Is torque possible without any pivot point? In the picture, AB is a rod. The resultant of force P=10N and Q=5N works at point C which is P+Q. When I apply both forces simultaneously, the rod will rotate and accelerate. 5N from both end will contribute accelerating toward a straight path and the rest 5N of P at point A will contribute to the rotation.

Now my question is, can I interpret that rotation as Torque though there seems to be no defined pivot point? What is the role of the resultant at C? Shouldn't is contribute to both acceleration to the straight path and rotation? Is point C comparable with center of mass(com) in that context?


 A: You can calculate torque about any desired point; that point is not required to be pivoted. In $\vec{\tau} = \vec{r} \times \vec{F}$, vector $\vec{r}$ is the position vector of point where force $\vec{F}$ acts with respect to the point, say, $P$ considered for calculating torque.

5N from both end will contribute accelerating toward a straight path and the rest 5N of P at point A will contribute to the rotation.

It is not like that. First consider, linear acceleration. You need to take resultant $\vec{P}+\vec{Q}$ and divide its magnitude by mass of the rod. It gives linear acceleration of center of mass.
Next, consider angular acceleration. Calculate net torque due to both forces about the center of mass $C$. Dividing magnitude of resultant torque by moment of inertia will give angular acceleration of the rod.
Note that in above, it has been assumed that all the forces lie in the same plane as your example indicates.
A: I'm assuming that the only forces actually applied to the bar are P and Q as I show in FIG 1 below. Then you are saying equivalent result force is P + Q = 15 located at point C. But you didn’t go far enough.
The equivalent force system to FIG 1 below is FIG 2, with the net force applied at the center of mass (COM assuming bar is uniform) to account for the linear acceleration of the COM of the bar, plus a force couple (torque) to account for the rotational acceleration about the COM due to the torque contributions of the forces at P and Q. That net torque $\tau_{net}$ is, counter clockwise torque being positive torque,
$$\tau_{net}=-5\biggl(\frac{AB}{2}\biggr) + 10\biggl (\frac{AB}{2}\biggr) = +2.5AB$$
Hope this helps.

A: The acceleration of the center of mass (CM) is determined by the net external force.
The net torque is the change in the angular momentum if you use the CM as the point about which to evaluate torque and angular momentum; this is true even if the CM is accelerating.  This is not true using a point other than the CM for torque and angular momentum if the point is accelerating.
So the overall motion is most easily evaluated as:
(1) translational acceleration of the CM with
(2) rotational motion about the CM.
You do not need to use any "resultant force".
Addition based on your later comment specifying location of point C.
You added "5.AC=10.BC. Distance of C from 5N is twice as it is from 10N."
Then AC = 2/3 d where d is length of rod; C is d/6 to right of CM. Total torque from resultant force wrt CM is 15 * d/6 = 15/6 d = 5/2 d. This equals total torque from the actual forces: 5 * -d/2 + 10* d/2 = 5 d/2. So, C is with respect to the CM in terms of equivalent torque. That is the resultant force equivalent to the actual forces with torques evaluated wrt the CM is 15N applied at d/6 to right of CM.
You can solve the motion using either the actual forces/locations or the resultant force at location C.
Using resultant force at C, evaluate motion as translational acceleration of CM plus rotational motion about CM as other answers discuss. (If a point is not fixed or is not the CM (can be moving) net torque is not equal to the change in angular momentum. So pick CM for rotational evaluation for this problem.)  CM will move upwards and rod will rotate counterclockwise wrt CM.  This agrees with earlier answer by @Bob D (except for minor math error that you pointed out.)
