# Torque of a distributed load: Integrating with dF vs using resultant force

We have a rod that rotates around an fixed point A, which coincides with the end of the rod. The mass distribution along the rod is uniform. We know that the torque generated by the force field at point A is:

$\int_c \vec{r} \times \vec{dF}$

I know that from reading different texts, at least when dF is constant throughout the rod and the mass distribution is homogenous, that:

$\int_c \vec{r} \times \vec{dF}$ = $\int_c \vec{r} \times \int_c\vec{dF}$ = $(\int_c \vec{r}) \times \vec{R}$ = $\vec{G} \times \vec{R}$

Vector $\vec{r}$ is the generic position vector of points throughout the rod, $\vec{dF}$ is the force exerted by the force field at each point of the rod, C is the rod that we integrate over, $\vec{R}$ is the sum of every $\vec{dF}$ across the rod (resultant force), and $\vec{G}$ is the position vector of the center of mass.

My questions are as follows:

1. Is $\int_c \vec{r} \times \vec{dF}$ = $\vec{G} \times \vec{R}$ under all circumstances, and if not, what are the conditions that allow this identity to be true?
2. Could you derive the above identity? I do not understand when $\int_c \vec{r} \times \vec{dF}$ = $\int_c \vec{r} \times \int_c\vec{dF}$ is allowed (I believe this is not always true, when is it?). I also do not understand how $\int_c \vec{r}$ = $\vec{G}$.

Any help will be appreciated.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind May 23 '17 at 14:02
• If the conversation I moved led you to an answer to your question, please consider adding it as an actual answer to this question for posterity. – ACuriousMind May 23 '17 at 14:03