Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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

share|improve this question
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
add comment

1 Answer

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 $$\tau_{C} = q F_D = mgd\sin\theta \frac{1}{1+(p/q)}.$$

share|improve this answer
Thanks I understand, thats great. –  Robotbugs Oct 3 '12 at 2:06
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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