# 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?

• Are P and Q the only applied forces. What exactly do you mean by the "resultant force"; is this one force equivalent to P and Q in terms of total force and total torque? Why do you need to even use the resultant force? Sep 27, 2022 at 16:29
• I have exactly the same comment as John Darby. Sep 27, 2022 at 16:58
• Dear @JohnDarby and BobD , my textbook tells that the resultant of both force is P+Q and it works at point C. That triggers to ask this question. May be they are considering no torque here. Sep 28, 2022 at 1:12
• By resultant force is typically meant a force that provides the same net force and torque as the actual applied forces. But the point about which the torque is to be evaluated needs to be specified; what is this point, the center of mass? Sep 28, 2022 at 2:35
• @JohnDarby, the point is C in the main question where 5.AC=10.BC. Distance of C from 5N is twice as it is from 10N. Sep 28, 2022 at 2:40

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.

• What if I think that in terms of energy as I create energy by applying forces. If you claim that all 15N is being used to make translational motion then is there any energy left to make angular motion? My mind says no energy left. I am claiming that 10N is responsible for translational motion and (5*AB/2)=2.5AB N is for rotation. Why and how I am wrong? You did a simple mistake! Net torque will be 2.5AB, not 5AB. Sep 28, 2022 at 1:32
• Regarding your question about energy, see my response to your comment on the answer by @Aarone below. Sep 28, 2022 at 2:11
• I agree with @Bob D (after correcting simple math error). Also, based on the extra information you provided, your point C is specified using torques wrt center of mass; see my answer. So 15 N at point C- d/6 to right of CM where d is rod length- is a resultant force equivalent to the actual applied forces. Sep 28, 2022 at 3:47
• @JohnDarby Thanks for pointing out math error. Corrected. Sep 28, 2022 at 6:09

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.

• Net torque = the change in angular momentum is always true using the CM, even if the CM is accelerating. This is not true using a point other than the CM that is moving; in this case an additional term must be considered. See Symon, Mechanics for the details. Sep 27, 2022 at 18:03
• @JohnDarby Thank you for correcting. I have edited my answer accordingly.
– t2m
Sep 27, 2022 at 18:15
• @JohnDarby, What if I think that in terms of energy as I create energy by applying forces. If you claim that all 15N is being used to make translational motion then is there any energy left to make angular motion. My mind says no energy left. Again, if I split the forces into two parts, why the net effect should change? It should remain the same, right? Sep 28, 2022 at 1:21
• @Isteak Ahamed Imon. Total kinetic energy (KE) is translational energy of center of mass (CM) plus rotational energy about CM. [Goldstein, Classical Mechanics] Translational KE due to net force; rotational KE due to net torques with respect to CM. For specific net force, translational KE is same regardless of points where forces are applied; but depending on points of application rotational KE can vary from zero to maximum value. Continued below Sep 28, 2022 at 2:08
• Reason same forces can result in different total KE is forces are assumed constant in magnitude, but more effort (work to maintain forces) is required to keep forces constant the greater the total KE of the body. Your question about energy is very good! Sep 28, 2022 at 2:09

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".