Let's say I have a robot arm with a force-torque sensor on the wrist, between the final motor and the gripper. If the arm and wrist are stationary, then what would the reading on the sensor be?

I know this might sound like a silly question, but I don't understand which of the following would be true:

  • The reading would be g (gravity), since the only force acting on the sensor is gravity.
  • The reading would be g + w, where w is the weight of the gripper, because both gravity and the gripper's weight are acting on the sensor.
  • The reading would be 0, because whilst g + w are acting down on the sensor, since the arm is stationary, there must be an equal an opposite upward force applied by the robot, which exactly cancels out the downwards force.

Which of these is correct? And consequently, why are the other two not correct?



1 Answer 1


The answer is $\vec{G}$! and also any $\vec{\tau}$ torque caused by it, if the axis connecting the geometrical center of the sensor (assuming it measures torques relative to its geometrical center) to the centre of gravity of the gripper.

$\vec{G} = m * \vec{g}$

where $m$ is the mass of the gripper, $g$ is the gravitational ACCELERATION, not force and G is the gravitational pull (force) on the gripper.

  • $\begingroup$ Thank you! So, ignoring torque created by gravity, we can say that the reading will be m * g. Forgive me if this is a silly question, but isn't there also an upwards force coming from the arm to hold the gripper in place, which should be equal to m * g? Why wouldn't this cancel out the force of gravity, such that the sensor reads 0? $\endgroup$ Commented Mar 1, 2021 at 20:50
  • 1
    $\begingroup$ No. If the sensor reads 0 it is in freefall, as in there is no force and no reaction force. Imagine the sensor as a membrane. The gravitational pull is pulling the middle of the membrane down and the base (side) is kept in place by the reaction force. You are right, action an reaction cancel out, but exactly action and reaction allows us to measure. If you hold a kitchen scale in your hand and press it against your other hand, you can measure the force. If you do not press it against anything and exert a force on it it will move and you cannot measure the force. $\endgroup$
    – 50k4
    Commented Mar 1, 2021 at 21:53
  • $\begingroup$ Ok, thank you! Newton's laws confused me with these kind of things... if every action has an equal and opposite reaction, how is there ever any acceleration...?! Anyway, I will have a think about this! $\endgroup$ Commented Mar 1, 2021 at 23:02

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