Question 6 of 7: Static joint torques and singular configurations of a two-link planar manipulator (20 marks)

Question 6: Static joint torques and singular configurations of a two-link planar manipulator (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The $2\times 2$ tip Jacobian of a planar RR arm, in the shorthand $s_1 = \sin\theta_1$, $c_{12} = \cos(\theta_1+\theta_2)$, and a tip force of magnitude 10 directed along $\hat{X}_0$.

Find. (a) the joint torque vector that produces that force, and (b) all configurations at which the Jacobian is singular, with their physical meaning.

Approach. Use the force–torque duality $\tau = J^{T}F$ from virtual work for part (a); for part (b) evaluate $\det J$ symbolically and find where it vanishes.

1. Establish the statics relation. With the manipulator at rest, the virtual work done by the joint torques must equal the virtual work done by the tip wrench for every virtual displacement: $\tau^{T}\delta\Theta = F^{T}\delta\chi$. Since $\delta\chi = J\,\delta\Theta$ for all $\delta\Theta$, $$ \tau = {}^{0}J^{T}(\Theta)\ {}^{0}F. $$ Note the transpose, not the inverse: the mapping from a tip force to joint torques never requires inverting $J$, so it stays valid even at the singular poses found in part (b).
2. Substitute the given force (part a). With ${}^{0}F = [10,\ 0]^{T}$ only the first column of $J$ survives the transpose product: $$ \tau = \begin{bmatrix} -l_1s_1-l_2s_{12} & l_1c_1+l_2c_{12} \\ -l_2s_{12} & l_2c_{12} \end{bmatrix}\begin{bmatrix}10\\0\end{bmatrix} = \boxed{\begin{bmatrix} -10\,(l_1s_1+l_2s_{12}) \\ -10\,l_2s_{12} \end{bmatrix}}. $$ Both torques are negative whenever the tip is above the base line ($s_1, s_{12} \gt 0$): pushing outward along $+\hat{X}_0$ tends to fold the arm downward, and the motors must resist that.
3. Read the result physically, with numbers. Each torque is $-10$ times the moment arm of the applied force about its own joint: $l_1s_1+l_2s_{12}$ is the vertical height of the tip above joint 1, and $l_2s_{12}$ its height above joint 2. Taking $l_1 = 0.800$ m, $l_2 = 0.600$ m, $\theta_1 = 30^\circ$ and $\theta_2 = 60^\circ$ (so $\theta_1+\theta_2 = 90^\circ$), $$ \tau_1 = -10\,(0.800\times 0.5 + 0.600\times 1) = -10.00\ \text{N}\cdot\text{m},\qquad \tau_2 = -10\,(0.600\times 1) = -6.00\ \text{N}\cdot\text{m}. $$
4. Form the determinant (part b). Expanding and using the angle-difference identity, $$ \det{}^{0}J = (-l_1s_1-l_2s_{12})(l_2c_{12}) - (-l_2s_{12})(l_1c_1+l_2c_{12}) $$$$ = -l_1l_2s_1c_{12} + l_1l_2c_1s_{12} = l_1l_2\sin\!\big[(\theta_1+\theta_2)-\theta_1\big] = \boxed{l_1l_2\sin\theta_2}. $$ The $l_2^{2}s_{12}c_{12}$ terms cancel, and $\theta_1$ drops out entirely.
5. Locate the singularities. With $l_1, l_2 \neq 0$ the determinant vanishes only when $\sin\theta_2 = 0$, i.e. $$ \boxed{\theta_2 = 0^\circ \ \text{ or } \ \theta_2 = 180^\circ} $$ for any value of $\theta_1$. There are therefore no isolated singular points but two whole one-parameter families, one for each $\theta_2$ value, traced out as the shoulder sweeps.
Figure 6.2 — the two singular families. Straight ($\theta_2 = 0$) puts the tip on the outer workspace boundary; folded ($\theta_2 = 180^\circ$) puts it on the inner boundary. In both, the two Jacobian columns are parallel and no tip velocity can be produced along the line of the arm.
6. Interpret each family. At $\theta_2 = 0$ the arm is fully extended and the tip lies on the outer boundary circle of radius $l_1+l_2$; at $\theta_2 = 180^\circ$ it is folded back onto itself and the tip lies on the inner boundary circle of radius $|l_1-l_2|$ (a single point at the base when $l_1 = l_2$). In both cases both Jacobian columns become parallel, the rank drops to one, and the reachable tip-velocity set collapses from the plane to a line: no tip velocity is possible along the direction of the outstretched arm. Dually, the arm can resist an arbitrarily large force along that direction with zero joint torque — which is exactly why the statics of part (a) remain well posed here while the inverse-velocity problem does not.
7. Note the practical consequence. Near, but not at, a singularity the inverse $\dot{\Theta} = J^{-1}\dot{\chi}$ demands joint rates that grow as $1/\sin\theta_2$, so a modest Cartesian speed commands an impossible elbow speed. Trajectories are therefore planned to keep $\theta_2$ away from $0^\circ$ and $180^\circ$, or the manipulator is commanded in joint space through the crossing.