a particle to have angular momentum, it does not really need to be attached to some axis and be rotating, but for example a projectile flying by can have angular momentum. A thrown stone has angular momentum with respect to some fixed point that is not in its way.
Anyway, here's the explanation:
Since I don't know how to put pictures here, I'll explain the picture with coordinates (you might want to draw it on paper). There is a stick that has length R. Its left end is in location (0,0), and the right end in (R,0). For simplicity, assume that it is very light elsewhere, but in location (r,0) it has mass m. This means that its mass is consenrtrated in a one point, just to make calculations easier. Then somebody exerts a force F in the right end of the stick, pointing upwards. That means the force vector is (0,F). Then we ask, that what is the angular acceleration of the stick? Or equivalently, what is the acceleration of the mass m in point (r,0)?
If the stick has angular acceleration \alpha, then the mass m in point (r,0) is accelerating with an acceleration a_m=r\alpha. So if we know a_m, we can solve \alpha.
The force F isn't coming from thin air, but there must be some object right below the location (R,0), that is causing this force. By the Newton's laws, the stick is then exerting a force (0,-F) on this object. If the mass of this object is M, then this object is instantaneously accelerating downwards with an acceleration
<br />
a_M=-\frac{F}{M}<br />
This means that its angular momentum with respect to the origo is changing with a time rate
<br />
\dot{L}_M = -RMa_M = -RF<br />
(Because
<br />
L_M = RM v \quad\implies\quad \dot{L}_M = RM\dot{v}<br />
the dot means derivative with respect to time.)
At the same time the angular momentum of the mass m in point (r,0) is changing with a time rate
<br />
\dot{L}_M = rm a_m<br />
where a_m is its acceleration. But since the total angular momentum must not be changing, this numbers must cancel, and we get
<br />
rma_m = RF<br />
which is the same thing as
<br />
\alpha mr^2 = RF<br />
Now, without rigorous explanations, here's how this result can be generalized. This time the moment of inertia of the stick was I=mr^2. If the mass distribution had been something different, then the momentum of inertia would be different, but the result would still be
<br />
\alpha I = RF<br />
Also if we had had several locations R_1,R_2,\ldots,R_n along the stick, and all experiencing corresponding forces F_1,F_2,\ldots F_n, then using similar arguments as were used now, we would have got
<br />
\alpha I = \sum_{k=1}^{n} R_k F_k<br />
From this you see, that if you want the stick to remain still, then must have \alpha=0 which is the same thing as
<br />
\sum_{k=1}^{n} R_k F_k = 0<br />
This is just saying, that the torques must cancel. If I understood correctly, this was what you were after.
If you didn't understand from where some equations popped out, just mention. They can be explained in better detail too, but it would have made the post too long.