# Robotics Kinematics and Dynamics/Serial Manipulator Statics

## Propagation of Force and Torque

This section deals with solving for the joint torques of serial manipulator, needed to keep it in equilibrium. This will require writing down the force and torque equations for each link.

The forces and torques acting on a single manipulator link.

For link ${\displaystyle i}$, as depicted in the figure on the right, this results in:

${\displaystyle _{i}f_{i}-_{i}f_{i+1}=0\,}$

and:

${\displaystyle _{i}n_{i}-\,_{i}n_{i+1}-\,_{i}P_{i+1}\times \,_{i+1}f_{i+1}=0}$

In this last equation, ${\displaystyle P_{i+1}}$ is the vector pointing from the origin of frame ${\displaystyle {i}}$ to that of frame ${\displaystyle {i+1}}$.

The above can also be written as:

${\displaystyle _{i}f_{i}=\,_{i}^{i+1}R\,\,_{i+1}f_{i+1}}$

and:

${\displaystyle _{i}n_{i}=\,_{i}^{i+1}R\,\,_{i+1}n_{i+1}+\,_{i}P_{i+1}\times \,_{i+1}f_{i+1}}$

The above formulas form the expressions of static force propagation from link to link. For the system to be in equilibrum, the component of ${\displaystyle n_{i}}$ along the Z-axis ${\displaystyle (0,0,{\hat {k}}_{i})}$ must equal the joint torque ${\displaystyle \tau }$. Hence the expression for the joint torque:

${\displaystyle \tau _{i}=\,_{i}n_{i}^{T}\,_{i}{\hat {k}}_{i}}$

### Example: the Two-Link Planar Manipulator

The two-link planar manipulator.

Suppose a force vector ${\displaystyle (F_{x},F_{y})}$ is applied to the end-effector of the manipulator on the right. A joint torque is taken positive if it increases the joint angle and negative otherwise. Applying the above equations leads to:

${\displaystyle _{2}F_{2}={\begin{pmatrix}F_{x}\\F_{y}\\0\end{pmatrix}}}$
${\displaystyle _{2}n_{2}=L_{2}{\hat {i}}_{2}\times {\begin{pmatrix}F_{x}\\F_{y}\\0\end{pmatrix}}={\begin{pmatrix}0\\0\\L_{2}F_{y}\end{pmatrix}}}$
${\displaystyle _{1}F_{1}={\begin{pmatrix}c_{2}&-s_{2}&0\\s_{2}&c_{2}&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}F_{x}\\F_{y}\\0\end{pmatrix}}={\begin{pmatrix}c_{2}F_{x}-s_{2}F_{y}\\s_{2}F_{x}+c_{2}F_{y}\\0\end{pmatrix}}}$
${\displaystyle _{1}n_{1}={\begin{pmatrix}c_{1}&-s_{1}&0\\s_{1}&c_{1}&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0\\0\\L_{2}F_{y}\end{pmatrix}}+L_{1}{\hat {i}}_{1}\times \,_{1}F_{1}={\begin{pmatrix}0\\0\\L_{1}s_{2}F_{x}+(L_{1}c_{2}+L_{2})F_{y}\end{pmatrix}}}$

Hence, the joint torques required for the system to be in equilibrium are:

${\displaystyle {\begin{pmatrix}\tau _{1}\\\tau _{2}\end{pmatrix}}={\begin{pmatrix}L_{1}s_{2}&L_{1}c_{2}+L_{2}\\0&L_{2}\end{pmatrix}}{\begin{pmatrix}F_{x}\\F_{y}\end{pmatrix}}}$

## Virtual Work and Jacobians

In general, work is the dot-product of a force or torque and a displacement, or angular displacement. According to the principle of virtual work, let these displacements become infinitesimally small:

${\displaystyle F\cdot \delta x=\tau \cdot \delta \theta }$,

where the left-hand side of the equation expresses the virtual work in the Cartesian, end-effector workspace, and the right-hand side corresponds with the virtual work in the joint displacements. The both of them must, of course, be the same.

The same expression may also be written as:

${\displaystyle F^{T}\delta x=\tau ^{T}\delta \theta }$

As the definition of the Jacobian is ${\displaystyle \delta x=J\delta \theta }$, the above equation becomes:

${\displaystyle F^{T}J\delta \theta =\tau ^{T}\delta \theta }$

This must be true for all ${\displaystyle \delta \theta }$, so:

${\displaystyle F^{T}J=\tau ^{T}}$,

or:

${\displaystyle \tau =J^{T}F}$

In the above example about the two-link planar manipulator, the matrix is indeed the transpose of the Jacobian, expressed with respect to frame ${\displaystyle \{3\}}$. With respect to the base frame ${\displaystyle \{0\}}$, the following expression is valid:

${\displaystyle \tau =\,_{0}J^{T}\,_{0}F}$