Varignon's theorem (mechanics)

Consider a set of ${\displaystyle N}$ force vectors ${\displaystyle \mathbf {f} _{1},\mathbf {f} _{2},...,\mathbf {f} _{N}}$ that concur at a point ${\displaystyle \mathbf {O} }$ in space. Their resultant is:

${\displaystyle \mathbf {F} =\sum _{i=1}^{N}\mathbf {f} _{i}}$.

The torque of each vector with respect to some other point ${\displaystyle \mathbf {O} _{1}}$ is

${\displaystyle \mathbf {\mathrm {T} } _{O_{1}}^{\mathbf {f} _{i}}=(\mathbf {O} -\mathbf {O} _{1})\times \mathbf {f} _{i}}$.

Adding up the torques and pulling out the common factor ${\displaystyle (\mathbf {O} -\mathbf {O_{1}} )}$, one sees that the result may be expressed solely in terms of ${\displaystyle \mathbf {F} }$, and is in fact the torque of ${\displaystyle \mathbf {F} }$ with respect to the point ${\displaystyle \mathbf {O} _{1}}$:

${\displaystyle \sum _{i=1}^{N}\mathbf {\mathrm {T} } _{O_{1}}^{\mathbf {f} _{i}}=(\mathbf {O} -\mathbf {O} _{1})\times \left(\sum _{i=1}^{N}\mathbf {f} _{i}\right)=(\mathbf {O} -\mathbf {O} _{1})\times \mathbf {F} =\mathbf {\mathrm {T} } _{O_{1}}^{\mathbf {F} }}$.

Proving the theorem, i.e. that the sum of torques about ${\displaystyle \mathbf {O} _{1}}$ is the same as the torque of the sum of the forces about the same point.