I have the following set up:

enter image description here

I want to understand how to compute the force that is being applied to the F_knife point given the Force/Torque registered by the sensor at T_Sensor. The position vector of the point F_Knife relative to T_Sensor is [0.21, 1.58, 0]. For simplicity, I want to assume that there is no gravity so there is only an external force applied to the point F_Knife in any direction. Also, I assume that the point where the gripper holds the knife is rigid.


1 Answer 1


If everything is stationary, then the sum of linear forces must be zero and the sum of torques must also be zero. If they're not zero, then you have a non-zero net force or torque, which means there will be some linear or angular acceleration, respectively.

If you have some transform $T_{knife}$ that expresses the position and orientation (pose) of your knife frame, then you can convert applied forces in the knife frame back to the sensor frame by using the inverse transform $T_{knife}^{-1}$.

Once everything is in the sensor frame, the resulting linear forces must be equal (or opposite, depending on the polarity of your sensor). $F_x$ must equal $F_x$, etc., or again there is some non-zero force in the x-direction, and you'd have linear acceleration in the x-direction.

Find the torque, $\vec{r} \times \vec{F}_{applied}$, where $\vec{r}$ is the location of the knife frame in sensor coordinates, and again your sensor must detect equal/opposite torques.

If the knife is not stationary then it gets more complicated, in that you must also take the motion into account. The end of the robot arm must be providing enough force and torque to get the knife to move despite the applied force. That is,

$$ F_{base} = m_{knife}a_{knife} - F_{applied} \\ \tau_{base} = I_{knife}\alpha_{knife} - \left(\vec{r} \times \vec{F}_{applied}\right) \\ $$

If the signs are confusing, welcome to the club lol. It really depends a lot on how you set your datum, what you choose as positive, etc. The key with all of this is to be consistent. The best way for me personally is to assume everything moves in a particular direction. If you get a negative answer, it's moving in the other direction.

Also, FYI, the first case, where sum of all forces and torques is zero, is a semester-long mechanical engineering class usually called "statics," and the second case, where $\Sigma F = ma$ and $\Sigma \tau = I\alpha$, is a second semester-long mechanical engineering class called "dynamics."

That's basically it, those two equations, and they're a combined academic year of study, and those classes are famously so difficult that they convince a very large portion of the students to quit mechanical engineering altogether.


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