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Take a look at this question below. The relationship between torques and wrenches are

$$ \tau = J_b^T(\theta) \mathcal{F}_b = J_s^T(\theta) \mathcal{F}_s $$ where the subscripts $\{b,s\}$ indicate in which frame the Jacobian matrix and the wrench are expressed. I can compute $J_b^T(\theta),J_s^T(\theta)$ but I'm not able to understand from the question in which frame the wrench expressed but I suspect it is in the frame $\{s\}$ since it says in the $\hat{x}_s$-direction.

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  • $\begingroup$ Which is the ${b}$ frame? Is it the ground? In that case ${s}={b}$ for this problem since the first joint angle is 0. Also, generalized forces are invariant to the frame so $J_s$ must be in the same frame as $\mathcal{F}_s$ and the same for $J_b$ and $\mathcal{F}_b$. $\endgroup$ Commented Jan 14, 2022 at 14:16
  • $\begingroup$ @JohnAlexiou the $b$ frame is the frame attached to the end-effector whereas the frame $s$ is fixed to the ground. $\endgroup$
    – CroCo
    Commented Jan 14, 2022 at 14:19
  • $\begingroup$ I think your question is really at which point is zero moment about $\boldsymbol{z}_s$ going to be at. It sounds from the question that the end effector has no moment applied to, but the subscripts make it a bit unclear. So the question is what to use for $\mathcal{F}_b$. $\endgroup$ Commented Jan 14, 2022 at 19:43

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Edited Answer

From the question, it seems the loading is applied at the end effector but its line of action along the $x_s$ direction.

I do prefer to resolve everything on the world inertial frame which makes all the coordinate transformations unnecessary.

The force

$$ \boldsymbol{F} = \pmatrix{5 \\ 0\\ 0} \text{N}$$

going through the first joint has a loading wrench is

$$ \mathcal{F}_s = \begin{bmatrix} \boldsymbol{F} \\ \boldsymbol{M} + \boldsymbol{r} \times \boldsymbol{F} \end{bmatrix} = \begin{bmatrix} \boldsymbol{F} \\ \hat{0} \end{bmatrix} $$

Similarly for the kinematics

The unit twists of the joints (the columns of the jacobian) are defined based on their locations

$$ \begin{aligned} \boldsymbol{r}_1 & = \pmatrix{ 0 \\0 \\ 0 }\text{m} & \mathcal{s}_1 & = \begin{bmatrix} \boldsymbol{r}_1 \times \boldsymbol{\hat{z}} \\ \boldsymbol{\hat{z}} \end{bmatrix} \\ \boldsymbol{r}_2 & = \pmatrix{ 1 \\ 0 \\ 0 }\text{m} & \mathcal{s}_2 & = \begin{bmatrix} \boldsymbol{r}_2 \times \boldsymbol{\hat{z}} \\ \boldsymbol{\hat{z}} \end{bmatrix} \\ \boldsymbol{r}_3 & = \pmatrix{ 1 + \tfrac{1}{\sqrt{2}} \\ \tfrac{1}{\sqrt{2}} \\ 0 }\text{m} & \mathcal{s}_3 & = \begin{bmatrix} \boldsymbol{r}_3 \times \boldsymbol{\hat{z}} \\ \boldsymbol{\hat{z}} \end{bmatrix} \\ \end{aligned}$$

and finally the Jacobian

$$ \mathcal{J}_s = \begin{bmatrix} \mathcal{s}_1 & \mathcal{s}_2 & \mathcal{s}_3 \end{bmatrix}$$

This makes the generalized forces

$$ \boldsymbol{\tau} = \mathcal{J}_s^\top \mathcal{F}_s = \pmatrix{ 0 \\ 0 \\ \tfrac{5}{\sqrt{2}} } $$

Let's see, if we can explain the results. From the wording of the question, the line of action of the force goes through the first joint, along the x-axis and thus it seems perfectly reasonable that no force component will cause any equipollent torque on the first two joints for them to counteract.

Thus it seems reasonable that only the 3rd joint is "energized" and the direction of the torque is positive. This is consistent with the required wrench acting counter-clockwise with respect to this joint.

fig

This torque would try to unwind the mechanism which means since it is fixed on one end, the end effector will push to the right.

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  • $\begingroup$ If the wrench is expressed in the body frame, then you can't multiply it with a Jacobian matrix that is expressed in the spatial frame. $\endgroup$
    – CroCo
    Commented Jan 18, 2022 at 0:00
  • $\begingroup$ I've got $\tau = \begin{bmatrix} 0\\ -\frac{5}{\sqrt{2}}\\0\end{bmatrix}$ for $J_b(\theta)$ and $\mathcal{F}_b$. $\endgroup$
    – CroCo
    Commented Jan 18, 2022 at 0:02
  • $\begingroup$ My result makes sense because if the robot needs to generate a force in the $\hat{x}_s$ at this particular configuration, then it is obvious from the figure, the only joint needs to generate torque is joint 2. $\endgroup$
    – CroCo
    Commented Jan 18, 2022 at 0:08
  • $\begingroup$ @CroCo Your result would also generate a force in the global y direction and a moment about the global z axis, though, whereas John's solution cancels out the force in the global y direction and the moment about the global z axis created by the second joint using the first joint so that the robot only generates a force in the global x direction. $\endgroup$ Commented Jan 21, 2022 at 17:25
  • $\begingroup$ @BrandonJ.DeHart which one is correct? $\endgroup$
    – CroCo
    Commented Jan 21, 2022 at 17:32

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