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I posted this on math stack exchange but realize it is more a physics question.

I have a structure which is set up as shown in the image. A weight hangs from point A with mass $m$. Joint B is free to rotate. Joint C connects the two parts ABC and CD. It is intended to be a fixed joint so that the two parts do not move relative to each other. The end D is free to slide along the ground plane.

The problem is to calculate the torque that is trying to rotate the two parts at joint C. I need to know if the joint will be strong enough or if it will fail when too much mass $m$ is applied. In actual fact the joint C is actually a motor and I need to know if its stall torque is enough to ensure it can hold the structure stable given a designated mass $m$.

Missing from the diagram is the angle $\theta$ which is the angle between the vertical and line AB. I know that the mass produces a torque around B of $dmg \sin \theta$. I can't quite work out the torque which is trying to open up the angle BCD.

Jointed structure

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physics.stackexchange? – user02138 Oct 2 '12 at 22:36
Hmm, I didnt know about that forum, but now that I check it appears that new users to physics stack exchange cant post images. – Robotbugs Oct 2 '12 at 23:04
Thanks to the person who edited this to add the diagram from math stack exchange. – Robotbugs Oct 3 '12 at 0:01
up vote 1 down vote accepted

The total force acting on the structure $ABCD$ consists of:

  1. The upward force $\hat{F}_D = F_D \hat{e}_z$ exerted by the ground at point $D$;
  2. The downward force $-mg \hat{e}_z$ exerted by gravity at point $A$;
  3. The force $\hat{F}_B$ exerted by joint $B$.

These contributions must sum to zero, so we have $\hat{F}_B = (mg - F_D)\hat{e}_z$. The total torque acting on the same structure around point $C$ consists of:

  1. The torque from the ground at point $D$, which is $q F_D$ counterclockwise;
  2. The torque from gravity at point $A$, which is $mg(d\sin\theta - p)$ clockwise;
  3. The torque from joint $B$, which is $p F_B = p(mg - F_D)$ clockwise.

These contributions must also sum to zero, from which we conclude that $$ -qF_D + mgd\sin\theta - pF_D = mgd\sin\theta - (p+q)F_D = 0, $$ or $$ F_D = \frac{mgd\sin\theta}{p+q}. $$ Finally, this means that the external torque on part $CD$ around joint $C$, which joint $C$ must counteract, is just $$\tau_{C} = q F_D = mgd\sin\theta \frac{1}{1+(p/q)}.$$

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Thanks I understand, thats great. – Robotbugs Oct 3 '12 at 2:06

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