# Singular Configuration - Balanced Wrench F?

I've been asked to show that if a manipular is at a singular configuration, then the End Effector wrench can be balanced without any joint torques.

I know that t = J*F where (t = torques, F= end effector forces, J is the Jacobian). Hence F = (pinv(J))*t. Hence, if J is singular, then we cannot invert it and find F, as the inverse of J is effectively infinity.

Here however, I basically have been asked to show that there is 0 torque for any end effector wrench F. Which is similar to Ax=0

How would I proceed?

• Why does singular imply Ax=0? Shouldn’t that be the derivative? – SteveO Feb 18 '20 at 1:26
• Also , should not be transpose(J) * F = \tau ? – jdios Feb 18 '20 at 8:03
• What is $Ax=0$ ? is it like state space stuff? $\dot{x} = Ax + B$ ? Or is it the Task-space inertia matrix? When a robot is singular it loses a DoF to my understanding. Given how you mention pseudo inverse $J^\dagger$ should I suspect that your robot is kinematically redundant? If that's the case - then maybe what you are talking about relates to the null space? – Spaceman Feb 22 '20 at 23:06