0
$\begingroup$

I have a (hypothetical) robot with two legs, each with 2 segments, with a joint at the body of the robot and in the 'knee' of each leg, as in this (poorly-drawn) diagram: enter image description here (Sorry about the quality and lighting and everything)

I want to know how much torque is applied at joint(s) B by the force of gravity, and therefore how much torque needs to be applied by a motor in the joint to exactly counteract the force of gravity and keep everything stationary.

It would be nice to get a general answer for the following ranges of the variables:

$$\hspace{1cm}90^\circ<A\leq 180^\circ, (270-A)^\circ\leq B \leq 180^\circ \hspace{1cm}0^\circ \leq C\leq90^\circ$$ $$0\ \text{kg}<M\leq100\ \text{kg}\hspace{1cm}0\ \text{m}<X,Y\leq 2\ \text{m} $$

The legs can be assumed to be weightless, and everything is at rest.

$\endgroup$
1
  • $\begingroup$ I edited the question to reflect physically relevant values for the angles. In the most obvious cases, $C$ should be $\leq 90^\circ$, while $180-A+180-B$ should be $\leq 90^\circ$, so we get a convex shape. $\endgroup$
    – Danu
    Oct 3, 2014 at 15:17

1 Answer 1

1
$\begingroup$

A quick diagram:

enter image description here

Key dimension here is the distance $d$. The weight of the robot $W = M \cdot g$ is carried equally by both legs, so we have a force $F$ along the lines $AC$ such that

$$F = \frac{W}{2 \cos\alpha}$$

This force results in a torque at point $B$ because $B$ is not on the line $AC$ - it is displaced by distance $d$, given by

$$d = Y \sin\beta$$

And finally, you combine to get

$$\Gamma_B = F\cdot d = \frac{M\cdot g\cdot Y}{2\ \cos \alpha \ \sin \beta}$$

I hope you can translate this to the dimensions / angles you were using in your diagram.

$\endgroup$
2
  • $\begingroup$ I edited the question just now, I hope it didn't mess with (the relevance of) your answer. If it did, feel free to re-edit! $\endgroup$
    – Danu
    Oct 3, 2014 at 15:18
  • $\begingroup$ @Danu - pretty sure it's OK. I am leaving the translation from my frame of reference to that OP as an exercise... $\endgroup$
    – Floris
    Oct 3, 2014 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.