If a force $\vec{F}$ acts at any point on a body with a distance vector $\vec{r}$ from the center of mass, then the resulting \textit{torque}\index{torque} is defined as the cross product of $\vec{r}$ and $\vec{F}$:
\begin{equation}
{\vec{D} = \vec{r}\times \vec{F}}
\end{equation}
The vector The torque is therefore perpendicular to $\vec{r}$ and $\vec{F}$. The magnitude of the torque can be written according to the definition of the cross product as
\begin{equation}
D = Fr\sin{\alpha}
\end{equation}
Here, $\alpha$ is the angle between the applied force and the distance vector. In order for a body to remain stationary, similar to translation, all $N$ applied torques must cancel each other out, i.e., the following must apply:
\begin{equation}\label{eq:def_torque}
\sum_{i=1}^N \vec{D}_i = \vec{0}
\end{equation}
This relationship can be impressively examined using a simple model of the human arm. The biceps is connected to the forearm via the distal biceps tendon and generally acts at an angle of $\alpha$ with a force of $\vec{F}$. The size of the angle depends on how far the arm is extended.
The distance between the point of application and the elbow joint is denoted by $l_1$ whereas $l_2$ is the distance between the elbow and the hand. If a weight of mass $m$ is located there, the torque $D = mgl_2$ acts on the joint. The torque of the muscle, on the other hand, is $D=F\sin\alpha$. Since all torques must cancel each other out in a state of equilibrium, the required muscle force is therefore
\begin{equation}
{F = \frac{mgl_2}{l_1\sin\alpha}}
\end{equation}
It is important to know that muscles cannot stretch on their own, but can only contract. The biceps is therefore called a flexor. To straighten the arm after bending, an extensor is needed, which in the case of the arm muscles is called the triceps. The well-known \textit{lever law}\index{lever law} is also derived from the torque theorem. This is essentially and in simpler form for easier memorization:
$$\text{Force times force arm = load times load arm}$$
For a scale, for example, to be in balance, the torques on both sides must cancel each other out. Two bodies are resting on a beam supported at its center of gravity. The balance remains in equilibrium exactly when the ratio of the weight forces $F_1$ and $F_2$ is equal to the inverse ratio of the distances from the axis of rotation:
\begin{equation}
{\frac{F_1}{F_2} = \frac{l_2}{l_1}}
\end{equation}
As can be seen from this equation, the center of gravity of the system is then located exactly above the axis of rotation, which must compensate for the weight force of the beam and the weights. The unloaded beam was mounted so that it could rotate at its center of gravity. By attaching different weights at different distances from the axis of rotation, the lever law can be demonstrated experimentally.